Let $A$ and $B$ be sets. Prove that $ \#(A \times B) = \#(B \times A)$.
What I have done:
There exist an element $m$ in $A$ such that the element also exists in $B$. If $\#A = \#B$, then $\#B = \#A$. Any two sets cross-multiplied must have equal sets of elements or else you cannot cross-multiply.
Also, there exist elements $(1,2,…n)\in A$ that is also $\in B$.
I need help using what I have, or maybe I shouldn't use this information for this proof…
Best Answer
The map $f:A\times B\to B\times A,(a,b)\to (b,a)$ is 1-1 and surjective.Let $(b',a')=(b,a)$ in $B\times A$. Then $b'=b$ and $a'=a$. This means that $(a',b')=(a,b)$ in $A\times B$ (1-1).
Also if $(b,a)\in B\times A$ then $f((a,b))=(b,a)$ (surjection).