Finite dimensional vector space.

Question

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.

Answer 1

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?

Answer 2

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.