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• CommentRowNumber1.
• CommentAuthorDaniel Luckhardt
• CommentTimeJul 27th 2016
• (edited Jul 27th 2016)

I updated separable space. I have two questions:

1. Is it possible to proof that a separable space is Lindelöf without any form of AC?
2. I do not understand how the theorem separable$\Leftrightarrow$second countable is subsumed by Theorem 2.
• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeJul 27th 2016
• (edited Jul 27th 2016)

So the statement was that the proof of Theorem 2 subsumes the proof of (1) $\Leftrightarrow$ (2).

Let’s see: the implication separable implies second-countable (for a metric space) was lightly indicated in the proof of Theorem 2.

In the other direction, the point was supposed to be that the hypothesis of having a countable base is stronger than the hypothesis that every open can be covered by a countable number of balls (so we may apply the argument after the first paragraph to conclude separability).

So let me address that point here, in case it’s not clear. Suppose a metric space $X$ has countable base $U_i$, and let $V$ be open. For each $x \in V$, choose a ball $B_r(x) \subseteq V$, and choose $U_i$ with $x \in U_i \subseteq B_r(x)$; let $\{U_{i_n}\}$ be the collection of all those $U_i$ chosen. Then each $U_{i_n}$ is contained in a ball $B_n \subseteq V$, and these countably many balls cover $V$.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJul 27th 2016

Re: question 1, I bet one could work the “separable $\Rightarrow$ Lindelof” proof backwards to cook up a family of separable spaces whose Lindelof-ness implies countable choice.

If true, this makes me suspect that “every open cover admits a countable subcover” may not be the right formulation of “Lindelof” in the absence of countable choice…

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeJul 27th 2016

Hmm, going backwards to CC is a trickier problem than I thought. First of all, the fact that the $V_\lambda$ form an open cover of a second-countable space is a nontrivial restriction on the family we are choosing elements of; if nothing else it seems to be a cardinality restriction, so we probably wouldn’t be able to get full CC. Second of all, the last part of the proof doesn’t actually require a choice of one $\lambda$ for each $i$ such that $U_i\subseteq V_\lambda$; it’s enough to be able to choose countably many of them, since the countable union of countable sets is countable. (I’m still assuming LEM here everywhere, in case anyone cares.)

So it seems that separable$\Rightarrow$Lindelof is almost certainly significantly weaker than countable choice. I still think it’s probably not provable without some form of choice (e.g. not provable in ZF), but giving a counterexample showing that seems much harder than I thought at first it would be.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJul 27th 2016

There’s a good chance I’m not following something, but the fact that a countable union of countable sets is countable certainly requires countable choice.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeJul 27th 2016

I should have said the countable disjoint union of countable sets. I don’t think that requires countable choice, does it?

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeJul 27th 2016

I thought it did. You have to choose an enumeration for each of the countable sets in the countable family, and that’s a countable number of choices to be made.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeJul 27th 2016

You’re right, of course.

Maybe what I should have said is that we don’t need the projection $\bigsqcup_i \{ \lambda \mid U_i \subseteq V_\lambda \} \to I$ to have a section; what we need is a surjection $J\to I$ that factors through this projection, where $J$ is countable.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeJul 27th 2016

So, some sort of “countable presentation axiom” like “every countable set is covered by a countable projective set” ought to suffice. Does that seem right?

1. Doesn’t the existence of a surjection $J\to I$ gives you a section?: choose a bijection from $J$ to $\mathbb{N}$; assign to each $i\in I$ the smallest element of its preimage.

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeJul 27th 2016

Daniel, not necessarily, the bijection could mess things up a lot. Important to remember which bijection!

2. David, I only need some well-ordering on $J$. I thought this does exist for any countable set also in ZF? (At least this article seems to indicate that)

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeJul 27th 2016
• (edited Jul 27th 2016)

I think Daniel is right. Just transport the well-ordering on $\mathbb{N}$ along a bijection $J \to \mathbb{N}$ to a well-ordering on $J$. Once you have a well-ordering on $J$, you can just discard that bijection and use the well-ordering to finish.

• CommentRowNumber14.
• CommentAuthorDavidRoberts
• CommentTimeJul 28th 2016

Ah, I see :-) Wasn’t thinking in terms of well-orderings.

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeJul 28th 2016

Clearly I’m not really concentrating on this discussion. Thanks guys. So maybe the following argument I had in mind does work to prove something.

Let $R$ be a separable metric space, and let $\{ S_i \}_{i\in \mathbb{N}}$ be a countable family of nonempty sets such that we have a dependent function $g : \prod_{i:\mathbb{N}} S_i \to \mathcal{O}(R)$ such that each $g_i : S_i \to \mathcal{O}(R)$ is injective, where $\mathcal{O}(R)$ denotes the set of open subsets of $R$. This is certainly a nontrivial restriction on the $S_i$, but it seems very unlikely to me that such a restriction would suffice to imply the existence of a choice function without some additional axiom.

Now let $X=\mathbb{N}\sqcup R$, where $\mathbb{N}$ has the discrete topology. This is again separable. However, I claim that if it is Lindelof, then $\{ S_i \}_{i\in \mathbb{N}}$ has a choice function. Consider the open cover consisting of the sets $V_{i,s} = \{i\} \sqcup g_i(s)$ and also the set $R$ itself. Any countable subcover induces a countable subset of the disjoint union $\sum_{i:\mathbb{N}} S_i$ that contains at least one pair $(i,s)$ for each $i\in \mathbb{N}$ (otherwise it wouldn’t cover the point $\{i\}$). Now by Daniel’s argument we can well-order that countable subset and choose the least such pair $(i,s)$ for each $i$ to obtain a choice function $f:\prod_{i:\mathbb{N}} S_i$.