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!


2 Answers 2


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.


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 $A \Delta B = (A \setminus B) \cup (B \setminus A)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.