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12 votes
1 answer
838 views

Partition of unity without AC

Several existence theorems for partition of unity are known. For example (source), Proposition 3.1. If $(X,\tau)$ is a paracompact topological space, then for every open cover $\{U_i \subset X\}_{i \...
BonBon's user avatar
  • 223
9 votes
0 answers
156 views

Irreducible subcontinuum without Zorn's lemma

In continuum theory we frequently use the fact that two points in a continuum are contained in an irreducible subcontinuum. A continuum $X$ is a compact connected metric space. A subcontinuum $K\...
D.S. Lipham's user avatar
  • 3,105
4 votes
0 answers
141 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 $...
Lorenzo's user avatar
  • 2,236
7 votes
2 answers
714 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 ...
Lorenzo's user avatar
  • 2,236
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 ...
Lorenzo's user avatar
  • 2,236
5 votes
1 answer
266 views

Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of ...
saolof's user avatar
  • 1,843
15 votes
2 answers
333 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 ...
Noah Schweber's user avatar
7 votes
3 answers
897 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.
Michael's user avatar
  • 133
17 votes
1 answer
420 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 ...
Taras Banakh's user avatar
  • 41.1k
18 votes
0 answers
365 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) $\varphi(\emptyset)=\emptyset$,...
bof's user avatar
  • 12.7k
10 votes
0 answers
289 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; ...
Noah Schweber's user avatar
9 votes
0 answers
362 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 ...
Alessandro Codenotti's user avatar
5 votes
0 answers
236 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 ...
D.S. Lipham's user avatar
  • 3,105
3 votes
0 answers
202 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 ...
LCO's user avatar
  • 496
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, ...
N. de Rancourt's user avatar

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