alpha_i miscellaneous metaproperties

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixperneggerhttps://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[prabau]

Felix, thanks for being persistent. There was a gap in the argument for $\alpha_{1.5}$, which should be fixed now. Let me know what you think.

Please don't approve yet. There is one more thing I'd like to discuss, but will have to do it tomorrow.

[felixpernegger]

It should say "coming from case (i)" (I cant edit the mathse post due to min character requirement)

[felixpernegger]

you switch (i) and (ii) up again in the paragraph after

[felixpernegger]

Argument is very clear now and looks good to me! :)

[prabau]

Thanks for catching the switch. Initially I had (i) and (ii) interchanged and then I forgot to change things everywhere. Fixed.

[prabau]

The other thing I wanted to ask you is about the meta-property with the Kolmogorov quotient:
https://math.stackexchange.com/questions/5123855

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

You also gave an answer in https://math.stackexchange.com/a/5124226. Maybe it's ok, maybe not. I just cannot tell from what is written. Can you write down here a standalone, complete and precise formulation of the equivalent definition that you are using for "X is an $\alpha_1$ space" ? (with the same level of detail as what's in the question)

I want to try to understand what you had in mind.

[felixpernegger]

I gave an answer in https://math.stackexchange.com/a/5124727, maybe a little long but complete in my opinion. Someone actually downvoted it a few weeks ago. Did you see any problem with that answer?

I read (3 weeks ago) through it at it seemed fine to me. If I remember correctly, I didnt checkt he "\alpha_2" etc. is similar too precisely, but for all intents and purposes it should be okay.
(in my opinion its intuitively relatively obvious that its preserved by kolmgorov quotient so its probably fine to omit some details)

[felixpernegger]

@prabau FYI as you can see I added the awaiting-author label to some PRs. This is an important and useful label to see which open PRs should be reviewed now or are stall / require the author's action.
Once the author adressed the review, they should remove the label (and if changes have been reviewed add label again etc)

Some users may lack perms to apply labels as of now, but since almost all PRs are from people with perms, I think we dont have to set up something more elaborate

[prabau]

@StevenClontz FYI

(@felixpernegger also fyi: if you want Steven to read something (for example your last comment), you need to notify him with @StevenClontz)

[felixpernegger]

yes I know this, but I didnt think its necessary

[prabau]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

[prabau]

@felixpernegger Two things:
(1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote).
(2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

[felixpernegger]

@felixpernegger Two things: (1) if you agree that my answer in https://math.stackexchange.com/a/5124727 is fine, can you upvote it? (It looks bad if pi-base refers to a post/answer that does not even have a single upvote). (2) (as mentioned in #1617 (comment)) Regarding your answer in https://math.stackexchange.com/a/5124226, I am waiting for you to provide a standalone, precise formulation of the equivalent definition that you are using for "X is an α 1 space" (with the same level of detail as what's in the question). That would be very helpful.

We say some set $S$ converges to $x$ if for every nbhd $U$ of $x$$S \setminus U$ is finite. In particular every finite set converges to every point.

Now $x$ is $\alpha_1$ iff for every countable (not countably infinite!) collection $S_n$ of countable (!) sets converging to $x$, we can pick a subset $S$ of their union converging to $x$ such that $S_n \setminus S$ is finite for every $n$.

The equivalence is trivial since the only case which doesnt automatically hold is the original $\alpha_1$ formulation.
Now do the argument as I described it.

[felixpernegger]

But your versioon as I said can be generalised nicely and is very precise, so lets go with it

[prabau]

We say some set S converges to x if for every nbhd U of x S ∖ U is finite. In particular every finite set converges to every point.

Now x is α 1 iff for every countable (not countably infinite!) collection S n of countable (!) sets converging to x , we can pick a subset S of their union converging to x such that S n ∖ S is finite for every n .

The definition of a set converging to a point is a little strange, especially the fact that every finite set converges to every point. That's basically saying that that notion does not really have any content for finite sets.

Do you have any source for this definition of a set converging to a point?

[felixpernegger]

I dont understand why we have to argue about this, we can just take yours.

[prabau]

We don't need to argue about this. It is perfectly fine to take the standard definition as given by Tsaban-Zdomskyy.

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

By the way, I just realized, people who work with the alpha_i properties often formulate things in terms of sequences converging to a point, and not in terms of a set converging to a point (like you and I are doing here).
If a sequence is injective, it's equivalent to say the sequence converges and the range of the sequence converges (as a set). If the sequence is not injective, it's again not very clear and subject to various interpretations.

[prabau]

Thinking for myself again, take the example of X = the real numbers. One would say that the constant sequence with value $3$ converges to $5$, because its range is the singleton $\{3\}$, which converges to $5$. That sounds a weird conclusion about the convergence of a constant sequence in a Hausdorff space.

[felixpernegger]

But I want to take the opportunity that you have your attention on this to ask what was never clear to me in your approach before. So, is there a literature source for your version of a finite set converging to a point?

I came up with that "definition" in order to prove the $\alpha_1$ komogorov result more cleanly. So id be surprised if you find something

[prabau]

Ok. That's what I thought. So it's best to ignore it then and go with the Tsaban-Zdomskyy definition.

Moving this PR out of draft so it can be officially reviewed.

[felixpernegger]

LGTM. Good we can finally finish this.

[prabau]

@felixpernegger Yeah, glad this finally got done. And thank you for taking the time to review everything.

[StevenClontz]

yes I know this, but I didnt think its necessary

Well, Steven is the main maintainer and final authority on this site, so he should definitely be notified of procedural changes.

Thanks @prabau -- just wrapped a weeklong workshop and actually had a few hours to triage the email backlog. 😅

I like the awaiting-author label idea. Thanks @felixpernegger