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

$$F^{I}$$ 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.

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.
• Why is $F^I = F_{\mathrm{fin}}^I$ if $I$ is finite? Jan 18, 2019 at 21:34
• If $I$ has an integer-valued cardinality (e.g. is finite), then what we might refer to as the "support" of a function $f\colon I\rightarrow F$ must have a (smaller) integer-valued cardinality (being a subset of $I$).
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.