-5
votes
1answer
512 views

Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice [closed]

A topological space is called quasi-compact if every open cover of it has a finite subcover. Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. The usual proof that $Z$ is quasi-compact uses a ...
-5
votes
1answer
145 views

Can we prove that a bounded closed subset of $\mathbb R^n$ is compact without Axiom of Choice? [duplicate]

Can we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice? This is a related question which was closed.
5
votes
1answer
107 views

Compact metric spaces is second countable and axiom of countable choice

Why we need axiom of countable choice to prove following theorem: every compact metric spaces is second countable? In which step it's "hidden"? Thank you for any help.
4
votes
1answer
105 views

Question regarding disjoint unions, sequential compactness, and Dedekind-finiteness

I have proved the following two results: $[\mathsf{ZF}]$ The disjoint union of a Dedekind-finite family of sequentially compact topological spaces is again sequentially compact. ...
5
votes
0answers
117 views

Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?

The answer to Cardinality of a locally compact space without isolated point shows that AC is required to show that if $X$ is a compact Hausdorff space with no isolated points then $|X| \ge ...
2
votes
1answer
107 views

Tychonoff Theorem and the axiom of choice

How to show that The Tychonoff Theorem and the axiom of choice are equivalent? Here I want to collect ways to prove it. Thanks for your help.
3
votes
1answer
101 views

Tychonoff Theorem in the Realm of $\neg AC$

It's widely know that the Tychonoff Theorem is equivalent to the Axiom of Choice; thus, assuming the negation of the axiom of choice, I'd like to know if there is a canonical example of a collection ...
14
votes
1answer
299 views

Cardinality of a locally compact Hausdorff space without isolated points

I am interested in the following result: Theorem. A locally compact Hausdorff topological space $X$ without isolated points has at least cardinality $\mathfrak{c}$. To prove it, one can find two ...
4
votes
1answer
99 views

Is compact metric space separable in ZF?

Reference; http://www.samos.aegean.gr/math/kker/papers/CompactMetric.pdf The paper says "Compact metric space is separable" is unprovable in ZF$^0$( That is, ZF without axiom of regularity). And i ...
2
votes
2answers
195 views

Is the negation of Gödel's completeness theorem consistent with $ZF$ without AC?

The proof of compactness and completeness of $\mathscr{FOL}$ (with Hilbert system) used Zorn's lemma. And Zorn's lemma is equivalent to the Axiom of choice in $ZF$. So my question is can they be ...
3
votes
3answers
288 views

If a metric space has the limit point property, is it separable? (ZF + AC$_\omega$)

If a metric space has the limit point property, is it separable? (ZF + AC$_\omega$) I'm struggling with this problem for a week. I'm talking about this in Metric space. Here's the part of ...
4
votes
1answer
250 views

Is a countable product of compact intervals in $\mathbf R$ compact (without using the AC)?

Let $\{I_n=[a_n,b_n]\}_{n\in\mathbf N}$ be a countable collection of closed, bounded intervals in $\mathbf R$. Is the infinite Cartesian product $$\prod_{n=1}^\infty I_n$$ compact without using the ...
1
vote
2answers
248 views

Heine-Borel Theorem ($\mathbb{R}^k$) (in ZF)

Heine-Borel Theorem; If $E \subset \mathbb{R}^k$, then $E$ is compact iff $E$ is closed and bounded. I have proved 'closed and bounded⇒compact' and 'compact⇒bounded'. (There exists $r\in \mathbb{R}$ ...
4
votes
3answers
312 views

Axiom of choice and compactness.

I was answering a question recently that dealt with compactness in general topological spaces, and how compactness fails to be equivalent with sequential compactness unlike in metric spaces. The only ...