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Let $X$ be any set, $F$ be any field, and $F^X=\{f\colon X\to F\}$ (the set of all functions from $X$ to $F$ with usual point wise addition and multiplication by scalars). Why is $F^X$ finite dimensional if and only if $X$ is finite?

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    Please use some mark-up. Also, please note that most mathematicians will not identify $X$ with $x$, so it's important to not use them interchangeably.2011-08-30

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For each $x\in X$, let $\mathbf{e}_x\colon X\to F$ be given by $\mathbf{e}_x(y) = \left\{\begin{array}{ll} 1 & \text{if }x=y,\\ 0 & \text{if }x\neq y. \end{array}\right.$

Then $\{\mathbf{e}_x \mid x\in X\}$ is linearly independent. Given $x_1,\ldots,x_n\in X$, pairwise distinct, if $\alpha_1\mathbf{e}_{x_1}+\cdots \alpha_n\mathbf{e}_{x_n}=\mathbf{0},$ then evaluating at $x_i$ gives $\alpha_i=0$.

If $X$ is infinite, this proves that $F^X$ has an infinite linearly independent subset, so $F^X$ is infinite dimensional.

If $X$ is finite, $X=\{x_1,\ldots,x_n\}$, and $f\in F^X$, then note that $f = f(x_1)\mathbf{e}_{x_1} + \cdots + f(x_n)\mathbf{e}_{x_{n}},$ so the set spans $F^X$. Thus, $F^X$ has a basis with $n$ elements, hence is finite dimensional.

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    Awesome, perfe$c$t, I had a rough idea but was struggling, that's sorted it all out -thanks!2011-08-30
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The set of elements $1_x $, where $ 1_x(y) = \cases{1, & y = x \\ 0, & otherwise} $ will always be a basis for FX.

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    It won't be a basis if $X$ is infinite, but it will be linearly independent.2011-08-30