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axiom of countable choice
The Axiom of Countable Choice (CC) is a weak form of the Axiom of Choice. It states that every countable set of nonempty sets has a choice function.
ZF+CC (that is, the Zermelo-Fraenkel axioms together with the Axiom of Countable Choice) suffices to prove that the union of countably many countable sets is countable. It also suffices to prove that every infinite set has a countably infinite subset, and that a set is infinite if and only if there is a bijection between and a proper subset of .
Defines:
countable choice
Keywords:
choice
Synonym:
countable axiom of choice, countable AC
Type of Math Object:
Definition
Major Section:
Reference
Mathematics Subject Classification
03E25 Axiom of choice and related propositions- Forums
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Comments
is proof available?
Can anyone give a proof of these statements about countable union and about infinite sets using CC? I could prove that infinite sets have countably infinite subsets using Dependent Choice only but DC implies CC, so it's 'too' strong for me.
Re: is proof available?
scineram
> Can anyone give a proof of these statements
> about countable union and about infinite sets
> using CC?
The set N x N (where N is the natural numbers) is countably infinite (no choice needed, since it's easy to construct an explicit bijection with N). Suppose S_1, S_2, S_3, ... is a sequence of countable sets. For each n, let T_n = S_n \ ( S_1 union ... union S_{n-1} ). For each n there is an injection i_n: T_n -> {n} x N. (We are choosing infinitely many injections here, so this uses CC.) Combining these injections gives us an injection from the union of the T_n into N x N. So the union of the T_n (which is the same as the union of the S_n) is countable.
Now suppose that X is an infinite set. For each natural number n there is subset S_n of X of cardinality n. (We are choosing infinitely many subsets here, so this uses CC.) The union of the S_n is countable (by the previous paragraph) and is clearly infinite. So every infinite set has a countably infinite subset.
Re: is proof available?
Many thanks. It's good to be sure about trivial statements without relying on false assumptions (full AC):D