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How would we prove that infinite sets have at least a cardinality of aleph naught?

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This is the definition of $\aleph_0$, no? The smallest infinite cardinal. –  Berci Sep 30 '12 at 19:56
    
How would we prove the definition? –  Derpstar Sep 30 '12 at 19:58
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What are your definitions of "infinite" and "aleph naught"? –  Chris Eagle Sep 30 '12 at 19:59
    
"infinite" set means there are an infinite number of elements in the set. "aleph naught" means the cardinality of the natural numbers. –  Derpstar Sep 30 '12 at 20:00
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This seems related: Why is $\omega$ the smallest $\infty$? –  Martin Sleziak Sep 30 '12 at 20:01

1 Answer 1

I am not sure to which extent this fulfills OPs needs, but perhaps it is useful to have this somewhere for reference. (I guess this can be found, in some form, in several other questions at this site.)

Several definitions of finite and infinite sets are used in mathematics. The following result, taken from H. Herrlich: Axiom of Choice, p.44, shows that one of them, called Dedekind-infinite, is equivalent to having cardinality at least $\aleph_0$. You can find a detailed proof there.

Definition 4.1. A set X is called Dedekind–infinite or just D–infinite provided that there exists a proper subset $Y$ of $X$ with $|X| = |Y|$; otherwise $X$ is called Dedekind–finite or just D–finite.

Proposition 4.2. Equivalent are:
(1) $X$ is D-infinite;
(2) $|X|=|X|+1$;
(3) $\aleph_0 \le |X|$.

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