All Questions
20 questions
9
votes
2
answers
3k
views
Compact Hausdorff spaces without isolated points in ZF
$S$ is uncountable := $\vert\mathbb{N}\vert<\vert S\vert$
$S$ is noncountable := $\vert S\vert \not\leq \vert\mathbb{N}\vert$
$(X,T)$ is a nice space := $(X,T)$ is a compact Hausdorff space ...
- 2,031
15
votes
1
answer
1k
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 ...
- 185
16
votes
1
answer
1k
views
Does Urysohn's Lemma imply Dependent Choice?
It's widely known$^{1}$ that in the proof of Urysohn's Lemma (UL) one uses the Principle of Dependent Choice (DC). Inspired by the equivalence between DC and Baire's Category Theorem$^{2}$, I'd like ...
- 4,713
4
votes
0
answers
142
views
Consistency of a strange (choice-wise) set of reals, pt. 2
This is a follow-up on this question. Consider a set $X\subseteq \mathbb{R}$ such that
$X$ is not separable wrt its subspace topology
Every countable family of non-empty pairwise disjoint subsets of $...
- 32.5k
7
votes
2
answers
722
views
Consistency of a strange (choice-wise) set of reals
Consider a set $X\subseteq \mathbb{R}$ such that
$X$ is not separable wrt its subspace topology
For all $r\in\mathbb{R}$ there exists a sequence $(x_n)_{n\in\omega} \subset X$ converging to $r$
In a ...
- 2,286
6
votes
0
answers
210
views
Failure of Baire's grand theorem when the hypothesis is weakened to separable metric space
The statement of Baire grand theorem gives a characterization of Baire class 1 functions between a completely metrizable separable space (aka Polish space) and a separable metrizable space. The ...
- 2,286
15
votes
2
answers
341
views
Do we need full choice to "efficiently" use (sub)bases?
This question was previously asked and bountied at MSE without success.
Suppose $(X,\tau)$ is a topological space, $B$ is a base for $\tau$, and $U\in \tau$ is an open set. Consider the following two ...
- 20.5k
8
votes
3
answers
937
views
BCT equivalent to DC
Do you know where I can find proof of equivalence Baire Category Theorem and DC (Axiom of Dependent Choice)? It is well known fact but I can't find appropriate literature with the proof.
- 4,713
17
votes
1
answer
429
views
Axiom of Countable Choice and meager sets
Let us recall that the Axiom of Countable Choice (denoted by ACC) says that the countable product $\prod_{n\in\omega}X_n$ of nonempty sets $X_n$ is nonempty.
It is easy to see that ACC implies that ...
- 1,869
18
votes
0
answers
370
views
Čech functions and the axiom of choice
A Čech closure function on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$,...
- 13.4k
10
votes
0
answers
293
views
Undetermined Banach-Mazur games: beyond DC
This question is a follow-up to this one; see that question for the definition of Banach-Mazur games. There James Hanson showed that ZF+DC proves that there is an undetermined Banach-Mazur game; ...
- 20.5k
9
votes
0
answers
367
views
A $\mathsf{ZF}$ example of two Baire spaces whose product is not Baire?
Motivated by this question I'm looking for a pair of Baire topological spaces whose product is not Baire and whose construction does not need the axiom of choice.
The example of such spaces I'm ...
- 39.8k
5
votes
0
answers
237
views
Polish transversals
A subset of $X$ an indecomposable continuum $Y$ is called a composant transversal if $X$ has exactly one point from each composant of $Y$.
So a continuum has a composant transversal precisely when ...
- 3,317
3
votes
0
answers
209
views
Compactification of Tychonoff spaces without full axiom of choice
If $X$ is a Tychonoff space, then using the Tychonoff theorem and thus the full axiom of choice, it follows that $X$ admits a Hausdorff compactification.
My question is : what remains true if we do ...
- 506
35
votes
1
answer
1k
views
Chromatic number of a topological space
Here is a question I asked myself years ago. Since it is not really in my field, I hope to find some (partial) answers here... Since it was unclear, I precise that I am looking for an answer in ZFC, ...
- 15.2k
7
votes
1
answer
531
views
Totally bounded spaces and axiom of choice
Wikipedia article on totally bounded spaces states "... the completion of a totally bounded space might not be compact in the absence of choice." Where is the axiom of choice used, and do you need it ...
- 39.8k
26
votes
2
answers
5k
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:
...
CommunityBot
- 1
24
votes
0
answers
2k
views
Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice
While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then ...
- 3,104
6
votes
3
answers
281
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 ...
- 6,210
8
votes
3
answers
1k
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 ...
- 6,210