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Let $D$ be any set and define $\mathcal{P}(D)=\{A:A\subseteq D\}$ to be the power set of $D$.

For any $A,B\in\mathcal{P}(D)$, define the symmetric difference operation $$A*B=(A\setminus B)\cup (B\setminus A).$$ It is pretty easily shown that $G(D)=(\mathcal{P}(D),*)$ is an abelian group, and its identity element is the empty set $\varnothing$. I will include a proof at the end of the question.

Anyway, consider any subset $A\in\mathcal P(D)$ and observe that $$A*A=(A\setminus A)\cup (A\setminus A)=\varnothing\cup\varnothing=\varnothing.$$ Thus every element of $G(D)$ has order $2$. Because of this, when $D=\{d_1,d_2,...,d_N\}$ is finite, we can say two things.

Firstly, $G(D)$ is generated by the singleton subsets of $D$: $$G(D)=\left\{(a_1\{d_1\})*(a_2\{d_2\})*\cdots *(a_N\{d_N\}):a_i\in\Bbb Z/2\Bbb Z\right\},$$ where we define $0\{d_i\}=\varnothing$ and $1\{d_i\}=\{d_i\}$, and $(a\{d\})*(b\{d\})=(a+_2b)\{d\}$, with $+_2$ being addition modulo $2$.

Secondly, we can define an isomorphism $\phi:\mathcal P(D)\to (\Bbb Z/2\Bbb Z)^N$ by the map $$\phi:(a_1\{d_1\})*(a_2\{d_2\})*\cdots *(a_N\{d_N\})\mapsto (a_1,a_2,...,a_N).$$ This is an isomorphism because $$\begin{align} \phi\left(A*B\right)&=\phi\Big((a_1+_2b_1)\{d_1\}*\cdots*(a_N+_2b_N)\{d_N\}\Big)\\ &=\Big(a_1+_2b_1,...,a_N+_2b_N\Big)\\ &=\Big(a_1,...,a_N\Big)+\Big(b_1,...,b_N\Big)\\ &=\phi(A)+\phi(B). \end{align}$$ Here, $A=(a_1\{d_1\})*\cdots*(a_N\{d_N\})$ and $B=(b_1\{d_1\})*\cdots *(b_N\{d_N\})$.

The upshot is that $G(D)\cong (\Bbb Z/2\Bbb Z)^{|D|}$ when $D$ is finite.

My question is, what happens when $D=\Bbb Z$ or $D=\Bbb R$? Or more generally, when $D$ is not finite? Can we still decompose $G(D)$ into copies of $\Bbb Z/2\Bbb Z$?

So far, I have made little progress. If $D$ is countably infinite (in bijection with $\Bbb Z$), we can enumerate the generators $\{\{d_1\},\{d_2\},...\}$, and its clear that every finite subset of $D$ can be written as a finite combination of the generators it contains. However, I am not sure how to deal with infinite subsets of $D$, because I don't know how we can appropriately take the $*$ operation infinitely many times.

When $D$ is uncountably infinite (such as $\Bbb R$ or $\Bbb C$), I have no idea how to proceed. Any ideas?

Here is a proof that $G(D)=(\mathcal P(D), *)$ is an abelian group.

If $A,B\in\mathcal P(D)$, we have $A\setminus B, B\setminus A\in\mathcal P(D)$, so that $A*B\in\mathcal P(D)$. That is, $G(D)$ is closed under $*$.

We have that $*$ is commutative, because $$B*A=(B\setminus A)\cup (A\setminus B)=(A\setminus B)\cup (B\setminus A)= A*B.$$

The identity element is $\varnothing$ because $$A*\varnothing=(A\setminus\varnothing)\cup(\varnothing\setminus A)=A\cup\varnothing=A,$$ and every element is its own inverse as shown above.

clathratus
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1 Answers1

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Associate to every subset $A$ of $D$ its indicator function $\chi_A\colon D\to 2=\{0,1\}$, given by $$\chi_A(x) = \left\{\begin{array}{ll} 1 &\text{if }x\in A\\ 0 &\text{otherwise.} \end{array}\right.$$ We know that $P(D)$ is naturally isomorphic to the set $2^D=\{\chi_A\mid A\in P(D)\}$, by associating each subset with its indicator function.

We can also identify $2$ with the field with $2$-elements, and thus consider $2^D$ as being the set $\mathbb{F}_2^D$ of functions from the set $D$ to the field $\mathbb{F}_2$. This set has a natural structure as an $\mathbb{F}_2$-vector space by pointwise addition. This addition corresponds precisely to the symmetric difference. That is, $\chi_A + \chi_B = \chi_{(A\Delta B)}$. This is essentially your observation, as an $\mathbb{F}_2$-vector space structure is equivalent to a $\mathbb{Z}/2\mathbb{Z}$ action.

Since this is a vector space, assuming the Axiom of Choice it has a basis $\mathcal{B}$. That basis consists of functions $\chi_B$ with the property that every subset of $D$ can be written uniquely as a symmetric difference of finitely many subsets from among those that occur in elements of $\mathcal{B}$. In fact, you can assume that your set $\mathcal{B}$ includes the singletons (which yield linearly independent indicator functions), since every linearly independent set can be extended to a basis. So assuming the Axiom of Choice, there is definitely an isomorphism of $(P(D),\Delta)$ with a direct sum of copies of $\mathbb{F}_2$, infinitely many when $D$ is infinite. So for example, $(P(\mathbb{Z}),\Delta)$ will be isomorphic to a direct sum of $2^{\aleph_0}$ copies of $\mathbb{Z}/2\mathbb{Z}$.

However, you cannot exhibit an explicit basis/isomorphism in general. This because it is consistent with $\mathsf{ZF}$ that there are groups of exponent $2$ that are not trivial but have trivial automorphism group, which means that they do not have an $\mathbb{F}_2$-basis.

You can't really define "infinite sums" in this vector space (there is no norm through which you can take limits). If you had a sequence of subsets $A_1,A_2,\ldots$ and attempted to define something like "partial sums", $$S_1 = A_1, S_2=A_1\Delta A_2,\ldots, S_n=A_1\Delta\cdots\Delta A_n,$$ you could consider $\limsup S_n$ and $\liminf S_n$, and define the "infinite symmetric difference" to be their common value if defined. But in "most" cases you will have that these two sets are not equal.

Arturo Magidin
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