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$A_n=\{1,2,3,\dots,n\}$ and $B$ is a nonempty set, define $$\prod_{a\in A_1}B:=B\;,$$ for $n>1$ we have $$\prod_{a\in A_n}B:=B\times\prod_{a\in A_{n-1}}B\;.$$ What exactly is $\prod_{a\in A_3}$ here and why is it equivalent to $\{f\mid f:A_3\to B\}$

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Please check to make sure that I interpreted everything correctly. – Brian M. Scott Feb 26 '13 at 18:15
I think that you omitted a $B$ from that product in the last line. Also, what do you mean by equivalent? That there is a bijection between the two sets? – Brian M. Scott Feb 26 '13 at 18:21

By definition

$$\prod_{a\in A_3}B=B\times \prod_{a\in A_2}B=B\times\left(B\times\prod_{a\in A_1}B\right)=B\times(B\times B)\;.$$

For convenience I’ll call this set $S$. The members of $S$ are all objects of the form $\big\langle b_1,\langle b_2,b_3\rangle\big\rangle$ with $b_1,b_2,b_3\in B$.

Let $F$ be the set of functions from $A_3$ to $B$. Each $f\in F$ is a subset of $A_3\times B$ of the form

$$\big\{\langle 1,b_1\rangle,\langle 2,b_2\rangle,\langle 3,b_3\rangle\big\}$$

for some $b_1,b_2,b_3\in B$. Now it shouldn’t be too hard to see how to pair up elements of $S$ and $T$: we want to pair each $\big\langle b_1,\langle b_2,b_3\rangle\big\rangle\in S$ with $\big\{\langle 1,b_1\rangle,\langle 2,b_2\rangle,\langle 3,b_3\rangle\big\}\in F$ and vice versa. In other words, we want to look at the function

$$\varphi:S\to F:\big\langle b_1,\langle b_2,b_3\rangle\big\rangle\mapsto\big\{\langle 1,b_1\rangle,\langle 2,b_2\rangle,\langle 3,b_3\rangle\big\}\;.$$

Now that we’ve identified the function $\varphi$, it’s straightforward to verify that $\varphi$ is a bijection.

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Thanks for your help. – user64030 Feb 27 '13 at 4:18
@user64030: You’re welcome. – Brian M. Scott Feb 27 '13 at 5:01

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