4
votes
1answer
51 views

Does the converse of Tychonoff's theorem hinge on the axiom of choice?

Tychonoff's theorem:$\phantom{---}$ If $A$ is a non-empty index set and $X_{\alpha}$ is a non-empty compact topological space for every $\alpha\in A$, then $X\equiv\times_{\alpha\in A} X_{\alpha}$ is ...
3
votes
2answers
66 views

Clarifications on proof of compactness theorem

I've been reading through the following proof of compactness theorem: http://www.princeton.edu/~hhalvors/teaching/phi312_s2013/compactness.pdf One thing that struck me is that this proof seems to ...
1
vote
1answer
49 views

Closed subsets of $\beta \mathbb R$

Definitions. Suppose $X$ is a topological space. $w(X)=\min\{|\mathcal B|:\mathcal B$ is a base for $X\}+\omega$ $e(X)=\sup\{|D|:D\subseteq X$ is closed and discrete$\}+\omega$ $K(X)$ is the ...
3
votes
2answers
70 views

Finite-case symmetry leads to infinite-case asymmetry

Formulas for sines or cosines of sums superficially appear to have a certain symmetry, specifically it looks as if sine and cosine play something like symmetrical roles: $$ \begin{align} & ...
5
votes
1answer
164 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.
5
votes
1answer
121 views

Is $\{0,1\}^{\omega_1}$ sequentially compact?

It is claimed that an uncountable product of $[0,1]$ is not sequentially compact, e.g. in Wikipedia (I think replacing $[0,1]$ by $\{0,1\}$ doesn't make much difference). However, the constructions I ...
4
votes
1answer
114 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. ...
1
vote
1answer
100 views

Show that $X$ is separable.

I'm working through Kunen's Set Theory and I'm not sure how to proceed on part of one exercise. Let $X$ be compact Hausdorff and $\mathbb{O}_X$ be the poset of nonempty open sets of $X$ ordered by ...
4
votes
0answers
77 views

Is $X$ pseudocompact

The following example with a little modified from the handbook of set theoretic topology, Page 574: Let $\kappa$ be any cardinal for which there exists a family $\{H_\alpha: \alpha < \kappa\}$ ...
14
votes
2answers
293 views

Is there a “tree-like” proof of compactness theorem in the case of uncountably many variables?

I like proofs using trees and König's lemma, since they are very visual. One of the applications of König's lemma you can show to students is proving compactness theorem for propositional calculus, ...
0
votes
1answer
101 views

Is this space countably compact

Let $X$ be a Tychonoff countably compact space and $A$ is a subapce of $X$ such that for any countable $B \subset A$ we have $\overline{B} \subset A$. My question is this: Is this subspace $A$ ...
3
votes
1answer
105 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 ...
1
vote
1answer
116 views

A question on linearly Lindelöf space

A space $X$ is linearly lindelöf if for every open cover of $X$, linearly ordered by the subset relation, has a countable subcover. Question is this: How could we see that $X$ is linearly lindelöf ...
3
votes
3answers
308 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
269 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 ...
7
votes
1answer
379 views

Addition on ultrafilters is non-commutative

I'm reading a book on ultrafilters and it tells me that showing addition on ultrafilters over the naturals (addition on the set of ultrafilters '$\beta \mathbb{N}$', defined in the standard way) is ...
3
votes
1answer
202 views

Set of ultrafilters $\beta \mathbb{N} - \mathbb{N}$ is not separable

I want to show the set of ultrafilters $\beta \mathbb{N} - \mathbb{N}$ (where $\beta \mathbb{N}$ is all ultrafilters on $ \mathbb{N}$) is not separable. I know we can take as a base of $\beta ...