A topological space $X$ is said to be Lindelöf if for every cover $\mathcal O$ of $X$ by open subsets of $X$, there is some $\mathcal C\subseteq\mathcal O$ such that $\mathcal C$ is a countable cover of $X.$ (If Axiom of Countable Choice holds, this is equivalent to saying that every open cover of $X$ admits a countable open refinement.) A space is hereditarily Lindelöf if every subspace of $X$ is Lindelöf.

It is easily shown that closed subspaces of Lindelöf spaces are again Lindelöf (similar to showing that closed subspaces of compact spaces are again compact).

I'm trying to show the following

Proposition: $[\mathsf{ZF} +$ Countable Choice$]$ A space is hereditarily Lindelöf if and only if every open subspace is Lindelöf.

One implication is trivial, but I'm stuck on the other. (I know how to prove it if we know that every open subspace is $\sigma$-compact, but I'm not sure how we can conclude this.)

What I've got so far:

Let $Y$ be an arbitrary (non-empty proper) subspace of $X$, $\mathcal O$ a cover of $Y$ by relatively open subsets of $Y$, $U$ the interior of $Y$ (with respect to the topology on $X$). Then considering $U$ as a subspace of $Y$ (so a subspace of $X$, since $Y$ is a subspace of $X$), $\mathcal O':=\{U\cap V:V\in\mathcal O\}$ is a cover of $U$ by relatively open subsets of $U,$ so since $U$ is an open subspace of $X$, then $U$ is Lindelöf, so $\mathcal O'$ admits a countable subset $\mathcal C_0$ such that $\mathcal C_0$ covers $U.$ $\mathcal C_0$ is readily an open refinement of $\mathcal O$ (since $U$ is an open subspace of $Y$) by definition of $\mathcal O'$--though of course, we can't conclude that $\mathcal C_0$ covers $Y$, since we didn't assume that $Y$ is open in $X$ (so that $Y=U$).

What I'd like to do at this point:

Letting $\mathcal O''$ be the set of all open subsets $V$ of $X$ such that $Y\cap V\in\mathcal O,$ I would like to prove the following

Conjecture: The boundary of $Y$ is contained in $\bigcup\mathcal O''.$

If I can manage that, then I can see how to use Lindelöf-ness of closed subspaces together with Countable Choice to induce a countable open refinement $\mathcal C_1$ of $\mathcal O$ such that $Y\setminus U\subseteq\bigcup\mathcal C_1,$ whence $\mathcal C:=\mathcal C_0\cup\mathcal C_1$ is a countable open refinement of $\mathcal O$ that covers $Y$.

Unfortunately, I am stymied in my attempts to show that for boundary points $z$ of $Y$ with $z\notin Y,$ there is an open subset $V$ of $X$ with $z\in V$ and $Y\cap V\in\mathcal O$. I'm beginning to suspect that my conjecture may not be true. If it were true, then as an alternate approach, we could simply cover the closure of $Y$ by $\mathcal O'',$ which would induce a relatively open cover of the closure of $Y$, which (by Lindelöf-ness of closed subspaces) we could then reduce to a countable subcover, which would induce a countable subcover of $\mathcal O.$

Primary Question:

If my Proposition and Conjecture hold, can anyone can give me any hints how I might show this (or, if you prefer, confirmation of my Conjecture's truth, together with hints at any simpler/more elegant approaches)? If my Proposition holds but my Conjecture doesn't, can anyone prompt an alternate way that I should proceed instead (and for bonus points, a counterexample to my Conjecture)? If a stronger Choice principle is required for my Proposition to hold, how much Choice is sufficient, and how might I proceed?

Secondary Question:

It seems that Countable Choice is necessary for my Proposition to hold, but is it possible to prove it with less Choice than that?

  • $\begingroup$ What happens if $Y$ has an empty interior? $\endgroup$ – Asaf Karagila Sep 17 '13 at 9:33
  • $\begingroup$ In that case, the boundary would be precisely the closure, and my approach would amount to the alternate approach I noticed while typing out the question (that I'm still not sure would actually work). $\endgroup$ – Cameron Buie Sep 17 '13 at 12:45

Let $V=\bigcup\{U: U\in \mathcal O''\}$ (using your notation). Observe that $\mathcal O''$ is an open cover of $V$ and $V$ it is open. Therefore, by our assumption, there is a countable subcover $\mathcal V$ of $V$. It follows that $\{U\cap Y: U\in \mathcal V\}$ is a countable open cover of $Y$.

  • $\begingroup$ Oh, good grief! So simple, and requiring no Choice at all. $\endgroup$ – Cameron Buie Sep 16 '13 at 5:13

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