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It is a well-known fact that $\aleph_0 = \vert \mathbf{N} \vert$ is the smallest infinite cardinal number. But I'm wondering why; does anyone know a proof?

Thanks!

Steven
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2 Answers2

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You can show show the following:

Every infinite set $\mathcal{A}$ has a subset $\mathcal{B}$, so $\mathcal{B}\subseteq\mathcal{A}$, which has the same cardinality as $\mathbb{N}$, which is $\aleph_0 = \vert \mathbf{N} \vert=\vert \mathcal{B} \vert$.

This proves that $\mathbb{N}$ has the smallest infinite cardinal number.

user190080
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Every cardinal is an ordinal, and $ω$ is the first ordinal after the finite ordinals, hence $ω$ is the first infinite ordinal and hence the first infinite cardinal since it is a cardinal.

user21820
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  • Thank you for the answer, but that's not exactly what I was looking for. – Steven Aug 12 '15 at 12:49
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    @Steven: But it is exactly the correct way to understand cardinals. You didn't specify what you knew already and the definition of cardinals in set theory must involve ordinals. The other answer does not even answer the question, which is about whether $\mathbb{N}$ is the smallest cardinal, not whether $\mathbb{N}$ has the smallest infinite cardinality. – user21820 Aug 12 '15 at 13:01
  • @Steven: Because $\aleph_0$ is by definition $ω$, which is just another name for $\mathbb{N}$ in set theory. – user21820 Aug 12 '15 at 13:03