# Boundaries in topological space

1. Oct 24, 2005

I posted this on another forum, but had no response. Maybe because it's too stupid the bother with? Anyway....
Say I have a set X and a topology T on X so that T = {X {} A} i.e A is an open subset of T. Then the complement of A is Ac = X - A, which is closed.
Now the interior of A, int(A) is the largest open set (or the union of all open sets) contained in A which is A, and the closure of A, cl(A) is the smallest closed set in {X {} Ac} containing A which is X. So if the boundary of A
bd(A) = cl(A) - int(A), we have that bd(A) = X - A = Ac.
Similarly, the closure of Ac is the smallest closed set containing Ac, which is Ac = X - A. So, using the alternative definition for the boundary of A,
bd(A) = cl(A) intersect cl(Ac) = X intersect (X - A) which is X - A = Ac.
This argument applies also for the trivial and discrete topologies, where the boundaries are respectivley X and {}. I've also tried it out on a number of arbitrary topologies of my own devising, and the answer is always the same, the boundary of A is the complement of A. Surely it's not right, though?

2. Oct 24, 2005

### George Jones

Staff Emeritus
Let X = R^2 with the standard topology and A ={(x,y) | x^2 + y^2 < 1}. Then the boundary of A is bd(A) = {(x,y) | x^2 + y^2 = 1}, but X - A = {(x,y) | x^2 + y^2 >= 1}.

Regards,
George

3. Oct 24, 2005

### HallsofIvy

Staff Emeritus
Am I missing something? How did you get from cl(A) to "X"?

I don't understand the notation {X {} A}. And don't you mean X is an open subset of A rather than T? I A is intended to be a subset of T, then it doesn't make sense to talk about the complement of A being X- A.

It is if A is dense in X which is what you seem to be saying when you replace cl(A) with X.

4. Oct 24, 2005

### George Jones

Staff Emeritus
I think Adriadne is trying to say that if A is any subset of set X, then T = {X, {}, A} is a topology for X. But I could be wrong. I not sure how one gets from T to arbitary topologies.

Regards,
George

5. Oct 24, 2005

### HallsofIvy

Staff Emeritus
Oh, I see! Thanks. In that case, the only open sets are X, {}, and A itself and so the only closed sets are X, {} and Ac, the complement of A.

Now what Ariadne said makes sense. The closure of A is, indeed, X and the interior of A is A itself. The "boundary" of A, defined as cl(A)- Int(A) is simply Ac. Yes, in this topology, the boundary of A is Ac.

However, she also make reference to the "discrete" topology in which all sets are open and all sets are closed. The boundary of any set A is cl(A)- int(A)= A- A= {}, not the complement of A.

In the "indiscrete" topology, where the only open sets are X and {}, the only closed sets are also X and {}. The closure of A is X, the interior or A is {} so the boundary of any set A is X-{}= X, not the complement of A.

6. Oct 24, 2005

George, no. A might possibly be subset X (but I don't think it need be - consider the quotient topology for e.g.)
The point is that the topology T on X is a set of sets, like the set X = {a,b,c}, a possible topology T on X = {X {} {a} {b,c}}. Here, evidently {b,c} may be subset X but a is an element of X, not a subset, but {a} is a subset of T.
So I was in no way suggesting that in my OP that A (or here {a}) subset X. However, once I define T on X, I can insist that A (here {a}) is subset T.
Am I being moronic again?
Hurkyl (EDIT: Eeek! Sorry Halls, got names muddled) I thank you for your response, I was half through and called away. Later

Last edited: Oct 24, 2005
7. Oct 24, 2005

Yep, figured that later, thanks
Yep, got that too, also after.
Thank you so much.

8. Oct 24, 2005

### George Jones

Staff Emeritus
Yes. That's what I wrote down. If, by T = {X {} A}, you didn't mean T = {X, {}, A}, then what did you mean?

Please put all the commas in - this type of detail is important in mathematics. Do you mean T = {X, {}, {a}, {b,c}}?

No, {a} is a subset of X. {a} is a member/element of T, not a subset of T. {a} is a subset (of X) in T.

I am somewhat confused.

Regards,
George

9. Oct 24, 2005

Hey are you serious? If I have a set X with elements a, b and c, I'm going to to write X = {a,b,c}. a is an element of X, not a subset, that makes no sense.
Remember, T is a set of sets, {a} is only a set ( a singleton) in T, not X.

10. Oct 24, 2005

### George Jones

Staff Emeritus
Let X = {a, b, c} and T = {X, {}, {a}, {b,c}}.

Take my statements one at a time. With which do you agree? With which do you disagree.

1) {a} is a subset of X.

2) {a} is a member/element of T.

3) {a} is not a subset of T.

4) {a} is a subset (of X) in T.

I say that:

a is a member of X;

{a} is a subset of X;

{a} is a member of T;

{{a}} is a subset of T.

Regards,
George

11. Oct 24, 2005

No, a (a point) is an element in X. {a} is a singleton set, not a point
Yes it is, in your topology
No, see (1)
Yes
No, a is an element in X
Yes, an element in T
Emphatically no (double braces indicate set of sets. T contains as its fewest elements X and {}) As above, {a} may or may not be an element of T, it's certainly allowed in some topologies.
How did I do?

Last edited: Oct 24, 2005
12. Oct 24, 2005

### HallsofIvy

Staff Emeritus
YES. The question was whether {a} was a subset of X, not a. Since a is an element in X, {a} is a subset of X

Last edited: Oct 25, 2005
13. Oct 25, 2005

Sorry guys, I was being really stupid. Of course elements of T on X are subsets of X (they are also subsets of T, as T is a set of sets, as I said). What Iwas getting at was that the partition of X for the topology may be quite different from the partition which generates "natural" subsets, but I may be wrong about that too.
Here's what I was thinking (I think it's just about worth saying). If there is a topology T on X, and A is a proper subset of X, then there is an induced topology on A, I'll call it T(A), the open sets of which are formed from the intersection of A with open sets in X. Now it matters not whether A is open in T on X or not, just because of the way the complements work, A and {} are always open and closed in T(A), as required.
That's quite a different situation to what we were discussing. I think it better not to talk about subsets the way I was back then. Sorry

14. Oct 27, 2005

### George Jones

Staff Emeritus
I'm not quite sure what you mean here.

Yes.

Agreed.

Regards,
George