The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.
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About a paper of Zermelo
This about the famous article
Zermelo, E., Beweis, daß jede Menge wohlgeordnet werden kann, Math. Ann. 59 (4), 514–516 (1904),
available here. Edit: Springer link to the ...
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2answers
572 views
Defining cardinality in the absence of choice
Under ZFC we can define cardinality $|A|$ for any set $A$ as
$$
|A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}.
$$
This is because the axiom of choice allows any ...
16
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2answers
748 views
For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
how to prove the following conclusion:
[for any infinite set $S$,there exists a bijection $f:S\to S \times S$] implies the Axiom of choice.
Can you give a proof without the theory of ordinal ...
12
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2answers
508 views
Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$.
Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< ...
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2answers
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Axiom of choice and automorphisms of vector spaces
A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order 1 or 2." I think that the straightforward solution uses that a exponent two group is ...
18
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5answers
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Advantage of accepting the axiom of choice
What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)?
It seems that there is no clear reason to prefer over other axioms..
Thanks for help.
8
votes
1answer
629 views
Nonnegative linear functionals over $l^\infty$
My purpose is a clarification of the role of the axiom of choice in constructing limits for bounded sequences. Namely, we want a linear functional of norm 1 defined on the space of all bounded complex ...
4
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3answers
333 views
Finite choice without AC
Can anyone explain how we choose one sock from each of finitely many pairs without the axiom of choice? I mean the following quote:
To choose one sock from each of infinitely many pairs of socks ...
18
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4answers
569 views
Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?
I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of ...
11
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1answer
386 views
There's non-Aleph transfinite cardinals without the axiom of choice?
I can't find anything on this anywhere. The book I'm largely using at the moment is based around ZFC, so it makes no mention of anything other than the Aleph numbers, but according to Wikipedia on the ...
7
votes
4answers
621 views
Foundation for analysis without axiom of choice?
Let's say I consider the Banach–Tarski paradox unacceptable, meaning that I would rather do all my mathematics without using the axiom of choice. As my foundation, I would presumably have to use ZF, ...
8
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1answer
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Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly
While reading David Williams's "Probability with Martingales", the following statement caught my fancy:
Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel ...
15
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2answers
561 views
Continuity and the Axiom of Choice
In my introductory Analysis course, we learned two definitions of continuity.
$(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ ...
15
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1answer
598 views
Infinite Set is Disjoint Union of Two Infinite Sets
A finite set is a set such that there exists a bijection from it to some finite ordinal. An infinite set is a set that is not finite.
In ZF, can you prove that every infinite set is the union of two ...
7
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2answers
632 views
Lying-over theorem without Axiom of Choice
This question is motivated by this and this.
Can the following proposition be proved without Axiom of Choice?
Proposition:
Let $k$ be a field.
Let $A$ and $B$ be commutative algebras without ...
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2answers
293 views
Advantage of accepting non-measurable sets
What would be the advantage of accepting non-measurable sets?
I personally feel that non-measurable sets only exist because of infamous Banach-Tarski paradox...
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4answers
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What are the Axiom of Choice and Axiom of Determinacy?
Would someone please explain:
What does the Axiom of Choice mean, intuitively?
What does the Axiom of Determinancy mean, intuitively, and how does it contradict the Axiom of Choice?
as simple ...
21
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4answers
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Is Banach-Alaoglu equivalent to AC?
The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem.
As I have ...
9
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2answers
633 views
Uncountable subset with uncountable complement, without the Axiom of Choice
Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is ...
13
votes
4answers
643 views
Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
I know the usual proof of the existence of an algebraic closure for any field using Zorn's Lemma. The answer to this previous question makes it clear that in general, some nonconstructive axiom (not ...
7
votes
1answer
193 views
Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?
$\mathbb R$ is an uncountably dimensional vector space over $\mathbb Q.$ We can define as many endomorphisms of this vector space as we want by picking their values on the elements of the basis. ...
9
votes
3answers
703 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 , ...
9
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2answers
772 views
Axiom of Choice Examples
In the wikipedia article, two examples are given which use/ do not use the axiom of choice. They are:
Given an infinite pair of socks, one needs AC to pick one sock out of each pair.
Given an ...
4
votes
2answers
255 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 ...
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3answers
247 views
Finding a choice function without the choice axiom
Is there a way to define a choice function on the set of subsets of $\{0,1\}\times\{0,1\}\times\ldots = \prod_{n \in \mathbb N} \{0,1\}$ in ZF? I know that $\prod_{n \in \mathbb N} \{0,1\}$ is ...
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votes
2answers
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Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?
Is there an explicit isomorphism between $L^\infty[0,1]$ and
$\ell^\infty$?
In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ ...
20
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5answers
686 views
Axiom of choice - to use or not to use
I was wondering if there are examples of results in mathematics that were first proven using axiom of choice and later someone found a proof of the result without using the axiom of choice.
9
votes
1answer
715 views
Axiom of choice and calculus
I thought many results in calculus need axiom choice.
For example, I thought one needs AC to prove that a bounded sequence in the real line has a convergent subsequence.
Recently I was taught that one ...
16
votes
3answers
608 views
Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$?
Assuming the axiom of choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the ...
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votes
2answers
282 views
Where is the Axiom of choice used?
In Reid's commutative algebra, there is a proof of equivalent conditions of Noetherian rings, especially (1) The set of ideals of $A$ has the a.c.c. $\Rightarrow$ (2) Every ideal in $A$ is finitely ...
7
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1answer
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Is the axiom of choice needed to show that $a^2=a$?
A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement ...
6
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1answer
252 views
Is Zorn's lemma required to prove the existence of a maximal atlas on a manifold?
In the definition of smooth manifolds, complex manifolds, and similar constructions, one starts by defining a property on neighborhoods in the space, specifying how they relate on overlapping ...
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9answers
527 views
Motivating implications of the axiom of choice?
What are some motivating consequences of the axiom of choice (or its omission)? I know that weak forms of choice are sometimes required for interesting results like Banach-Tarski; what are some ...
3
votes
2answers
322 views
How do I choose an element from a non-empty set?
Suppose I have a non-empty set $A$.
How do I choose an element $x\in A$?
More precisely, I believe I would like to find a formula $P(x,y)$ of ZF such that for every non-empty set $y$ there is ...
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2answers
211 views
Non-aleph infinite cardinals
I'm now confused with a concept of $\aleph$.
1.$\aleph$ is a cardinal number that is well-ordered in ZF.(Defined as an initial ordinal that is equipotent with). Does that mean $\aleph_x$ in ZF may ...
7
votes
2answers
263 views
axiom of choice: cardinality of general disjoint union
I have this exercise involving the axiom of choice, but I don't understand where it's needed: Let $(X_i)_{i \in I}$ and $(Y_i)_{i \in I}$ be pairwise disjoint sets with $|X_i| = |Y_i|$. Prove, using ...
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6answers
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Axiom of Choice and finite sets
So I am relatively familiar with the Axiom of Choice and a few of its equivalent forms (Zorn's Lemma, Surjective implies right invertible, etc.) but I have never actually taken a set theory course.
I ...
2
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2answers
617 views
Nilpotency of the Jacobson radical of an Artinian ring without Axiom of Choice
Let $A$ be a commutative ring.
Suppose $A$ has a composition series as an $A$-module.
EDIT
Since $A$ has a composition series, $A$ has a maximal ideal.
Let $J$ be the intersection of all the maximal ...
10
votes
2answers
276 views
Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
Background:
This question came up in my homework (but was not a homework problem). The problem was proving one direction of the Heine-Borel theorem. As with all proofs of compactness, one begins ...
7
votes
1answer
437 views
Can one construct a non-measurable set without Axiom of choice?
Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice?
Related question on MO says it is ...
2
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2answers
203 views
Every Hilbert space has an orthonomal basis - using Zorn's Lemma
The problem is to prove that every Hilbert space has a orthonormal basis. We are given Zorn's Lemma, which is taken as an axiom of set theory:
Lemma If X is a nonempty partially ordered set with the ...
2
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2answers
153 views
Bourbaki-Witt fixed point theorem: two questions
Consider the following theorem:
Let $f\colon E \to E$ have the propery that $f(x)\geq x$, where $(E,\leq)$ is a non-void partially ordered set with the property that every totally ordered subset of ...
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2answers
301 views
A question about cardinal arithmetics without the Axiom of Choice
Is multiplication of infinite cardinals defined in ZF without Choice?
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3answers
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Finite dimensional subspaces of a linear space
Suppose $V$ is an infinite dimensional vector space. I do not want to assume the axiom of choice, so I will define a vector space $V$ to be infinite dimensional if there is a proper subspace ...
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1answer
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Dedekind's theorem on a weakly Artinian integrally closed domain without Axiom of Choice
Let $A$ be a commutative ring.
Let $f$ be any non-zero element of $A$.
Suppose that $A/fA$ has a composition series as an $A$-module.
Then we say $A$ is a weakly Artinian ring (this may not be a ...
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2answers
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Is there a well-ordering of the reals , measurable or not?
I just stumbled on these two claims:
"Nice" well-orderings of the reals
and, in the answer, no well-ordering of the reals is Lebesgue measurable.
And I am surprised. Is there a ...
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4answers
987 views
Does every set have a group structure?
I know that there is no vector space having precisely $6$ elements. Does every set have a group structure?
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0answers
457 views
Explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$
Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on ...
14
votes
4answers
483 views
Why is the axiom of choice separated from the other axioms?
I don't know much about set theory or foundational mathematics, this question arose just out of curiosity. As far as I know, the widely accepted axioms of set theory is the Zermelo-Fraenkel axioms ...
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1answer
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Given an injection $\mathbb{N}\to\mathcal{P}(X)$, how can we construct a surjection $X\to\mathbb{N}$?
I goofed on my earlier post, here Given a surjection $f:\mathbb{N}\to\mathcal{P}(X)$, how can one construct an injection $X\to\mathbb{N}$?
I am trying to show that there is an injection ...
