5
votes
2answers
154 views

A few standard results (on metrizability and relative separation strength) without Choice?

I've been going back over some results from Munkres's Topology, and I'm curious about some things. (I originally posted this on M.SE, but I think it is probably a better fit here.) I know that Choice ...
8
votes
3answers
776 views

Axiom of Choice and continuous functions

Do you know if the following statement is an equivalent form of the axiom of choice or not? If $X$ is a compact metric space, then every continuous function $f: X \longrightarrow \mathbb{R}$ is ...
5
votes
3answers
226 views

Well-ordering with a topological property

Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it. An ...
8
votes
0answers
497 views

In ZF, when is a disjoint union of metrizable spaces metrizable?

It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of metric spaces is metrizable, simply by rescaling or chopping off the individual metrics to have diameter at most one, and ...
1
vote
1answer
355 views

Counterexemple to Urysohn's lemma in a topos without denombrable choice ?

Hello ! The Urysohn's Lemma assert that in every topological spaces which is normal two closed subset may be separated by a real valued function. It's proof use axiom of countable choice (but not the ...
3
votes
1answer
378 views

How much choice do we need for regularity of product of regular spaces ?

It is usually stated that the (possibly uncountable) product of regular topological spaces is regular. However the only proof that I know of this fact seems to use the full axiom of choice : See ...
1
vote
1answer
556 views

Is “second-countable implies separable” equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for ...
15
votes
1answer
1k views

Does Arzelà-Ascoli require choice?

Inspired by a recent Math.SE question entitled Where do we need the axiom of choice in Riemannian geometry?, I was thinking of the Arzelà--Ascoli theorem. Let's state a very simple version: ...
6
votes
2answers
1k views

Compact Hausdorff spaces without isolated points in ZF

S is uncountable := |$\mathbb{N}$| < |S| S is noncountable := |S| $\not\leq |\mathbb{N}|$ (X,$T$) is a nice space := (X,$T$) is a compact Hausdorff space without isolated points Does [ ZF / ...
3
votes
2answers
723 views

Is it still impossible to partition the plane into Jordan curves without choice?

It is an easy exercise to show that the Euclidean plane cannot be partitioned into round circles (note however that it is possible to do so for $\mathbb{R}^3$). It seems almost obvious that it is not ...