# Alternative proof of the cardinality of the set of all mappings

Can anyone please tell me if there is any other proof for the cardinality of all mappings, that is not by induction, i.e., not this one (http://www.proofwiki.org/wiki/Cardinality_of_Set_of_All_Mappings) ?

Thanks

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You could just define $\vert T\vert^{\vert S\vert}$ to be $\left\vert T^S\right\vert$ and you're done! Actually for the non-finite case, that's probably the best way. –  George Lowther Mar 18 '11 at 1:50
If $A=\{a_1, a_2, \ldots, a_n\}$ has $n$ elements and $B$ has $m$ elements, a mapping $f$ from $A$ to $B$ is defined uniquely by choosing $f(a_1)$ ($m$ options), then $f(a_2)$ ($m$ options again), and so on until you choose $f(a_n)$ (as always, $m$ options). By the product rule, there are $m \times m \times \cdots \times m$ ($n$ factors) overall, namely, $m^n$.