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Here's the problem:

Let $F$ be field, $X$ an infinite set and $F^{X}$ be the set of all functions $f:X\rightarrow F$. Then $F^{X}$ is a vector space over $F$ (with $(f+g)(x)=f(x)+g(x)$ and $(rf)(x)=rf(x)$ ).

The thing puzzling me is how to prove the equation $\dim_{F}F^{X}=|F|^{|X|}$ because I find it difficult to determine some basis of $F^{X}$ directly. Anyway, any and all help is appreciated. Thanks.

user26857
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Jay
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  • Unless $X$ is countable, I think explicitly writing down a basis is going to be a difficult task. – David Wheeler Apr 02 '15 at 13:50
  • II think so. Moreover I notice that $F^{X}\cong V^{}$ where the $V=\sum F$ with index set $X$ is a free vector space over $F$. Then $\dim_{F}F^{X}=\dim_{F}V^{}$ and we may prove that $\dim_{F}V^{*}=|F|^{|X|}$? Even if my conjecture above is correct, I also try to find a proof which is both direct and simple. – Jay Apr 02 '15 at 14:16
  • Maybe this could help you. – user26857 Apr 02 '15 at 15:46
  • Well, I think the problem can be solved by this method. Thanks,@user26857 – Jay Apr 03 '15 at 04:32

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