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Natural numbers are famously a way to build up "the rest" of the numbers: integers as pairs of natural numbers modulo the correct equivalence relation, and similarly for rational numbers, etc.

The powerset of some set $A$ has a similar structure to the natural numbers, in that $(\mathbb{N}, +, 0)$ is a monoid, and so is $(\mathcal{P}(A), \cup, \emptyset)$, but neither has a subtraction that is always nicely behaved. However, one could imagine extending $\mathcal{P}(A)$ in an analogous way to how we obtain the integers, by defining an equivalence relation

$$(X, Y) \sim (Z, W) \iff X \cup W = Z \cup Y$$

Essentially making $(X, Y)$ into $X \setminus Y$, but without losing information in the case that $Y \subsetneq X$, just as we can represent an integer $a - b$ as $(a,b)$ without losing information when $b > a$ (assuming we use saturating subtraction for natural numbers).

At any rate, my question is, does this structure have a name, and if so, does the study of it lead anywhere interesting?

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Contra your claim, the relation $\sim$ is not an equivalence relation: consider $\alpha=(\{1,2\},\{1,2\})$, let $\beta=(\{1,2\},\{1\})$ and let $\gamma=(\{1\},\{1,2\})$. Then we have $\alpha\sim \beta$ and $\alpha\sim\gamma$ but $\beta\not\sim\gamma$.

The issue is that whenever $Z\cup W\subseteq X=Y$ we have $(X,Y)\sim (Z,W)$ for silly reasons: $X\cup W=X$, $Z\cup Y=Y$, and $X=Y$.

In fact, more generally we have $(\mathbb{N},\mathbb{N})\sim(X,Y)$ for every $X,Y\subseteq\mathbb{N}$. So if we try to fix things by looking at the transitive closure of $\sim$ instead, everything trivializes.


EDIT: We can fix this specific problem by restricting attention to the set $Disj_2(\mathbb{N})$ of disjoint pairs of sets of natural numbers. However, this causes two new issues.

First, there's no point in bringing an equivalence relation into the picture anymore: if $A\cup Y=B\cup X$ and $A\cap B=X\cap Y=\emptyset$ then $A=X$ and $B=Y$. For example, we have $A\subseteq A\cup Y=B\cup X$ so $A\subseteq X$ since $A\cap B=\emptyset$, and similarly $X\subseteq A$; and by symmetry, we also have $B\subseteq Y$ and $Y\subseteq B$.

More importantly, we now need to be careful about our arithmetic operations: "coordinatewise union" is no longer defined on $Disj_2(\mathbb{N})$ since it doesn't preserve disjointness! Instead, the best addition analogue seems to be $$(A,B)\oplus (X,Y)=((A\setminus Y)\cup (X\setminus B), (Y\setminus A)\cup(B\setminus X)).$$ This unfortunately isn't too well-behaved: while it is commutative and has identity and inverses, it is not associative. This is because it is idempotent: $(A,B)\oplus (A,B)=(A,B)$.

There is a natural way to think of the structure $(Disj_2(\mathbb{N}), \oplus)$, however. Intuitively, we start with the set $M$ of all multisets of natural numbers where multiplicities are allowed to be arbitrary integers; this is a group under "multiset union" $\underline{\cup}$, and really is just a messy way of describing the product group $\prod_\mathbb{N}\mathbb{Z}$. We think of $(A,B)\in Disj_2(\mathbb{N})$ as representing the multiset containing one of each $a\in A$, negative one of each $b\in B$, and zero of every other number. Then $(Disj_2(\mathbb{N}), \oplus)$ can be gotten from the group $(M,\underline{\cup})$ by "truncating" each multiset to allow only the multiplicities $-1,0$, and $1$, with $\oplus$ being the analogous "truncation" of $\underline{\cup}$.

More abstractly, given any group (or indeed magma) $(G,*)$ and any function $\mathfrak{F}:G\rightarrow G$, we can build a new structure $(G_\mathfrak{F},*_\mathfrak{F})$ defined by setting $$G_\mathfrak{F}=ran(\mathfrak{F}),\quad g*_\mathfrak{F}h=\mathfrak{F}(g*h).$$ Unfortunately, this isn't a very nice construction in general, as we've seen in the case above; indeed, I don't think it has a name at all.

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  • $\begingroup$ Ah. Thanks, I should've checked more than a couple cases... What if we only consider this relation on pairs of disjoint sets? I believe this makes it into an equivalence relation (I checked it with the proof assistant Isabelle this time...hopefully didn't make any mistakes there.) $\endgroup$ – Reed Oei Jul 9 at 22:07
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    $\begingroup$ @ReedOei I've deleted my comments and incorporated them into a new edit. $\endgroup$ – Noah Schweber Jul 9 at 23:01
  • $\begingroup$ @MishaLavrov No, I don't think so. We have $(A,B)\oplus (B,A)=(\emptyset,\emptyset)$ and $(X,Y)\oplus (\emptyset,\emptyset)=(\emptyset,\emptyset)\oplus (X,Y)=(X,Y)$ for all appropriate $A,B,X,Y$. So $(\emptyset,\emptyset)$ is the identity, "flipping" yields inverses, and we get nonassociativity since no idempotent operation with identity and inverses on a set with more than one element can be associative. Or am I having a stupid moment? $\endgroup$ – Noah Schweber Jul 10 at 6:53
  • $\begingroup$ No, I'm having one! (In my defense, it was 2am.) I saw how idempotent operations could prove that we have no inverses, but I didn't notice that I used associativity to do it. $\endgroup$ – Misha Lavrov Jul 10 at 14:13

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