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.
13
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1answer
662 views
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 ...
14
votes
2answers
640 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
votes
2answers
788 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
votes
2answers
533 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}< ...
38
votes
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
votes
5answers
952 views
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.
9
votes
1answer
665 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
votes
3answers
352 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
votes
4answers
596 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
votes
1answer
396 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 ...
8
votes
4answers
698 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, ...
15
votes
2answers
612 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 $z_n \to a$ ...
8
votes
1answer
337 views
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 ...
31
votes
2answers
1k views
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 $L^r$ ...
21
votes
4answers
746 views
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 ...
