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let $X$ and $Y$ be sets and $Y^X$ the set of function $f:X\to Y$. How can we interpret $Y^X$ as the cartesian product $\prod_{x\in X}Y_x$ where $Y_x=Y$ for each $x\in X$?

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    They are the same. No interpretation is needed.2012-02-23
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    I’m not really sure what your question is: by definition that Cartesian product is the set of functions from $X$ to $Y$.2012-02-23
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    for example suppose $X$ is finite. then we have the bijection $$\prod_{x\in X}Y_x\to Y^X$$ defined by sending a tuple $$(y_1,y_2,...,y_n)$$ maps to the map $f$ that sends $f(x_1)=y_1,...,f(x_n)=y_n$ is that correct?2012-02-23
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    Yes, that is correct.2012-02-23

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The elements in the Cartesian product $\prod_{x\in X}Y_x$ are sequences indexed by $X$ whose elements are members of $Y$, namely $\langle y_x\mid x\in X\rangle$.

Such sequence is naturally isomorphic to $\{\langle x,y_x\rangle\mid x\in X\}$, which is exactly a function from $X$ to $Y$.

This means that there is a very natural way to identify $\prod_{x\in X} Y_x$ with $Y^X$.

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    That natural isomorphism is the identity: a sequence indexed by $X$ whose elements are members of $Y$ **is** a function from $X$ to $Y$.2012-02-23
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    @Brian: Indeed if we think on the set theoretical universe, in particular the one of ZF, in which products cannot be arbitrarily defined except for functions. However, on the more elementary and intuitive level sequences are not *exactly* functions, they are tuples (possibly infinitely long).2012-02-23
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    Depends on who’s teaching them, and in what context: in any course in which I’ve had reason to teach infinite Cartesian products, I’ve also taught sequences as functions.2012-02-23
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    @Brian: But what if $X$ is finite? I do agree that in a context where products come up, it is a good idea to teach them as functions. Apparently not everyone do that (otherwise this question would not have come up...) :-)2012-02-23
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    Well, no matter how you teach products, sooner or later you'll be confronted with the question of why $(X \times Y) \times Z$ is "only" naturally isomorphic to $X \times (Y \times Z)$ and not actually equal...2012-02-23
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    @Zhen: Also $X\times Y\times Z$ can be thrown into that mix. :-)2012-02-23