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 :-)

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- Dec 21st 2008, 06:18 AMJohn GQuestion 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 :-) - Dec 23rd 2008, 09:35 PMGammaRight 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 $\displaystyle k1=|A|\geq |B|=k2$ and $\displaystyle m1=|C|\geq |D|=m2$ then there must exist injective functions $\displaystyle f_1:A \rightarrow B$ and $\displaystyle f_2:C \rightarrow D$. But then it is pretty obvious that $\displaystyle f:A\times C \rightarrow B \times D$ defined by $\displaystyle f((a,c))=(f_1(a),f_2(c))$ would also be injective. Just write out definitions and it is clear.