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

1. ## 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

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 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

$A\triangle B\,,\,\,A\cap B$ are in A , so also is $(A\triangle B)\triangle (A\cap B)=A\cup B$ .

Tonio

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.
e
I feel your pain. I just finished this book, and while it's amazing and the problems do get better the first couple chapters are rife with computational nightmares. How did you like proving that the symmetric difference is a associative?

P.S. When you start getting to higher chapters you may want to look at my blog.

Some of the problems may look familiar

4. Originally Posted by tonio
$A\triangle B\,,\,\,A\cap B$ are in A , so also is $(A\triangle B)\triangle (A\cap B)=A\cup B$ .

Tonio
Yeah, I figured it out almost immediately after I posted this thread. I felt a bit silly for pondering it for so long. But thank you for replying. I really appreciate it.

Originally Posted by Drexel28
I feel your pain. I just finished this book, and while it's amazing and the problems do get better the first couple chapters are rife with computational nightmares. How did you like proving that the symmetric difference is a associative?

P.S. When you start getting to higher chapters you may want to look at my blog.

Some of the problems may look familiar
Haha, I'm glad that I'm not the only one who just used the brute force method of proving that. I just substituted based on the definitions and then kept pushing the symbols around until both were in (what amounts to) disjunctive normal form. And then I had to figure out why the right-hand side ended up with the intersection of A and B as one of the sets in the union and how to get rid of it (which was sort of like a trick question). What's really funny is that if you just picture three circles in the universe and mentally take the symmetric difference both ways, it's obvious that you end up with the same "picture".

I did find this quote very interesting:

Originally Posted by GF Simmons
No mathematics, however abstract it may appear, is ever carried on without the help of mental images of some kind, and these are often nebulous, personal, and difficult to describe.
This jumped out at me, because a while ago I posted in another math forum asking how geometry could go from something visual (Euclidean, projective, differential) to something purely symbolic (algebraic). Some snide poster told me that I was being naive to expect to be able to "picture everything", and that if algebraic geometry looked daunting to me it was probably because I'd reached my "abstraction limit". Hmmm...I think I'll go with Simmons on this one!

This jumped out at me, because a while ago I posted in another math forum asking how geometry could go from something visual (Euclidean, projective, differential) to something purely symbolic (algebraic). Some snide poster told me that I was being naive to expect to be able to "picture everything", and that if algebraic geometry looked daunting to me it was probably because I'd reached my "abstraction limit". Hmmm...I think I'll go with Simmons on this one!
Haha, while I do agree that you can't picture everything. Try to envision a space where everything is exactly one unit apart.

That said, I do try to have a "feeling" for everything. A picture is nice, but an intuitive feeling of why a theorem is true is pure gold.

6. I got the impression that he meant some sort of mental imagery, not necessarily a literal picture of a particular mathematical idea or object. I get what you're saying about feeling. I "feel" the truth of, for example, the binomial theorem, and even though I don't have a literal picture of it, I imagine (a+b)^n being written out as (a + b)(a + b)...(a + b) and then I can "see" that the multiplication is the sum of all of the different ways of choosing which terms to multiply together.

Similarly, I would guess that understanding higher-dimensional spaces involves some sort of "nebulous, personal" imagery which is somehow based on three-dimensional space. Literal pictures, no; "imagery", yes.