Let $A$ be a non empty set. Let $P(A)$ denote the power set of $A$.

$P(A)$ can be given a group structure in multiple ways.

1. Using Disjoint union a group operation Source

may be worth noting in this context that the power set of $X$ is a group under the operation of disjoint union, with the empty set as identity. if the complement of $A \subset X$ is $A'$ then the disjoint union of $A$ and $B$ is: $$ A \circ B = (A \cap B') \cup (A' \cap B) $$ proving associativity is a nice exercise, though it follows much more simply if the disjoint union is viewed as ordinary addition (mod 2) in $\mathbb{F}_2^X$

2. Using Symmetric Difference as Group Operation Source

The power set $G$ of any set $A$ becomes an abelian group under the operation of symmetric difference:

  • Why abelian? Easy to justify, just use the definition in $(1)$ above: it's defined in a way that $g_1 \triangle g_2$ means exactly the same set as $g_2 \triangle g_1$, for any two $g \in G$.

  • As you note, the symmetric difference on $G$ is associative, which can be shown using the definition in $(1)$, by showing for any $f, g, > h \in G, (f\; \triangle\; g) \triangle \;h = f\;\triangle\; (g > \;\triangle\; h)$.

  • The empty set is the identity of the group (it would be good to justify this this, too), and

  • every element in this group is its own inverse. (Can you justify this, as well? Just show for any $g_i \in G, g_i\;\triangle \; g_i = > \varnothing$).

Question: Can we define a ring structure on power set of $A$.


I believe the "disjoint union" alluded to by David Holden means "disjunctive union" which is a synonym for the symmetric difference. So, you aren't listing multiple ways, really.

Can we define a ring structure on power set of A.

As the wiki clearly states: The power set of any set becomes a Boolean ring with symmetric difference as the addition of the ring and intersection as the multiplication of the ring.

This is the standard ring structure on the powerset of a set. It may have other ring structures, but none would arise so naturally as this one, and would probably not have meaningful connection to the fact the underlying set is a powerset.

  • $\begingroup$ This is offtopic but what do people(mathematician) mean when they say that this "structure arises naturally". Is it related to category theory or it is just an expression. $\endgroup$ – StammeringMathematician Oct 13 '18 at 15:53
  • $\begingroup$ @StammeringMathematician There is a precise meaning in category theory, but here I just mean it as an expression of natural language. I just mean that when looked at from a certain perspective, it's fairly obvious, and there aren't really any other competing structures, so it jumps out at us. $\endgroup$ – rschwieb Oct 13 '18 at 16:01

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.