# Tagged Questions

**5**

votes

**2**answers

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

**3**answers

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

**3**answers

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

**0**answers

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

**1**answer

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

**1**answer

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

**1**answer

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

**1**answer

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

**2**answers

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

**2**answers

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 ...