# All infinite sets have a cardinality of at least aleph naught

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
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
This seems related: Why is $\omega$ the smallest $\infty$? – Martin Sleziak Sep 30 '12 at 20:01

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.
(1) $X$ is D-infinite;
(2) $|X|=|X|+1$;
(3) $\aleph_0 \le |X|$.