Let $F$ be a field and define for any set $I$ the vector space $F^I = \{f \colon I \to F \}$ with addition $(f+g)(i) = f(i) + f(i)$ and rescaling $(\alpha f)(i) = \alpha f(i), \forall f,g\in F^I, \alpha \in F, i \in I$
From $F^I$ define
$$\bigoplus_{i\in I} F \colon=\{f\in F^I \colon f(i) = 0_{F} \text{ Except for finitely many } i \in I \} $$
I'm having trouble understanding this construct, more precisely what is means for a function $f \in \bigoplus_{i\in I} F$ to be zero except for finetely many $i$:s. This might be a bit soft but what does it mean and how should I think about it? Is it just a requirement that we must have some nonzero values?
Can someone perhaps give a good example of function in $\bigoplus_{i\in I}F$ for some set $I$ and a field $F$ or a nonexample of a function in $f\in K^I$ such that $f\notin \bigoplus_{i\in I} F$
The requirement means that the set $\{ i \in I : f(i) \neq 0_F \}$ is finite. To put in other words, $f$ takes a nonzero value only a finite number of times.
For example consider the case $I = \mathbb{N}$. Then the function $f \in F^\mathbb{N}$ that has $f(0) = 1_F$ and $f(n) = 0_F$ for $n \ge 1$ is in $\bigoplus_{i \in \mathbb{N}} F$, because the set $\{ i \in \mathbb{N} : f(i) \neq 0_F \} = \{0\}$ is indeed finite.
But the function $g \in F^\mathbb{N}$ given by $g(n) = 1_F$ for all $n \in \mathbb{N}$ is not in $\bigoplus_{i \in \mathbb{N}} F$, because $\{i \in \mathbb{N} : g(i) \neq 0_F\}$ is $\mathbb{N}$, which is infinite.