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  "body": "\n## Statement\n\nAssume the [[axiom of choice|0HTGB13FE2]].\nLet $A$ be an [[infinite set|01MKW4R283]].\nThen there is an [[injective function|0PVG1CRAJN]] $f \\colon \\mathbb{N} \\to A$, so $A$ has a [[countably infinite|0SEZ3JGZP7]] subset $f(\\mathbb{N})$.\n\n## Remarks\n\nThe proof chooses $a_0 \\in A$, then $a_1 \\in A \\setminus \\{a_0\\}$, and so on. Each set $A \\setminus \\{a_0, \\dots, a_n\\}$ is nonempty because $A$ is infinite.\nThe axiom of countable choice is enough, but some choice is needed: in ZF alone, the statement cannot be proved.\nIt follows that $\\aleph_0$ is the least infinite cardinal.\n\n## Explanations\n\n### L1 human\n\nIn an infinite set you can always pick one more element.\nPicking forever gives a sequence of different elements, which is a copy of $\\mathbb{N}$ inside the set.\n",
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