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Results tagged with axiom-of-choice
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user 61536
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.
32
votes
Accepted
Chromatic number of a topological space
The chromatic number $\chi(X)$ of a topological space $X$ is related to the separation dimension $t(X)$ introduced and studied by Steinke.
The separation dimension $t(X)$ is defined inductively:
$\ …
- 41.8k
17
votes
1
answer
423
views
Axiom of Countable Choice and meager sets
Let us recall that the Axiom of Countable Choice (denoted by ACC) says that the countable product $\prod_{n\in\omega}X_n$ of nonempty sets $X_n$ is nonempty.
It is easy to see that ACC implies that fo …
- 41.8k
10
votes
1
answer
260
views
The partial preorder on $\mathbb N$ generated by the finite axioms of choice
Let $\mathsf C_n$ denotes the statement:
for any family $\mathcal F$ of $n$-element sets there exists a choice function (i.e., a function $f:\mathcal F\to\bigcup\mathcal F$ such that $f(F)\in F$ for …
- 41.8k
5
votes
1
answer
321
views
Is Axiom of Choice equivalent to its version for families of sets, indexed by ordinals? [duplicate]
Is Axiom of Choice equivalent to the following statement?
Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function …
- 41.8k
3
votes
1
answer
936
views
Implications of the existence of a pair of surjective functions, without Axiom of Choice
The classical Cantor-Schroder-Bernstein Theorem says that there exists a bijective function $X\leftrightarrow Y$ if and only if there exist injective functions $X\hookrightarrow Y$ and $Y\hookrightarr …
- 41.8k