1. ## Question in cardinality

k1, k2, m1, m2 are cardinal numbers.
I need to prove that if k2>=k1 and m2>=m1, then k2m2>=k1m1
I know that i need to prove that this function f:AXC->BxD is injective, but then what?
Thanks for any help :-)

2. ## Right Track

I mean I think you are already pretty much done. I am guessing from your notation this is what you mean. You know since $k1=|A|\geq |B|=k2$ and $m1=|C|\geq |D|=m2$ then there must exist injective functions $f_1:A \rightarrow B$ and $f_2:C \rightarrow D$. But then it is pretty obvious that $f:A\times C \rightarrow B \times D$ defined by $f((a,c))=(f_1(a),f_2(c))$ would also be injective. Just write out definitions and it is clear.