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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 ( ?


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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
up vote 3 down vote accepted

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, $m^n$.

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