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I am a bit confused on the following argument, the part that is highlighted: enter image description here

What does it mean that $\epsilon \cdot \aleph_0 = \epsilon$, does that mean that there exist a bijection between an infinite set and a countable copy of it? Can anyone give a proof?

MkSn1999
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    This is a standard fact about cardinal arithmetic: the product $\alpha \cdot \beta$ of two infinite cardinals $\alpha$ and $\beta$ is equal to the larger of $\alpha$ and $\beta$. – Rob Arthan Jul 15 '21 at 22:49
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    It means if $S$ is infinite and $T$ is countably infinite, then $S\times T$ has the same cardinality as $S.$ – Thomas Andrews Jul 15 '21 at 22:50
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    @MkSn1999 Note that this is obvious for any cardinality that can be written as $\varepsilon \cdot \aleph_0$, since we already know that $\aleph_0 \cdot \aleph_0 = \aleph_0$. So it shouldn't be that surprising. – Erick Wong Jul 15 '21 at 23:25
  • Thanks, can someone refer me to a proof? – MkSn1999 Jul 16 '21 at 00:10
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    I've tried to find an online proof, but everything I found either stated the result without proof, or was garbage. The first steps are easy. Assume $0 \ne \alpha \le \beta$. Then it is simple to show that $\beta \le \alpha \cdot \beta \le \beta \cdot \beta$. So it remains to show that $\beta^2 = \beta$. For that, let $\mu$ be the smallest ordinal with $|\mu| = \beta$. Use the well-ordering of $\mu$ to create a bijection between $\mu$ and $\mu \times \mu$ similar to the proof that $\Bbb N \times \Bbb N$ and $\Bbb N$ are bijective. (Note that this result requires the axiom of choice.) – Paul Sinclair Jul 16 '21 at 16:42
  • @PaulSinclair Thanks for the proof. To understand the part $\beta^2 = \beta$, do I have to use ordinals or is there a way to work around it? I am not quite familiar with the concept, could you suggest some readings? Thanks a lot! – MkSn1999 Jul 19 '21 at 23:11
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    Sorry, if I had a resource available I could had directed you to, I would have done so. That was why I was looking on-line. Failing that, I tried to recall as much as I could about how to prove it, and reproduced that. Unfortunately the rest I do not recall (if I ever even fully learned it). This was as far as I could go. Which is also why it is a comment, not an answer. I only gave the easy part. The hard part is what is left. – Paul Sinclair Jul 19 '21 at 23:48

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