# Set of finite subsets as vector space: Double dual?

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 $S$.

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+1. I changed \mbox{} to \text{}, and I wonder why I sometimes see \mbox{} used for that purpose? Is there some help page on stackexchange or some web site somewhere that says that's how it should be done? If you're using $\TeX$ or LaTeX in the normal way, as opposed to the way it's used on web sites, \mbox{} simply doesn't have that same effect, but is used for an altogether different purpose. Therefore this usage misleads people about $\TeX$ usage. – Michael Hardy Jul 7 '12 at 22:51
@MichaelHardy: \text is an amsmath macro. As far as I know, without amsmath the only easy way to locally enter text mode in math is via \mbox. Therefore anybody who has done a lot of $\text{LaTeX}$ before learning about the amsmath commands (as I did) will have developed the habit of using \mbox here (as well as some other hard to change habits, like using eqnarray instead of align). – celtschk Jul 7 '12 at 23:08
Is there some version of TeX or LaTeX in which \mbox{} has that effect, other than on web sites? – Michael Hardy Jul 7 '12 at 23:33
@MichaelHardy: Is there any version where \mbox doesn't have that effect? (Note that \mbox does also reset the font size, so you cannot use it in subscripts without creating ugly effects, but that just means you wouldn't use it e.g. in indices). – celtschk Jul 7 '12 at 23:55

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 $\dim(P(S))$. )

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Thank you also for your additional explanation of why the construction works. – celtschk Jul 7 '12 at 23:23