Math 3345
Section 16 Exercise 9
Let A, B, C, D be sets such that A is equinumerous to C and B is equinumerous to D. Show that AxB is equinumerous to CxD.
I believe I have the correct answer for when all A, B, C, D are finite. I think it works but I'm not sure if it's correct for when the sets are infinite.
My answer:
I'm using the '||' symbols to denote "the cardinality of"/"the length of", and 'x' to represent Cartesian product (I apologize, I'm new to formatting on the site)
A equinumerous to C, therefore |A| = |C|
B equinumerous to D, therefore |B| = |D|
|AxB| = (|A|)(|B|) = (|C|)(|B|) = (|C|)(|D|) = |CxD|
Therefore, since |AxB| = |CxD|, then AxB is equinumerous to CxD.
Best Answer
As Asa Karagila said your argument is not right since essentially is what you have to prove (at least in the infinite case)
Try the following:
As $|A|=|C|$ then there exists a bijection $f:A\rightarrow C$. Analogous there exist a bijection $g:B \rightarrow D$.
The function $f\times g :A\times B\rightarrow C\times D$ defined as $(f\times g) (x,y)=(f(x),g(y))$ is a bijection. Prove it.
Finally conclude $|A\times B|=|C\times D|$.