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This is a homework question, so I am looking for a hint rather than an answer.

If $S$ is finite, then $\mathbb{F}^S$ is a finite dimensional vector space. (Note: $\mathbb{F}^S$ is the set of functions from $S$ to $\mathbb{F}$.

My proof goes like so:

1) $\mathbb{F}^S$ is a finite dimensional vector space if a list of vectors $f_1,f_2,\dots,f_n$ in $\mathbb{F}^S$ spans $\mathbb{F}^S$.

2) The list $f_1,f_2,\dots,f_n$ of vectors from $\mathbb{F}^S$ spans $\mathbb{F}^S$ if span$(f_1,f_2,\dots,f_n)=\mathbb{F}^S$.

Choose an arbitrary element $f$ from span$(f_1,f_2,\dots,f_n)$. Then we can write $f$ as

$$f=a_1f_1+a_2f_2+\cdots+a_nf_n,$$ where each of the $a_i$ belongs to $\mathbb{F}$. Since $\mathbb{F}^S$ is a vector space, then each element of the sum is also a vector belonging to $\mathbb{F}^S$ and thus so is the sum itself. So $f$ itself belongs to $\mathbb{F}^S$.

My problem: I never used the finiteness of $S$, so I have a feeling that this proof is flawed. Can somebody give me a hint as to where I am lacking or where I need to use the finiteness of $S$? Thanks.

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    $\begingroup$ You never actually proved the existence of the list of vectors $f_1,\dots , f_n$, and that's the whole point of the exercise. $\endgroup$ – uniquesolution Sep 6 '15 at 19:37
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Hint:

You assumed that the finite set $\{f_1,...,f_n\}$ spans $\mathbb F^S$ to begin with. In assuming this, you assumed the result you are trying to prove. The key is to find a finite set $\{f_1,...,f_n\}$ that spans $\mathbb F^n$ (that is, explicitly define each $f_i$). Can you think of such a set?

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  • $\begingroup$ My first thought would be to enumerate $S$ by $S=\left\lbrace s_1,s_2,\dots,s_n\right\rbrace$ and define the $f_i$ off of that. $\endgroup$ – nonremovable Sep 6 '15 at 19:41
  • $\begingroup$ That's a good start. In fact, you shouldn't need any more than $n$ functions to form a basis. $\endgroup$ – Alex S Sep 6 '15 at 19:46
  • $\begingroup$ I defined each of the $f_i$ as $f_i=1$ if $s=s_i$ and 0 otherwise. I think I got the answer now. Thanks for the hint. $\endgroup$ – nonremovable Sep 7 '15 at 0:10
  • $\begingroup$ That's the way I would have done it. $\endgroup$ – Alex S Sep 7 '15 at 0:16
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Since $S$ is finite, there exists a bijection with $[n]=\{1,2,3,\cdots,n\}$, say $g:S\rightarrow [n]$. Show that this induces an isomorphism between $\mathbb{F}^n\cong \mathbb{F}^{[n]}$ and $\mathbb{F}^S$. If you need further hints on how to produce this isomorphism, just ask.

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