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I was thinking about different rules for rings and fields and I decided to analise the following structure: $(\mathcal{P}(X), \cup, \cap)$, where "$\mathcal{P}(X)$" denotes the set of subsets of a given set $X$.

I noticed this followed all the axioms of commutative rings with identity, where the identity of "$\cap$" is "$X$", except for the inverse element of addition, so I decided to switch the "$\cup$" operation for something that could work.

My next try was with the following operation: $+ : \mathcal{P}(X) \times \mathcal{P}(X) \rightarrow \mathcal{P}(X)$ \begin{cases} A + B = A\cup B, \text{ if $A\cap B = \emptyset$} \\ A + B = \emptyset, \text{ if $A\cap B \neq \emptyset$} \end{cases}

In this case, I got the inverse element, and the operation is still commutative and has a neutral element (i.e. the empty set). But since there were many inverses, I assume this is not associative, though I haven't personally checked.

In short, I would really like an operation that, together with the intersection, turns $\mathcal{P}(X)$ into a ring. Does anyone know of any such operation, or know any reason why it should not exist?

Thanks in advance!

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Note that $$\mathcal{P}(X) \to {}^X\mathbb{F}_2 \\ A \mapsto \mathbf{1}_A,$$ which maps sets to indicator functions, is a bijection from the set $\mathcal{P}(X)$ to the ring $${}^X\mathbb{F}_2 = \{f : X \to \mathbb{F}_2\}.$$ It therefore induces a ring structure on $\mathcal{P}(X)$.

In particular, this induces $$A\cdot B := A\cap B \\ A + B := A\Delta B$$


If you want $A\cdot B := A\cup B$, then instead consider the map $A \mapsto \mathbf{1}_{X\setminus A}$ which induces, once again, $$A+B:=A \Delta B$$ (it seems we can't get away from it so easily).


If you want, try other bijections to find other "natural" ring structures, but those two are probably the most natural.

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If you want to use $\cap$ as the multiplication operation, a compatible choice for the addition operation is the symmetric difference $\Delta : \mathcal{P}(X) \times \mathcal{P}(X) \to \mathcal{P}(X)$ defined by $A \Delta B = (A \setminus B) \cup (B \setminus A)$.

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  • $\begingroup$ That’s perfect, thank you! @Chris, if, instead, I wanted to use $\cup$ as multiplication, is there an analogous operation to act as addition? I ask it here because I don’t really want to create the exact same post again with this slight tweak. $\endgroup$
    – Gauss
    Aug 13, 2020 at 21:59
  • $\begingroup$ It looks like Brian Moehring has included an answer to your followup in their answer (which also gives a really nice perspective on why the symmetric difference operation is a very natural "addition" when "multiplication" is intersection). $\endgroup$ Aug 14, 2020 at 16:21

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