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Results tagged with axiom-of-choice
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user 75
An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.
16
votes
Accepted
Artin Rings, Noetherian Rings, and the Axiom of Choice
Suppose $A$ is a nonzero artinian ring. Then the collection of all nonzero (possibly improper) ideals of $A$ has a minimal element, say $I\subseteq A$. Then $I$ has no nonzero proper submodules, and …
- 31.2k
13
votes
Accepted
Does "$|{\cal P}_2(X)| = |X|$ for $X$ infinite" imply ${\sf (AC)}$?
Yes, the usual proof that $|X|^2=|X|$ for all $X$ implies AC works for (S) as well. In detail, let $A$ be any infinite set, let $H$ be its Hartogs number (the least ordinal that does not inject into …
- 31.2k
8
votes
Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?
This theorem follows from Dependent Choice, and thus is strictly weaker than the Axiom of Choice. Here is a proof using only DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$. We w …
- 31.2k
5
votes
Accepted
Freeness of the group of principal ideals of a number field
To elaborate on Asaf's comment: the usual proof does not actually need the axiom of choice. The proof that a subgroup of a free abelian group is free actually shows (without using AC) that a subgroup …
- 31.2k
8
votes
Does k(X) have a k-basis for every set X, without AC?
As a generalization of David Speyer's argument, here is a proof that if $k(X)$ always has a basis, then the axiom of choice for finite sets of bounded cardinality holds. In fact, to get the axiom of …
- 31.2k