Just wanted some help with this little proof.:
Let X and Y be Finite Sets. Prove that |X^Y| = |X|^|Y|
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Possibly related: http://math.stackexchange.com/questions/1199078/proof-of-the-definition-of-cardinal-exponentiation – Asaf Karagila Mar 21 '15 at 23:25
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$X^Y$ is the set of functions $Y \to X$. Given a particular $y$ in $Y$, how many options are there for its image in $X$? What if you consider two points $y_1$ and $y_2$ in $Y$; how many ways can they be mapped to things in $X$? What if you consider all points in $Y$?
angryavian
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So for any particular y in Y there are |X| options for its image in X? and one less option for each additional point? And all points in Y have just the one option? I'm not sure what assumptions I can make on 'jectivity – user208628 Mar 21 '15 at 20:41
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@user208628 No, for each additional point, you still have $|X|$ options. Functions are allowed to map different points of $Y$ to same points of $X$. – angryavian Mar 21 '15 at 20:43
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i tried this from the X perspective, $x_1$ (if $|Y|=4$ for example) has $2^4$ possibilities, but calculating $x_2$ number of possibilities is now dependent on $x_1$. – JMP Mar 21 '15 at 21:10