Prove that if sets $A$ and $B$ are finite, then $|Atimes B| = |A|cdot|B|$.
0
$begingroup$
Prove that if sets $A$ and $B$ are finite, then $|Atimes B| = |A|cdot|B|$ . I started to show that if $B$ is a singleton ${b}$ , then $|Atimes B| = |Atimes {b}|=|A|.$ Then, if $B = {b_1, b_2, dots, b_n}$ for $ninmathbb{N}$ , so $Atimes B = bigcuplimits_{i=1}^{n} (Atimes {b_i})$ . My question is what should I do next? I though about showing that $A = {a_1, a_2,dots, a_m}$ , but I do not know how to "connect" it with $|Atimes B|$ and then with $|A|cdot|B|$ , to get the final result.
elementary-set-theory
share | cite | improve this question
edited Jan 12 at 0:15
whiskeyo
...