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Show each infinite set $S \subset \mathbb R$ contains a countably infinite subset.

I understand that if you remove an object from the set, it will still be infinite, and if we remove another object, it will be countably infinite (I think so at least). I'm very lost on this kind of proof. However if someone can help that would be appreciated.

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  • $\begingroup$ Each incite set $S$ contains a countable subset. Should $S$ be a subset of $\mathbb R$ or not. $\endgroup$ Commented Sep 8, 2020 at 17:53
  • $\begingroup$ @mathcounterexamples.net: I suppose you meant Each infinite set $\endgroup$ Commented Sep 8, 2020 at 18:00
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    $\begingroup$ Since all finite sets are countable, this is trivial as stated. I expect that what you really want to prove is that every infinite subset of $\Bbb R$ contains a countably infinite subset. $\endgroup$ Commented Sep 8, 2020 at 18:00
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    $\begingroup$ This is a restatement of the countable axiom of choice (aka $\omega$-AC). This equivalent to saying: $S$ has properly subset with the same cardinal as $S$. $\endgroup$ Commented Sep 8, 2020 at 18:00
  • $\begingroup$ Some authors use countable set to mean countably infinite set $\endgroup$ Commented Sep 8, 2020 at 18:04

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Clearly $S$ is nonempty so there is an element $a_1 \in S$. Then $S -\{a_1\}$ can’t be empty (as $S\neq \{a_1\} $since $S$ is not finite) and so we can pick $a_2 \in S -\{a_1\}$. Next pick $a_3 \in S - \{a_1, a_2\}$ inductively you will get a sequence and you are done!

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