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If $A$ is any infinite set, then how to prove $\mathrm{card}(A)=\mathrm{card}(A×A)$.

I have seen one proof in the book "Algebra" by Serge Lang (revised third edition, Appendix 2, Theorem 3.6), that is very difficult to understand. Can someone just suggests which book to see for it? Please explain idea too.

Vsotvep
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user582661
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  • It would help if you would tell which book, and what part was difficult to understand. – Vsotvep Feb 15 '20 at 11:33
  • I'm pretty sure that "Graduate texts in mathematics" is not the title of the book... – Vsotvep Feb 15 '20 at 11:52
  • @Vsotvep it is there only in appendix 2 – user582661 Feb 15 '20 at 12:08
  • "graduate texts in mathematics" is a series of (well...) graduate texts published by Springer. There are more than 300 books published in that series, twelve of which by Serge Lang – Vsotvep Feb 15 '20 at 12:22
  • @Vsotvep ohh sorry I forgot to mention it is algebra one – user582661 Feb 15 '20 at 12:23
  • "ohh sorry I forgot to mention it is algebra one " for heaven's sake!!! After repeated requests you can't be bothered to tell us the title?????? – David C. Ullrich Feb 15 '20 at 12:33
  • @DavidC.Ullrich I am extremely very sorry for it. I thought mentioning 'algebra' is not necessary.. and then I told about it in comment rather than editing. I am very sorry for it, Sir. – user582661 Feb 15 '20 at 12:44
  • @user582661 I have edited your question to add the reference in such a way that people can actually find the proof you have difficulties with. Now, it would still be most helpful if you could explain what part you have difficulties with. Since you're reading "Algebra", I presume you do not know what aleph numbers are, so my answer below might not make a lot of sense. I'll expand my answer a bit to explain what it is about. – Vsotvep Feb 15 '20 at 12:46
  • @Vsotvep Thank you so much Sir for your help. Now I understand it. – user582661 Feb 15 '20 at 13:03

1 Answers1

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Note that "$|A|=|A\times A|$ for all infinite sets $A$" is equivalent to the Axiom of Choice, this is a theorem by Tarski. So we will have to use the Axiom of Choice to prove the claim.

Here is how I would do it:

The Axiom of Choice is equivalent to the Well-Ordering Theorem, which states that it is possible to define a well-order on any set. If $A$ is a set, then $<$ is a well-order on $A$ if $<$ is a total order and if every nonempty subset of $A$ has a least element (with respect to $<$). I believe in Lang this is called inductively ordered.

Therefore, for every infinite set $A$, we can find a well-order $<$ of $A$. In fact, we can do even better: we can find a minimal well-order, in the sense that this minimal well-order has an order-embedding into any other well-order of $A$. This is what is meant with aleph numbers, in a rough way.

Now, the way to show that $A$ and $A\times A$ have the same cardinality, is by defining a well-order $<_{cw}$ of $A\times A$ using some minimal well-order $<$ of $A$, and then showing that $\langle A\times A,<_{cw}\rangle$ and $\langle A,<\rangle$ are order-isomorphic.

This is what is done in this question & answer.

Vsotvep
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