I have a question very similar to Finite ring of sets

For a given set U, the set R of subsets is a ring under the operations of symmetric difference ($\bigtriangleup$) and intersection ($\cap$).

Given a finite set V of subsets of U, the set W generated by V is a subring of R. I.e., W is the set with the following two properties:

1) $V \subset W$

2) If $a \in W$ and $b \in W$, then $a \cap b \in W$ and $a \bigtriangleup b \in W$.

My question is: Why is W finite?

This is clear when I draw a picture. If I start with a fixed set of subsets of U, and start drawing intersections and symmetric differences, and keep enlarging the set by the results, I eventually exhaust all the possibilities.

Clearly this is not true for general rings generated by a finite set. For example, given the finite set $\{1,-1\} \subset Z$, the generated set is all the integers.

One way to argue this is with a VERY BIG HAMMER which I'd like to avoid.

Assume we have two boolean expressions in Disjunctive Normal Form (DNF), I can express the intersection and symmetric differences of the two expressions also in DNF. And there is only a finite number, $2^{2^N}$, of possible DNFs of N variables.

I hope there's a simpler way to argue this other than resorting to DNF representations.


1 Answer 1


There are only finitely many sets that are intersections of sets in $V$ and their complements. Every set in $W$ must be a union of those.

(This straightforward argument may be just a rephrasing of the idea behind the BIG HAMMER you want to avoid.)

Edit in response to comment from the OP

Not doubting of course, but not understanding the reasoning.

Here's another kind of argument. Maybe it helps your intuition.

Each subset of $U$ is completely described by its characteristic function. The ring structure on subsets is isomorphic to the ring structure of the characteristic functions with pointwise operations mod $2$.

You are interested in a finitely generated subring of that ring. The finite number of characteristic functions $\chi_a$ and $1 - \chi_a$ for $a \in V$ can be distinguished by their values at a finite subset of $U$, so that will be true of the functions in the subring they generate.

This is a more formal way to say that the universe $U$ might as well be finite - in which case the question is trivial.

  • $\begingroup$ Why must every set in W be a union of intersections of sets in V and their complements? And yes, this indeed seems like a claim that every element of W has a unique DNF. $\endgroup$
    – Jim Newton
    Commented Dec 18, 2016 at 14:18
  • $\begingroup$ @JimNewton Because any calculation of an intersection or symmetric difference that starts with sets of that kind results in a set of that kind. I guess that's formally a structural induction, but I find it a convincing informal argument. $\endgroup$ Commented Dec 18, 2016 at 14:22
  • $\begingroup$ Hi Ethan, There is still something I'm missing. Not doubting of course, but not understanding the reasoning. Let R be a ring, and $V \subset R$ be finite. Define set W as the smallest set obeying the following two properties 1) $V \subset W$ 2) If $a \in W$ and $b \in W$, then $a+b \in W$ and $a \cdot b \in W$ It is not true in general that W is therefore finite. But in the case of the ring of subsets, W is finite. Perhaps it stems from the fact that $A \cap A = A$ and $A \bigtriangleup A = \emptyset$? $\endgroup$
    – Jim Newton
    Commented Dec 18, 2016 at 14:38
  • $\begingroup$ OK, I don't exactly know what characteristic functions on rings are. So I guess I need to go and research that. BTW, Ethan, thanks for the help. I'm working on an article wherein this is something I need to claim/prove/cite before going on to my actual thesis. If you send me your credentials, I'll be happy to site you in the article. $\endgroup$
    – Jim Newton
    Commented Dec 18, 2016 at 16:30
  • $\begingroup$ The characteristic function of a set is the function that's $1$ on the set and $0$ elsewhere. I think if you said your ring of sets is obviously finite no one would object. You can find my contact on my SE profile, but I don't need thanks in the article. $\endgroup$ Commented Dec 18, 2016 at 17:24

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