Is the Boundary of an Open Set Always Its Complement?

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Discussion Overview

The discussion revolves around the relationship between the boundary of an open set and its complement within the context of topology. Participants explore definitions and properties of open and closed sets, boundaries, and specific examples in various topological spaces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant argues that the boundary of an open set A is equal to its complement, based on definitions of closure and interior.
  • Another participant provides a counterexample using the set A = {(x,y) | x^2 + y^2 < 1} in R^2, showing that its boundary does not equal its complement.
  • Several participants discuss the implications of different topologies, such as discrete and indiscrete topologies, on the definitions of boundary, closure, and interior.
  • There is confusion regarding the notation and definitions used, particularly around the topology T and the relationship between sets and subsets.
  • Participants express uncertainty about the correctness of the initial claim and seek clarification on the definitions involved.
  • Some participants attempt to clarify the distinction between elements and subsets in the context of set theory and topology.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the boundary of an open set is always its complement. Multiple competing views and interpretations of definitions remain unresolved.

Contextual Notes

There are limitations in the discussion regarding the clarity of notation and definitions, as well as the assumptions made about the nature of sets and topologies.

Adriadne
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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?
 
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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
 
Adriadne said:
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.
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.

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?
It is if A is dense in X which is what you seem to be saying when you replace cl(A) with X.
 
HallsofIvy said:
And don't you mean X is an open subset of A rather than T?

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
 
George Jones said:
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
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.
 
George Jones said:
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
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
 
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HallsofIvy said:
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.
Yep, figured that later, thanks
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.
Yep, got that too, also after.
Thank you so much.
 
Adriadne said:
The point is that the topology T on X is a set of sets

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

Adriadne said:
like the set X = {a,b,c}, a possible topology T on X = {X {} {a} {b,c}}.

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

Adriadne said:
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.

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. :confused:

Regards,
George
 
George Jones said:
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.
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
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
George Jones said:
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.
No, a (a point) is an element in X. {a} is a singleton set, not a point
2) {a} is a member/element of T.
Yes (in your topology)
3) {a} is not a subset of T.
Yes it is, in your topology
4) {a} is a subset (of X) in T.
No, see (1)
I say that:
a is a member of X;
Yes
{a} is a subset of X;
No, a is an element in X
{a} is a member of T;
Yes, an element in T
{{a}} is a subset of 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?
 
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  • #12
Adriadne said:
"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."
No, a (a point) is an element in X. {a} is a singleton set, not a point
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
"2) {a} is a member/element of T."
Yes (in your topology)[\quote]
Good, we all agree on that!
"3) {a} is not a subset of T."
Yes it is, in your topology
No, it is not, {a} is a member of T, not a subset.
"4) {a} is a subset (of X) in T."
No, see (1)
YES, see (1). Precisely because a is an element of X, {x} is a subset of X (which subset is in T.)
"I say that:
a is a member of X;"
Yes
Once again, we all agree on that!
"{a} is a subset of X; "
No, a is an element in X
He didn't ASK about a! Precisely because a is an element of X, the set {a} is a subset of X.
"{a} is a member of T; "
Yes, an element in T
Good
"{{a}} is a subset of 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.
Emphatically YES. Precisely because {a} is a member of T, {{a}} is a subset of T.

If A is any set, and p is an element of A, the {p} is a set containing only elements of A, hence a subset of A!
How did I do?
I wouldn't presume to judge.
 
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  • #13
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
Adriadne said:
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.

I'm not quite sure what you mean here.

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.

Yes.

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.

Agreed.

Regards,
George
 

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