Skip to content

Conversation

@felixpernegger
Copy link
Collaborator

@felixpernegger felixpernegger commented Dec 27, 2025

Once #1556 and #1455 and this PR are merged, we only need

for this space (S35) to complete all ordinal spaces.

Copy link
Collaborator

@yhx-12243 yhx-12243 left a comment

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

P76: See Theorem 6.2 in {{zb:1375.54007}}.

felixpernegger and others added 2 commits December 28, 2025 01:32
Co-authored-by: yhx-12243 <yhx12243@gmail.com>
@felixpernegger
Copy link
Collaborator Author

felixpernegger commented Dec 28, 2025

Thanks, I included it in the PR. How did you find this by the way?

value: false
---

Consider the open cover $\mathcal{O}=\{[0,a)\mid a \in X\}$. If $X$ were weakly Lindelöf, we would find a countable $\mathcal{U}\subseteq \mathcal{O}$ such that $\bigcup \mathcal{U}$ dense. But then $\bigcup\mathcal{U}=X$, since otherwise we would find an $a \in X$ with $[a,\omega_1)\subseteq X \setminus \bigcup\mathcal{U}$. Therefore $X$ would be a countable union of countable sets and thus countable.
Copy link
Collaborator

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Suggested change
Consider the open cover $\mathcal{O}=\{[0,a)\mid a \in X\}$. If $X$ were weakly Lindelöf, we would find a countable $\mathcal{U}\subseteq \mathcal{O}$ such that $\bigcup \mathcal{U}$ dense. But then $\bigcup\mathcal{U}=X$, since otherwise we would find an $a \in X$ with $[a,\omega_1)\subseteq X \setminus \bigcup\mathcal{U}$. Therefore $X$ would be a countable union of countable sets and thus countable.
Consider the open cover $\mathcal{O}=\{[0,a)\mid a \in X\}$. If $X$ were weakly Lindelöf, we would find a countable $A\subseteq X$ such that $U = \bigcup \{[0, a) : a\in A\}$ is dense in $X$. But if $a_0 = \sup A < \omega_1$, then $U\subseteq [0, a]$, so that $U$ is contained in a proper closed set and is not dense.

@Moniker1998
Copy link
Collaborator

Moniker1998 commented Dec 31, 2025

That $\omega_1$ embeds into a $W$-group is a corollary of proposition 7.2 in Products of normal spaces by Przymusiński in Handbook of set-theoretic topology by Kunen and Vaughan. It embeds into a uncountable $\Sigma$-product of $\mathbb{R}$, so that into a $W$-group.

@Moniker1998
Copy link
Collaborator

In basic language, let $f:\omega_1+1\to \{0, 1\}^{\omega_1}$ be defined as $f(\alpha) = x^\alpha$ such that $x^\alpha_\beta = 1$ iff $\beta\in \alpha$. Then it's easy to check that $f$ is an embedding, simply take preimage of $\{x : x_\beta = 1\}$ and notice how it's clopen in $\omega_1+1$. So this is an embedding. And its restriction to $\omega_1$ is an embedding into the $\Sigma$-product of $\omega_1$ copies of $\{0, 1\}$, which is a $W$-group.

@felixpernegger
Copy link
Collaborator Author

This is super great thanks! I think Toronto property should be easy to add with some theorems (for ordinals) and then we completed all ordinals :)

@Moniker1998
Copy link
Collaborator

@StevenClontz @ccaruvana could one of you review the proximal property for this space?
It's about topological games and I'd rather an expert check if the result cited in that paper is correct

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Labels

Projects

None yet

Development

Successfully merging this pull request may close these issues.

4 participants