Prove that $b\cdot\aleph_0=b$, where $b$ is an infinite set. I tried by multiplying $1\le\aleph_0\le b$ with $b$ but then I have to prove that $b\cdot b=b$ and that seems harder than my original problem.
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The usual approach is to use Zorn to show that each infinite set is a disjoint union of infinite countable subsets. – Angina Seng Jun 24 '18 at 09:51
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I think the usual approach would actually be to prove that $bb=b$, which in turn can be proved by supplying $b\times b$ with a nice well-ordering which can only be of order type $b$ (by induction on $b$) – Maxime Ramzi Jun 24 '18 at 11:03
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For the proof of $b^2=b$ you can have a look at the answer posted here: About a paper of Zermelo. Some of other post linked there might be of interest, too. – Martin Sleziak Jun 28 '18 at 12:28