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Suppose $A$ and $B$ are sets and $A$ is finite. Prove that $A \sim B$ iff $B$ is also finite and $|A| = |B|$.


Notes on notation:

$A \sim B$ indicates that $A$ is equinumerous with $B$.

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1  
Looks like a proof by contradiction would be neatest here.... –  Simon Hayward Dec 4 '12 at 13:09
    
What does the relation ~ mean ? –  Amr Dec 4 '12 at 13:12
    
@Amr: I think it means there is a bijection $A \to B$. –  Clive Newstead Dec 4 '12 at 13:23
    
I think ~ is the equinumerous or equipollent relation on sets: whenever two of them have the same cardinality, which ammounts to the same as Cilve wrote. –  DonAntonio Dec 4 '12 at 13:37
    
What does $|A|=|B|$ mean? Usually it and $A\sim B$ are just typographical variants of the same thing, but if so there is nothing to prove. –  Henning Makholm Dec 4 '12 at 13:39

2 Answers 2

What do you mean by 'finite' here? Something like $A$ is finite iff $A$ is empty or there is an $m$ such that there is a bijection between $A$ and $\{1, 2, \ldots m\}$??

Assuming so, suppose $A$ and $B$ are both non-empty but finite, i.e. there is an $m$ and $n$ such that $A$ and $B$ are equinumerous with $\{1, 2, \ldots m\}$ and $\{1, 2, \ldots n\}$ respectively. Now what happens when $m = n$, when $m \neq n$???

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$$A\sim B\Longleftrightarrow \,\exists\, f:A\to B\,\,,\,f\,\,\text{is a bijection}\Longleftrightarrow |A|=|B| $$

so

$$A\,\,\text{is finite together with the above}\,\Longleftrightarrow |B|=|A|<\infty$$

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