Basic point-set topology question (closure of rings of sets under union)

Edit: I think I figured it out (immediately after going to bed, of course). Take the symmetric difference of A and B, and also take their intersection. Then take the symmetric difference of their symmetric difference and their intersection. I haven't verified this, but just thinking about it visually it seems to be clearly correct. Since the symmetric difference of A and B is disjoint with their intersection, the symmetric difference of those two sets is just the union of the original sets A and B. Is this right?

Hi,

I just got Introduction to Topology and Modern Analysis by Simmons and unfortunately no answers are given for any of the exercises. If it matters, this is not for a class.

*A ring of sets is a non-empty class ***A** of sets such that if A and B are in **A**, then the symmetric difference of A and B and the intersection of A and B are also in **A**. Show that [b]A must also contain the empty set, the union of A and B, and A - B.

I had no problem showing that the empty set must be in **A**, and I thought the rest would be equally trivial, but I've been pondering how to show that the union of A and B is in **A** and I just don't see how it follows. I mean, since the symmetric difference of A and B is in **A** and their intersection is also in **A**, you can get their union by combining those two sets; but that's just begging the question (you'd have to take the union of those two). In other words, suppose you have a universe containing two overlapping sets A and B. You could color (A - B) blue, (B - A) red, and the intersection of A and B yellow, and you'd have a class of sets consisting of four sets: one blue, one red, one yellow (their intersection), one red and blue (their symmetric difference). I don't see why the set consisting of the blue set, the red set, and the yellow set must be the fourth member of the class. If we took the union of the other three it would be, but again, that would be assuming what we're trying to prove.

I see that if we assume that A and B are disjoint, then their union must be in **A**, because their union equals their symmetric difference. I also see that if we take the symmetric difference of A and B twice in succession, we have the union of two sets; but not of A and B themselves.

Can someone please explain this to me?

Thanks,

Jesse