Set of finite subsets as vector space: Double dual?

Question

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 $S$.

Answer 1

As a vector space, $F(S)$ is a vector space over $\mathbb{F}_2$ with basis indexed by $S$ itself. So it has dimension $|S|$. It is isomorphic to the direct sum of $|S|$ many copies of $\mathbb{F}_2$, and so the dual space is isomorphic to the direct product of $|S|$ many copies of $\mathbb{F}_2$.

As a vector space, $P(S)$ is the direct product of $|S|$ many copies of $\mathbb{F}_2$, so it is indeed isomorphic to $F(S)^*$, and so $F(S)^{**} = P(S)^*$.

This just follows from the universal property: a linear map $$f\colon \bigoplus_{s\in S}(\mathbb{F}_2)_s\to \mathbb{F}_2$$ is completely determined by what happens to the basis. Specifying what happens to each basis element $1_s$ is equivalent to specifying an element of $\mathop{\prod}\limits_{s\in S}\mathbb{F}_2$, where the map $f$ corresponds to the tuple $(f(1_s))_s$. Conversely, any element $\mathbf{f}$ of the product defines a function $\mathop{\oplus}\limits_{s\in S}(\mathbb{F}_2)_s\to\mathbb{F}_2$ by mapping the basis element $1_s$ to $\mathbf{f}_s$.

Once you've identified $F(S)^*$ with $P(S)$, it follows that $F(S)^{**}$ can be identified with $P(S)^*$. However, in order to make a similar correspondence you would need to start with a basis for $P(S)$ (so as to express it as a direct sum); this is difficult to do (you need AC, I believe, to guarantee the existence of a basis).

Note that when $S$ is infinite, it is difficult to even express $F(S)^{**}$, let alone make it correspond to something "obvious". It takes a bit of cardinal arithmetic to even show that it cannot be isomorphic to $F(S)$.

(Yes, if $V$ is an infinite dimensional vector space, then $V^*$ is a vector space of dimension strictly larger than $V$: see for example this answer; in particular, if $S$ is infinite, then $P(S)$ is an infinite dimensional vector space of dimension strictly larger than $|S|$, and so $P(S)^*$ is itself of dimension strictly larger than $\dim(P(S))$. )