# Tagged Questions

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### 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 ...
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### Characterization of dense open subsets of the real numbers

Does the complement of every dense open subset of the real numbers have Lebesgue measure $0$? This is certainly not a characterization of dense open subsets of reals, since the complement of the ...
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### Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice

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 ...
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### Amount to choice necessary to prove instances of Tychonoff theorem

Let $I$ be a fixed nonempty set. I would like to know how much choice is necessary in order to prove that the product of any $I$-indexed family of compact topological spaces is compact (under the ...
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### How should one think about results that depend on AC?

I just encountered this: "(Theorem of A. H. Stone) Every metric space is paracompact... Existing proofs of this require the axiom of choice... It has been shown that neither ZF theory nor ZF ...
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### 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.
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### Is there anything wrong with this use of the axiom of choice?

I have a proof of the following theorem: Let A be a finite set and X be a perfectly normal topological space, and let $\{(A,\succsim_x): x\in X\}$ be a family of binary relations on $A$ satisfying ...
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### Can we prove that every ordered space is normal without choice?

In ZFC, every linear ordered space respect to the order topology is completely normal. I saw the this proof and proof of this statement in the book "Counterexamples of topology" (Example 39). But as I ...
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### Is it possible to prove product of basis is a basis for the box topology without AC?

Let $\{(X_i,\tau_i)\}$ be a collection of topological spaces and $\mathscr{B}_i$ be a basis for $\tau_i$. Let $\mathbb{B}=\{\prod p_i : p\in \prod \mathscr{B}_i\}$. Then is $\mathbb{B}$ a basis for ...
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### 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. ...
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### Showing that if every open subspace is Lindelöf, then the space is hereditarily Lindelöf.

Background: A topological space $X$ is said to be Lindelöf if for every cover $\mathcal O$ of $X$ by open subsets of $X$, there is some $\mathcal C\subseteq\mathcal O$ such that $\mathcal C$ is a ...
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### Using Zorn's Lemma-Topology

Question: A subset $W$ of the set $Z$ of integers is said to be closed under addition if given any elements $w$ and $w'$ of $W$, $w+w'\in W$. Prove that there is a maximal subset of $Z$ which is ...
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### Tychonoff's theorem and axiom of choice in Hausdorff spaces

Does anyone know if axiom of choice is nessesary in proving Tychonoff's theorem in a Hausdorff space? Thanks!
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### Relationships between a few strong separation axioms without Choice

Disambiguation: Let $\langle X,\tau\rangle$ be a topological space. I will say that two subsets $A,B$ of $X$ are separated if they are disjoint from each other's closures (their closures needn't be ...
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### How can one tell if a sequence is well-defined; is the axiom of choice needed?

This question is in the context of the following exercise: Let $X$ be a first countable topological space, let $A \subseteq X$, and let $x \in X$ with $x \in \overline{A}$. Then there exists a ...
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### Cluster point of a sequence $\{x_n\}$ is the limit of some subsequence - Axiom of Choice? [duplicate]

In a metric space, a cluster point of a sequence $\{x_n\}$ is the limit of some subsequence. The only proof that I know works like this: Construct a sequence $\delta _k \to 0$. For each $\delta _k$ ...
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### A few standard results (on metrizability and relative separation strength) without choice?

I've been going back over some results from Munkres's Topology, and I'm curious about some things.... I know that Choice principles have some connection to the separation axioms (in ZF, at ...
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### Some properties of Polish space

Let $X$ be a separable complete metric space. I wonder if following properties hold in ZF. Limit Compact ⇒ Compact Does there exist a function$f$ such that $f(E)$ is closed and $f(E)\subset E$, for ...
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### Constructing a choice function in a complete & separable metric space

Let $X$ be a complete & separable metric space. Let $\{E_i\}_{i\in I}$ be a collection of closed and nonempty sets in $X$. If $X$ is just a complete metric space, it seems not possible to ...
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### Analysis Proof without Axiom of Choice

Lemma: Given any function $f:M\to N$ where $M$ and $N$ are both metric spaces, $\lim_{x\to a}f(x)$ converges to $L$ only if given any function $\gamma:\mathbb{I}\to M$ (where $\mathbb{I}$ is the unit ...
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### Discontinuous Functions on the Real Line

I want to prove that given any function $g:\mathbb{Z} \to \mathbb{R}$ there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that its restriction to the integers is equal to g and such that it not ...
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### (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed.

Let $X$ be a metric space. Let $\{p_n\}$ be a sequence in $X$. Let $E$ be a set of all subsequential limits of $\{p_n\}$. How do i prove that $E$ is closed in ZF? Is there a well-ordering of ...
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### (ZF) Equivalent statements to Baire Category Theorem

So far, I have proved following two for a polish space $X$; 1.If $\{F_n\}$ is a family of closed subset of $X$, where $X=\bigcup_{n\in \omega} F_n$, then at least one $F_n$ has a nonempty inteior. ...
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### (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior

Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF. Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$. ...
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### 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 ...
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### (ZF) Dedekind infinite + Limit Point Compact ⇒ Separable

If every infinite subset has a limit point in a metric space $X$, then $X$ is separable (in ZF) Yesterday, i posted this question and got an answer that 'Limit Point Compact⇒Separable' is unprovable ...
### Maximal ideals in $C(X)$ and Axiom of Choice
The following result are true if we assume full axiom of choice: A. If $X$ is a compact Hausdorff space, then every maximal ideal of the ring $C(X)$ has the form $A_p=\{f\in C(X); f(p)=0\}$. B. If ...