All Questions
8 questions with no upvoted or accepted answers
24
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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 ...
user5810
18
votes
0
answers
370
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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
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293
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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
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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 ...
- 3,011
6
votes
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210
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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
5
votes
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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
4
votes
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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 $...
- 2,286
3
votes
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209
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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 ...
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