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

$$A\neq\emptyset,\; B\neq \emptyset, \;A\times B \;\text{finite} \rightarrow A\; \text{and} \;B$$ 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?

• Proof is ok, yes. Notice that $f(k), b_0$ was already a good function to use Commented Sep 17, 2019 at 13:10
• @FranciscoJoséLetterio Nice observation, thank you Commented Sep 17, 2019 at 13:12
• 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. Commented Sep 17, 2019 at 15:12
• 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. Commented Sep 17, 2019 at 15:25

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.