2
votes
1answer
66 views

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 ...
1
vote
1answer
32 views

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 $a$. ...
1
vote
0answers
31 views

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?
0
votes
2answers
54 views

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. ...
1
vote
2answers
122 views

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 ...
9
votes
3answers
137 views

Can $\mathbb{R}$ be written as an ascending union of proper additive subgroups?

Can the group $\mathbb{R}$ be written as countable ascending union of proper subgroups? (i.e. does there exists a series of proper subgroups $H_1\leq H_2\leq \cdots $ such that $\cup ...
3
votes
2answers
190 views

Why is the measure of the reals not zero?

I have followed the argument that rationals, being countable and ordered, can be covered by a convergent sequence of decreasing intervals. I am trying to understand why the same argument can’t be ...
0
votes
2answers
183 views

Why is the l.u.b. property equivalent to Cauchy-sequence convergence for $\mathbb{R}$?

Math people: I browsed some questions with similar titles and could not find a duplicate. I apologize if it is. If you read my question it will be obvious that I am not a logician, so please be ...
8
votes
3answers
243 views

Is Zorn's lemma necessary to show discontinuous $f\colon {\mathbb R} \to {\mathbb R}$ satisfying $f(x+y) = f(x) + f(y)$?

A UC Berkeley prelim exam problem asked whether an additive function $f\colon {\mathbb R} \to {\mathbb R}$, i.e. satisfying $f(x + y) = f(x) + f(y)$ must be continuous. The counterexample involved ...
0
votes
2answers
197 views

Can the axiom of choice be used to define the real numbers?

I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible. I ...
2
votes
3answers
120 views

Definition of limit and axiom of choice

In the definition of limit of a function ($\epsilon-\delta$ definition) we say certain statements such as for every $\epsilon>0$ there exist $\delta>0$ .... Now my question is, is a choice ...
3
votes
2answers
205 views

Axiom of Choice, Continuity and Intermediate Value Theorem

I am trying to understand a proof I read in Herrlich's book Axiom of Choice. For those who know the book, it is theorem 4.54 on page 74. The part I am interested in reads: (9) A function $f:X ...
6
votes
1answer
223 views

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 ...
2
votes
1answer
435 views

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 ...
1
vote
1answer
161 views

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 ...
4
votes
0answers
81 views

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, ...
5
votes
2answers
330 views

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 $f\in\mathscr{B}$ ...
2
votes
2answers
97 views

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 $\{f_{n_k}\}$ ...
1
vote
1answer
84 views

Convex and continuity (ZF)

Let $f:(a,b)\rightarrow \mathbb{R}$ be a continuous function. Suppose $\exists k\in (0,1)$ such that $\forall x,y\in (a,b), f(kx+(1-k)y)≦kf(x)+(1-k)f(y)$. Let $A=\{\lambda\in [0,1]|\forall x,y\in ...
2
votes
0answers
125 views

Connected subset of a separable metric space is separable?

Continuity and the Axiom of Choice I have proved a small generalization of Brian's argument, that is, "If $f:X\rightarrow Y$ is sequentially continuous on $X$ and $X$ is separable, then $f$ is ...
1
vote
1answer
109 views

Limit of function on connected set in $\mathbb{R}$

Let $C$ be a connected set in $\mathbb{R}$. Let $f:C\rightarrow \mathbb{R}$ be a function. Let $p$ be a limit point of $C$. Here, $\phi(q)$ : For every sequence $\{p_n\}$ in $C$ where $p_n ...
15
votes
5answers
1k views

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, ...
5
votes
2answers
405 views

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 ...
-2
votes
2answers
394 views

(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 ...
3
votes
3answers
827 views

(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 ...
1
vote
2answers
257 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 $r\in \mathbb{R}$ ...
0
votes
0answers
475 views

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 ...
5
votes
1answer
137 views

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 ...
6
votes
2answers
270 views

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: ...
4
votes
2answers
292 views

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 ...
4
votes
2answers
453 views

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 ...
10
votes
3answers
1k views

Axiom of choice, non-measurable sets, countable unions

I have been looking through several mathoverflow posts, especially these ones http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice , ...
2
votes
2answers
248 views

Dense subset of the plane that intersects every rational line at precisely one point?

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 ...