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Cardinality and Bijection

This was a question on my last test. The whole class got it wrong. I understand the falacy of my argument because there might be other functions f defined other than the bijection I stipulated. Can anyone help solve this? Thanks.

Suppose f: A$\displaystyle \rightarrow$B is a function between two finite sets A and B with the same cardinality. Prove that f is 1-1 iff f is onto.

Thanks, we were never given that definition

Quote:

Originally Posted by

**tonio** Let $\displaystyle A=\{a_1,...,a_n\}\,,\,\,B\{b_1,...,b_n\}$ :

1) Suppose f is 1-1 ==> the set $\displaystyle f(A)\subset B$ has as many elements as A (and as B) ==> $\displaystyle f(A)=B$, otherwise B is a finite set having a proper subset with its same cardinality and this is the definition of INFINITE sets ==> f is onto.

2) Suppose now that f is onto ==> $\displaystyle f(A)=B$; if $\displaystyle f(a_i)=f(a_j)\,\,and\,\,a_i\neq a_j$ then $\displaystyle Card(f(A))\leq n-1$, which is impossible since $\displaystyle Card(F(A))=Card (B)=n$, thus it must be that $\displaystyle a_i=a_j$ ==> f is 1-1.

Tonio

Thanks, we were never given that definition. The only definition in Rosen is "A set is said to be infinite if it is not finite."

This is the first math class I've ever taken that has not been fun! (Crying)