Some problem I've found while thinking about duals of vector spaces:
Be $S$ an arbitrary set. Denote by $F(S)$ the set of finite subsets of $S$, and by $P(S)$ its power set.
Now it is easy to see that $F(S)$ forms a vector space over $\mathbb{Z}/2\mathbb{Z}$ if you define vector addition as symmetric difference and multiplication with scalar by the simple relations $0A=\emptyset$ and $1A=A$. A particular basis of that vector space is $\{\{s\}: s\in S\}$.
From that, it is also easy to construct the dual space $F(S)^*$: Its members are simply given by the elements of $P(S)$ with the following application rule: If $\alpha\in P(S)$ and $v\in F(S)$, then $\alpha(v) = \begin{cases}1 & \text{if $\alpha\cap v$ has an odd number of elements}\\0 & \text{if $\alpha\cap v$ has an even number of elements}\end{cases}$ (or simply, express the number of elements in $\alpha\cup v$ in $\mathbb{Z}/2\mathbb{Z}$. It is also not hard to show that vector addition in $F(S)^*$ is again the symmetric difference. So one can say that, given the application rule above, $F(S)^* = P(S)$ Note that for finite $S$, $F(S)=P(S)$, while for infinite $S$, $F(S)$ is smaller than $P(S)$.
So far, so good. However, what about the double-dual $F(S)^{**} = P(S)^*$? Since for finite $S$, $P(S)=F(S)$, one would conclude that the very same construction works again. However for infinite $S$, it cannot work, because you cannot say whether an infinite set has an even or odd number of elements. Also, $P(S)$ already contains all subsets of $S$, and as far as I understand, $P(S)^*$ should then be larger than $P(S)$.
Therefore my question: Does there exist a simple representation of $F(S)^{**}=P(S)^*$, and if so, what does it look like? Ideally one should easily see the equivalence to $F(S)$ in the case of finite $S$.