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

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

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 (in your topology)
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

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 by a moderator: 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