Proving that if $A$ is a non-empty set, then $|A| ≤ |A \times A|$

Question

I just need some help with this problem.

Let $A$ be an non-empty set. Prove that $|A| \leq |A \times A|$. $A$ may or may not be infinite!

Intuitively, this statement makes sense. $A \times A$ must have at least as many elements in it as $A$ does, but how can I prove this statement? I can make a hypothetical set $A$ that consists of the elements say $\{1,2,3\}$ and illustrate that $A \times A$ obviously has a bigger cardinality than $A$ but I have trouble when it comes to proving this statement for all sets $A$, whether infinite or not...

Thank you!

@nicefella I edited your question to use correct tex codes. You might want to see the edits to learn how to do it yourself. — Ittay Weiss, Apr 01 '13 at 01:10

score 4 · Accepted Answer · answered Apr 01 '13 at 01:04

To furnish a proof you need to work with the definition. Now, $|A|\le |A\times A|$ has a very precise meaning. It means that there exists an injection $f:A\to A\times A$. So, try to construct such an injection. For inspiration, you can try to take a particular set $A$ and see if you can find a very naturally occurring such injection. Then go from the particular example to the general case.

By the way, the restriction that $A\ne \emptyset $ is not needed.

I've read "needed" where you wrote "not needed"...sorry. – Julien Apr 01 '13 at 01:15 — Julien, Apr 01 '13 at 01:15

score 3 · Answer 2 · answered Apr 01 '13 at 01:00

3

Hint: You need to show that there is an injection $A \to A \times A$. As $A$ is nonempty choose $a \in A$ and think about $\{a\} \times A$.

answered Apr 01 '13 at 01:00

Jim

31,547

6

Or you can simply consider the diagonal injection and not even bother about non emptyness. – Julien Apr 01 '13 at 01:09

Proving that if $A$ is a non-empty set, then $|A| ≤ |A \times A|$

2 Answers2

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