Locally orderable (part 4) - #1416

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prabau merged 17 commits into
mainfrom
P120-spaces3
Aug 27, 2025
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Locally orderable (part 4)#1416
prabau merged 17 commits into
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P120-spaces3

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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

@pzjp

pzjp commented Aug 25, 2025

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

@pzjp

pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

@prabau

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

prabau commented Aug 27, 2025

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

pzjp commented Aug 27, 2025

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
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prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

Merged
prabau merged 17 commits into
mainfrom
P120-spaces3
Aug 27, 2025
Merged

Locally orderable (part 4)#1416
prabau merged 17 commits into
mainfrom
P120-spaces3

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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

@prabau

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
@prabau

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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

@pzjp

pzjp commented Aug 25, 2025

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

@pzjp

pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

@prabau

prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

@prabau

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

@yhx-12243

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

prabau commented Aug 27, 2025

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
1 check passed
@prabau
prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

Merged
prabau merged 17 commits into
mainfrom
P120-spaces3
Aug 27, 2025
Merged

Locally orderable (part 4)#1416
prabau merged 17 commits into
mainfrom
P120-spaces3

Conversation

@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

@prabau

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

@pzjp

pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

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prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
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prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

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prabau merged 17 commits into
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P120-spaces3
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Locally orderable (part 4)#1416
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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

@pzjp

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

@pzjp

pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

@prabau

prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

@prabau

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

@prabau

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

@prabau

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

@yhx-12243

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
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prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

Merged
prabau merged 17 commits into
mainfrom
P120-spaces3
Aug 27, 2025
Merged

Locally orderable (part 4)#1416
prabau merged 17 commits into
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P120-spaces3

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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

@prabau

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
@prabau

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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
@prabau

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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

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I'm not convinced. Also do not see a mistake in the current version.

@prabau

prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

@prabau

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

@prabau

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

prabau commented Aug 27, 2025

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
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prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

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prabau merged 17 commits into
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P120-spaces3
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Locally orderable (part 4)#1416
prabau merged 17 commits into
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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

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pzjp commented Aug 25, 2025

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

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pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

@prabau

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

@yhx-12243

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

pzjp commented Aug 27, 2025

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
1 check passed
@prabau
prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

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prabau merged 17 commits into
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P120-spaces3
Aug 27, 2025
Merged

Locally orderable (part 4)#1416
prabau merged 17 commits into
mainfrom
P120-spaces3

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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

@prabau

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

@pzjp

pzjp commented Aug 25, 2025

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

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pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

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prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

@prabau

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

@yhx-12243

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

prabau commented Aug 27, 2025

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

pzjp commented Aug 27, 2025

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
1 check passed
@prabau
prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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Locally orderable (part 4) - #1416

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prabau merged 17 commits into
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P120-spaces3
Aug 27, 2025
Merged

Locally orderable (part 4)#1416
prabau merged 17 commits into
mainfrom
P120-spaces3

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@pzjp

@pzjppzjp commented Aug 23, 2025

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P120 decided for few more spaces.
For all other spaces I know the answer seem to become derivable in the future (see #1413, #1388).

Removed locally 1-Euclidean ones.

Some minor cleanup.

@prabau

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

If you merge main into your branch, that part will be taken care of.
(If you are not sure how to do that, please ask. Depending on your setup, that may be different ways to do it.)

@pzjp

pzjp commented Aug 23, 2025

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Is there a different way than a reverse direction PR?

@prabau

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It all depends on what system you use to edit the PR. But definitely no need to have another PR. One commit within the same PR should be enough.

If you use the command line, you can use git rebase main from the topic branch. If you use github.dev, I am not sure. If you use something else, Moniker told me once that there is a way, but I don't remember the details. Maybe other people will know better.

Maybe doing nothing will be fine too, as the merge will just ignore it.

@yhx-12243

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The removal of redundant P120 for S86 and S209 was already merged into main with an earlier PR.

The change of two PR of this part are same (compatible) so we needn't take care of this duplicate removal, just forward. (Or you can git merge main and it wouldn't report any conflict).

Comment threadspaces/S000076/properties/P000120.md Outdated
@prabau

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S59-P82: I am a little confused by the argument. What does "rational numbers in $d$" mean at the end?
And where would a contradiction come from then?

Also, to make sure I understand the previous part: To show the Euclidean closure of $A_N$ has a nonempty Euclidean interior, you are using the Baire category theorem on $U$ (which is homeomorphic to the reals, hence Baire), right? And that's done by taking the cover of $U$ consisting of all the $A_n$, together with all the singletons $\{z\}$ for $z\in U\cap\mathbb Q$ (those singletons are nowhere dense in the Euclidean topology). Hence the existence of $A_N$ as above ... Is that part of the argument what you had in mind?

Minor: typo "cathegory" -> "category"

@pzjp

pzjp commented Aug 25, 2025

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Rational numbers should be dense but they have positive distance $d$ from the points in the interior. Maybe it should read '...by rational numbers (with respect to $d$).'

@prabau

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Hmm. Ok for the meaning.

But still confused. The rational numbers are NOT dense in $U$, because irrational numbers have nbhds that contain only irrationals. So I don't see how irrational numbers need to be approximated by rational numbers.

Comment threadspaces/S000059/properties/P000010.md Outdated
Comment threadspaces/S000059/properties/P000082.md Outdated
Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
@prabau

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S71-P120: (Irregular lattice topology is locally orderable)
I don't think the equations in the last sentence are quite right. But it can be written in a slightly easier (although longer) way in terms of neighborhoods maybe. How about the following for ideas:

Given the order defined on $V_1$, we have to show that the corresponding order topology and the topology of X$ coincide.

(1) For points in $V_1\cap A$: each of them has a successor and a predecessor, so each such point is isolated in both topologies.

(2) A typical nbhd of the origin point $z=(0,0)$ in the order topology is an interval $J=(p,\to)$ for some $p\in V_1\cap A$. If $p\in V_n\setminus V_{n+1}$, then $J$ contains $V_{n+1}$ and hence $J$ is a nbhd of $z$ in the topology of $X$. Conversely, a typical nbhd of $z$ in the topology of $X$ is some $V_n$. If $p\in V_n$, then $(p,\to)\subseteq V_n$, so $V_n$ is a nbhd of $z$ in the order topology.

@pzjp

pzjp commented Aug 26, 2025

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I'm not convinced. Also do not see a mistake in the current version.

@prabau

prabau commented Aug 26, 2025

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S71: Thanks for the update. But I don't think the equation in item (2): $V_n=\bigcup_{k\geq 0}(\langle n,n+k\rangle,\rightarrow)$ is right (the right hand side is contained in $V_n$ but is missing the "vertical left side" of $V_n$).
But it does not matter. $V_n$ is open in the order topology, that's all we need to say.

@pzjp

pzjp commented Aug 26, 2025

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Points $\langle n+k, n\rangle$ are already contained in $(\langle n,n\rangle,{\rightarrow})$ since they are mapped to $\langle n, +k\rangle$ by the map $s$. Compare with $s(\langle n, n+k\rangle)=\langle n,-k\rangle$.

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Ah, yes. Now I see it. I somehow had confused the order of the coordinates in my mind.

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S185-P120 (one-point compactification of the metric fan is locally orderable):
I am not sure the proposed order on $\omega^2\cup\{\infty_i\}$ induces the right topology, as some points don't have immediate predecessors, so they are not isolated.

Some ideas: There is a homeomorphism of $X$ interchanging $\infty_x$ andd $\infty_y$. So it's enough to consider one of the infinity points and showing the (open) set $X\setminus\{\infty_x\}$ is locally metrizable. That subspace is S202.
So we can refer to that. The proof that S202 is locally orderable would have to be added.

For that, use a bijection from $\omega$ to $mathbb Z$, extend that for each row, define anti-lexicographic order on the image, now all points are isolated in the order, except the "infinity" point, etc.

@yhx-12243

yhx-12243 commented Aug 27, 2025

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

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@prabau S202|P133 (Metric fan with ω-many spines is LOTS) will be resolved in the next period of strongly zero-dimensional.

Our preliminary plan is to replace T462 with (Metrizable + Strongly zero-dimensional => LOTS). Remind #1414.

Good to know.

@pzjp Should we remove the proposed change for S185 for now? Then add something after this proposed T462 change is done.

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”).
Another choice is to postpone this change to the PR that change T462.

@prabau

prabau commented Aug 27, 2025

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The only other thing for me in this PR is
S102-P133 (Baire metric on $\mathbb R^\omega$ is LOTS).
I am getting lost in too many details that make it harder than it needs to be, I think. Could we write something more high level maybe? Something like:

Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.
So the neighbourhoods of $x$ are the same for the order topology and for the metric topology.

I may actually be off by one for the $1/n$ radius. One would need to check.
The annoying thing is the definition of https://topology.pi-base.org/spaces/S000102, if we take the index set to be $\omega$ as suggested, that starts at $n=0$. But in S&S, the index is specifically set to start at $i=1$, which simplifies things.

@prabau

prabau commented Aug 27, 2025

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T462 wouldn't help for S185. So we can just reference the result of S202 (although if these PR merged first, it will get a bad reference to somewhat “S202 ? P133”). Another choice is to postpone this change to the PR that change T462.

I know T462 will not help for S185. What I am suggesting is to not do anything for S185 right now. And then do it after (or at the same time also ok) as T462.
@yhx-12243 I think that's also what you are saying.

@pzjp

pzjp commented Aug 27, 2025

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P120 for S185 would be resolved by the "local version of T462". See #1413. Although I do not know when and in what form it will be added. Unlike the global result, this does not seem natural to split.

@pzjp

pzjp commented Aug 27, 2025

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Given a element $x=(x_i)_i \in X$, every $y\in B(x,1/n)$ shares the same prefix of length $n$ with $x$. So if $y&lt;x&lt;z$ with $y$ and $z$ having the same prefix of length $n$, then $B(x,1/(n+1))\subseteq (y,z)\subseteq B(x,1/n)$.

Actually $x$ may agree with $y$ or $z$ on a longer prefix than $y$ and $z$ agree with each other. Then you need smaller ball in order to keep it inside $(y,z)$. That's why I introduce $k$ and $l$.

@prabau
prabau merged commit 0d1e4c2 into mainAug 27, 2025
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prabau deleted the P120-spaces3 branch August 27, 2025 18:20
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