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I have the following problem:

Given a non-empty set $U$ and the power set $P(U)$. Proof $(P(U);\triangle, \cap)$ is a commutative ring. Which subset is the neutral and which one is the identity?

I have an idea of how to check if it's a ring (by checking all properties one by one) but I don't know how to deduce the neutral set and the identity.

Rolando González
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    Well, you need to propose a specific element of $P(U)$ to be the neutral element, and then verify it satisfies the properties. Is there a subset $A$ of $U$ such that $X\triangle A = X$ for all $X\subseteq A$? Is there a subset $B$ of $U$ such that $X\cap B=X$ for all $X\subseteq U$? – Arturo Magidin Nov 11 '19 at 02:13
  • Thanks Arturo, I've found that in the case of $X \triangle A = X$ it is the empty set. But can you give me any hint for the case of $X \cap B = X$. I've made some attempts but I think there's no way – Rolando González Nov 11 '19 at 02:21
  • Really? There's no subset of $U$ which, when intersected with any subset $B$ of $U$, will always give you $B$? – Arturo Magidin Nov 11 '19 at 02:27
  • I don't know how to express it correctly, but it is $U$ isn't it? The subset of $P(U)$ which contains all the elements of $U$, then when we do $B \cap U = B$. – Rolando González Nov 11 '19 at 02:35
  • Yes, $U$ works. So now you have specific elements of $P(U)$ that you can propose as neutral elements for $\triangle$ and as identity elements for $\cap$. So now just run through all the axioms and verify that they are satisfied. – Arturo Magidin Nov 11 '19 at 02:37
  • Thanks a lot and sorry for the inconvenience, have a Sunday evening – Rolando González Nov 11 '19 at 02:38

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