All Questions
Tagged with metric-spaces axiom-of-choice
9 questions
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 ...
- 307
12
votes
1
answer
879
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 \...
- 223
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.
- 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 $A \geq 1, B \geq 0$,...
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 ...
- 251
7
votes
0
answers
172
views
Choice and the Baire property in non-separable complete metric spaces
It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has ...
- 29.8k
3
votes
0
answers
150
views
Metric space has a basis countably locally finite
it is know that all metric space has a basis countably locally finite and this result is proved by using axiom of choice. Then, the natural question is: is possible to prove this result without using ...
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 ...
- 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 ...
- 17.6k