Prove that if $A$ is a set with infinite cardinality, then there exists a bijection between $A$ and $A^2$.

Question

Prove that if $A$ is a set with infinite cardinality, then there exists a bijection between $A$ and $A^2$.

I know that there exists a bijection between $\Bbb{R}$ and $\Bbb{R^2}$, but I don't think that technique can be generalized.

I think the proof of this may be highly non-trivial. Any help would be great.

What size of cardinality? You can certainly generalize for countably infinite sets. — graydad, Dec 12 '14 at 16:01
You could try well ordering the set and generalizing the countable bijection. — Matt Samuel, Dec 12 '14 at 16:50
This is equivalent to the axiom of choice. Incidentally, this question has been asked about a zillion times before, and will be closed as a duplicate in a few minutes. — Asaf Karagila, Dec 12 '14 at 18:11
This, including links, shows how to do it - without the axiom of choice - for ordinals. — Asaf Karagila, Dec 12 '14 at 18:25

score 1 · Answer 1 · answered Dec 12 '14 at 18:19

1

The only way I can think of proving this for uncountable sets is invoking the Cantor-Schroder-Bernstein theorem. http://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

You can easily find an injective function $f: A \to A\times A$ defined by something like $f(a) = (a,a)$. If you can in turn find an injection $g:A\times A \to A$ then you'll know the two sets have the same cardinality. I imagine finding $g$ will be difficult.

answered Dec 12 '14 at 18:19

graydad

14,325

This is equivalent to the axiom of choice. Of course that coming up with an injection back is "difficult". This is not quite an answer to the question (even if it were easy to find such injection). – Asaf Karagila Dec 12 '14 at 18:21

Prove that if $A$ is a set with infinite cardinality, then there exists a bijection between $A$ and $A^2$.

1 Answers1