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9 votes
1 answer
557 views

Is it possible to prove that any two points of a convex complete metric space are connected by some metric segment without the axiom of choice?

We say that a point $m$ is between points $p$ and $q$ of a metric space $(M, d)$ if $d(p, q) = d(p, m) + d(m, q)$ and $p ≠ m ≠ q$. A metric space $M$ is said to be metrically convex if given any two ...
Juan Atacama's user avatar
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.
Michael's user avatar
  • 143
5 votes
0 answers
296 views

For which classes of metric spaces can we prove that quasi-isometry is an equivalence relation in ZF?

Given two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, a map $\phi \colon (M_1, d_1) \to (M_2, d_2)$ is a large-scale Lipschitz essentially surjective map if there exist constants $A \geq 1, B \geq 0$,...
Carl-Fredrik Nyberg Brodda's user avatar
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 ...
C. Eratosthene's user avatar
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 ...
Ivan's user avatar
  • 249
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 ...
David Feldman's user avatar