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$A\neq\emptyset,\; B\neq \emptyset, \;A\times B \;\text{finite} \rightarrow A\; \text{and} \;B $ are both finite too.

Assume on contrary that one of $A$ or $B$ is infinite. WLOG let $A$ infinite and $B$ finite.

$A $ infinite $\rightarrow$ $\exists f:N\to A$ an injection

$B$ finite $\rightarrow \exists g:[n]\to B$ a bijection for some $n\in N$

Since $B\neq \emptyset$ $\rightarrow \exists b_0\in B$

then $h:N \to A\times B$ where

$k\mapsto (f(k),g(k)) $ if $k\in [n]$ and

$k\mapsto (f(k),b_0) $ if $k>n$

This map is injection so $A\times B$ becomes infinite contradiction

Is this proof okay?

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    $\begingroup$ Proof is ok, yes. Notice that $f(k), b_0$ was already a good function to use $\endgroup$
    – Francisco José Letterio
    Commented Sep 17, 2019 at 13:10
  • $\begingroup$ @FranciscoJoséLetterio Nice observation, thank you $\endgroup$
    – chesslad
    Commented Sep 17, 2019 at 13:12
  • $\begingroup$ The proof is not ok as written, though. One has to translate what you write to understand what you meant to say instead. For instance, the sentence starting with "then $h$" makes no sense. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 17, 2019 at 15:12
  • $\begingroup$ A slight nitpick. You didn't not consider if $A$ and $B$ are both infinite. Which is to say just assume wolog $A$ is infinite and assume nothing about $B$ except it is non empty. Otherwise your proof is fine but a bit stiff and overly formal. It'd be a lot simpler to say: let $b\in B$ then $f(x)=(x,b);f:A\to A\times\{b\}\subset A\times B$ is easily shown to be a bijection. So if $A$ is infinite we have an injection from an infinite set to a finite set. $\endgroup$
    – fleablood
    Commented Sep 17, 2019 at 15:25

1 Answer 1

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Your proof is fine. But you can do it directly: fix $a\in A;\ b\in B$ and note that the inclusions $i_b:A\to A\times B:x\mapsto (x,b)$ and $j_a:B\to A\times B:y\mapsto (a,y)$ are injections into a finite space.

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