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 Karagilasaid 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 $|A\times B|=|C\times D|$.