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I was unable to find a proof (let alone a short one) for the proposition that $$|A\times A| = |A|\text{ for any infinite set }A.$$

The function from right to left is trivial, but I cannot find one from left to right.

Would be happy for help.

modnar
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Gulzar
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    In the infinite case, this is $not$ a trivial result (as far as I know). Enderton has a proof that I found relatively easy to understand. – Matematleta Dec 28 '19 at 22:48
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    The statement that for any infinite set $A$, the cardinality of $A$ and $A\times A$ is the same, is equivalent to the Axiom of Choice (this is Tarski’s Theorem); so it is not surprising that you are having some difficulty proving it. You’ll need the Axiom of Choice to establish it in generality. – Arturo Magidin Dec 28 '19 at 22:49
  • Allright then, is it easier to show just for 'R' (real numbers vectors) – Gulzar Dec 28 '19 at 23:07
  • For $\mathbb{R}$, there are some ad hoc arguments you can make. For example, Cantor proved that $[0,1]\times[0,1]$ has the same cardinality as $[0,1]$ by "interweaving decimal representation" (map $(0.a_1a_2a_3\ldots,0.b_1b_2b_3\ldots)$ to $0.a_1b_1a_2b_2a_3b_3\ldots$). And of course, $\mathbb{R}$ is bijectable with $[0,1]$ and $\mathbb{R}\times\mathbb{R}$ with $[0,1]\times[0,1]$. – Arturo Magidin Dec 28 '19 at 23:17
  • @ArturoMagidin In fact Cantor did not interweave decimal representations at all. He first showed the irrationals in $[0,1]$ satisfied it by interweaving continued fractions (which is cleaner, as they're unique) and then arguing about uncountable sets minus countable sets etc. I've read the original papers in German (while reading the biography by Dauben) and getting it for reals instead of just the irrationals was quite a struggle for Cantor, technically. – Henno Brandsma Dec 29 '19 at 08:18
  • There are many questions on this site with answers to your question, both the general one and in the case of the reals. – Asaf Karagila Dec 29 '19 at 10:58
  • @AsafKaragila Helpful. – Gulzar Dec 29 '19 at 11:27
  • Sarcasm is equally helpful. I guess I was expecting you to use the search function on the website. Here, let me do that for you. – Asaf Karagila Dec 29 '19 at 11:57
  • I've marked three duplicates on this. There are more. Especially if you're using ordinals and the well-ordering theorem to prove the general statement. The keywords here are "Gödel pairing". Good luck. – Asaf Karagila Dec 29 '19 at 12:00
  • @HennoBrandsma: Thank you for the correction. – Arturo Magidin Dec 29 '19 at 17:49

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