Show that $F^{I}=F^{I}_{\text{fin}} \iff I$ finite, where $I$ is a set and $F$ is a field

Question

$F^{I}$ is the vectorspace of functions which map from $I$ to $F$ and $F^{I}_{\text{fin}}:= \{ f:I\rightarrow F |\text{ } \overset{\exists}{n\in\mathbb{N}_0,}\overset{\exists}{I_f\subseteq I}:\text{#} I_f = n \text{ and} \begin{cases}f(i)=0\text{ , for each }i\in I\backslash I_f\\f(i)\neq 0\text{ , for each }i\in I_f\end{cases}\}$

The context is that I have already proved that $B:=\{e_i:i\in I\}$ is linear independent in $F^{I}$ where

$$\forall_{i\in I}e_i:I\rightarrow F, e_{i}(j):=\delta_{ij}:=\begin{cases} 1,\text{ if }i=j\\0,\text{ if }i\neq j\end{cases}$$

And now I want to Show that for $F_{\text{fin}}^{I}$ $B$ is a base. For the special case it is noteworthy that $F_{\text{fin}}^{I}$ and $F^{I}$ describes the same vectorspace and thus have the same base if $I$ is finite.

Answer 1

We do not even need that $F$ is a field. If $I$ is finite, then we certainly have $F^I = F_{\mathrm{fin}}^I$. The inclusion $F_{\mathrm{fin}}^I \subset F^I$ always holds (by definition). To see the other inclusion, let $f\in F^I$. Then $$ \lbrace i\in I\,\vert \, f(i)\neq 0\rbrace\subset I, $$ and since the RHS is finite, the LHS must also be finite. Hence $f\in F_{\mathrm{fin}}^I$ by definition.

If $I$ is infinite, simply pick a nonzero element in $F$ (e.g. $1$) and consider the function which maps all elements of $I$ to $1$. This function does not have finite support, hence $F^I \neq F_{\mathrm{fin}}^I$, proving the contrapositive of the opposite implication, hence completing the proof.

Answer 2

The space $F^I_{\mathrm{fin}}$ can be more simply defined as $$ F^I_{\mathrm{fin}}=\bigl\{f\in F^I: |\{i\in I:f(i)\ne0\}|<\infty \bigr\} $$ Take $u\in F^I$ defined by $u(i)=1$, for every $i\in I$. Then $u\notin F^I_{\mathrm{fin}}$, unless $I$ is finite.