# Tagged Questions

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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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### Application of the axiom of choice

I would like to prove the following statement. If $A$ is a bounded, infinite subset of $\mathbb{R}$, then there is an element $a \in A$ such that $A-\{a\}$ contains a sequence which converges to $a$. ...
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### Continuous is sequentially continuous [duplicate]

Is it possible to prove $f: \mathbb{R} \to \mathbb{R}$ is continuous everywhere iff it is sequentially continuous everywhere without the axiom of choice?
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### Is this a basis for the dual space?

There is an example on Wikipedia that I don't understand and I'd appreciate some help. They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. ...
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### does the mean value theorem implicitly assume the axiom of choice?

in real analysis, including some of the theory of transcendental numbers, the mean value theorem is an essential tool. I was wondering if its dependence on the existential quantifier, $\exists$, at ...
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### Do proper dense subgroups of the real numbers have uncountable index

Just what it says on the tin. Let $G$ be a dense subgroup of $\mathbb{R}$; assume that $G \neq \mathbb{R}$. I know that the index of $G$ in $\mathbb{R}$ has to be infinite (since any subgroup of ...
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### Lebesgue measure, Borel sets and Axiom of choice

I cannot proceed my study on measure theory since it seems my measure theory is really unstable. I desperately need someone to briefly answer below 3 questions... **For convenience, i will write ...
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### Is there a clever way to avoid choice in Riesz Representation Theorem?

Rudin RCA p.43 Riesz Representation for LCH: Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. Then there exists a $\sigma$-algebra ...
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### When $K$ is compact, if $S\subset C_b(K)$ is closed,bounded and equicontinuous, then $S$ is compact? (ZF) [duplicate]

Let $K$ be a compact metric space and $S\subset C(K,\mathbb{C})$. Let $S$ be closed,bounded and equicontinuous. The usual proof for this is, using Arzela-Ascoli Theorem and Axiom of countable choice, ...
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### Why uniform closure $\mathscr{B}$ of an algebra $\mathscr{A}$ of bounded complex functions is uniformly closed?

Let $\mathscr{A}$ be an algebra of bounded complex functions. (Or if necessary, continuous and domain of functions is compact) Definition: $\mathscr{B}$ is uniformly closed iff $f\in\mathscr{B}$ ...
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### Is it possible to choose a subsequence countable times in ZF?

Rudin PMA p.157 I'm trying to prove; "If $\{f_n\}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{f_n\}$ has a subsequence $\{f_{n_k}\}$ such that $\{f_{n_k}\}$ ...
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### Importance of Axiom of Choice

First a quick question regarding the definition of the axiom of choice. Do the sets have to be mutually disjoint nonempty sets or just non-empty? One source states: "For any set X of nonempty sets, ...
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### What does a well ordering of $\mathbb{R}$ look like? [duplicate]

Possible Duplicate: Is there a known well ordering of the reals? I am having a hard time wrapping my head around what a well-ordering of $\mathbb{R}$ looks like. I have seen the ...
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### (ZF)subsequence convergent to a limit point of a sequence

Arthur's answer; (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed. Let $\{p_n\}$ be a sequence in a metric space $X$. Let $B=\{p_n|n\in\mathbb{N}\}$ and ...
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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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### 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 $r\in \mathbb{R}$ ...
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### Every k-cell is compact (in ZF)

This is the part of proof on my book. Let $I$ be a k-cell. Then for every $x\in I$ and $j\in k$, $a_j ≦ x(j) ≦ b_j$ for some $a_j, b_j \in \mathbb{R}$. Let {$G_{\alpha}$} be an open cover of I and ...
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### Does the specification of a general sequence require the Axiom of Choice?

Many results in elementary analysis require some form of the Axiom of Choice (often weaker forms, such as countable or dependent). My question is a bit more specific, regarding sequences. For ...
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### Is there a Lebesgue measurable choice function?

A mapping $f$ from $\mathbb R$ to $\mathbb R$ is called a choice function if, for any $x, y \ {\rm in}\ \mathbb R$, $f(x)-x \in\mathbb Q$ and $f(x)=f(y)$ whenever $x-y$ is rational. My questions is: ...
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### Open Sets of $\mathbb{R}^1$ and axiom of choice

In the proof of 'Every open set in $\mathbb{R}^1$ is a countable union of disjoint open intervals', we need to pick one rational representative from each of the intervals hence establish the ...
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### A concrete example of a choice function

I'm trying to understand a bit what lies behind the Axiom of Choice, and I was wondering, are there concrete examples of a choice function on the Borel set? The Borel set seems nice enough for a ...
It seems there should exist a non-measurable bijection $f: \mathbb{R}\rightarrow \mathbb{R}$. And thus we can obtain a non-measurable graph on $\mathbb{R}^2$ which intersects every horizontal or ...