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Prove: A point is in the boundary of A if every neighborhood of has non-empty: A, X/A

  • Thread starter Hodgey8806
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  • #1
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Homework Statement


I tried to state the whole thing, but here it is again:
Prove: A point is in the boundary of if and only if every neighborhood of it contains both a point of A and a point of X\A.


Homework Equations


Boundary def: X\(Int A [itex]\cup[/itex] Ext A)
Def of Int A: [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
Def of Ext A: X\[itex]\bar{A}[/itex] where [itex]\bar{A}[/itex] means closed (including boundary)
Def of [itex]\bar{A}[/itex]: [itex]\bigcap[/itex]{B[itex]\subseteq[/itex]X:B[itex]\supseteq[/itex]A and B is closed in X}


The Attempt at a Solution


Please bare with me as this is a little messy. I used a bunch of set theory rules:
Let p[itex]\in[/itex][itex]\delta[/itex]A
p[itex]\in[/itex]X\(Int A [itex]\cup[/itex] Ext A)
1) p[itex]\in[/itex](X\Int A) [itex]\wedge[/itex] 2)p[itex]\in[/itex](X\Ext A)
Thus breaking each one down separately:
1)p[itex]\in[/itex]X\[itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
p[itex]\in[/itex][itex]\bigcap[/itex]{X\C:C[itex]\subseteq[/itex]A and C is open in X}[itex]\subseteq[/itex]=X\A--(I realize now this could have been stated without these set properties.)
2)p[itex]\in[/itex]X\(X\[itex]\bar{A}[/itex])
p[itex]\in[/itex][itex]\bar{A}[/itex][itex]\supseteq[/itex]A
Thus every neighborhood of p contains both a point of A and a point of X\A.
Q.E.D.
 

Answers and Replies

  • #2
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First of all, this is only one leg of the proof, the "only if" part. You need to prove the "if" part as well.

Secondly, I do not see how it follows that any neighborhood contains anything - you proof does not say anything about the neighborhood at all (except let p be in one).
 
  • #3
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I did the other leg by contradiction; it seemed a bit easier.

I have shown that P is in the closed "neighborhood" of X\A and [itex]\bar{A}[/itex].
So if I union them together we end up with the entire set X.

I'm honestly a bit lost though on how to show the neighborhood portion of this.
Do I need to discuss a Ball of length R?
 
  • #4
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390


I think this one could also be done by contradiction. Assume there is a point in the boundary that has a neighborhood whose intersection with int or with ext is empty. Let's say that with int is empty: [itex]p \in V \subset X : V \cap int A = \emptyset[/itex] Then consider [itex]V \cap \partial A[/itex]. Can you see it is empty so we have a contradiction? Ditto for ext A.
 
  • #5
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I'm sorry, I'm not quite understanding it. It looks like we want to show p being in the neighborhood AND the boundary, and then we discover it's not in their intersection. Is that correct? I'm having trouble see how that is the case though.

These are my thoughts on that. We see that the intersection with interior must be empty. But if it is on the boundary, then the intersection with the boundary shouldn't be empty. And that is our contradiction. But, getting to their I'm have trouble seeing--sense it is empty with interior of A, are we trying to imply that it is empty with the boundary?
 
  • #6
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I would like to try one more method. However, I the very last bits may need clarification:
Assume for contradiction there does exist such a neighborhood that either V[itex]\cap[/itex]A = [itex]\phi[/itex] or V[itex]\cap[/itex]A=[itex]\phi[/itex].

Let p[itex]\in\delta[/itex]A
Thus p[itex]\in[/itex]X\A and p[itex]\in[/itex][itex]\bar{A}[/itex]
However, since X\A is closed, [itex]\exists[/itex]V[itex]\subseteq[/itex]A, thus V[itex]\cap[/itex]A≠[itex]\phi[/itex]
Also, since [itex]\bar{A}[/itex], [itex]\exists[/itex]V[itex]\subseteq[/itex][itex]\bar{A}[/itex], thus V[itex]\cap[/itex]X\A≠[itex]\phi[/itex]
Q.E.D.
 
  • #7
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390


If the intersection of the ENTIRE neighborhood with the boundary is empty, then it contains no point in common with the neighborhood. But it must contain p, which is in the boundary. This is the contradiction.

As for proving it is empty, just do the set theory gymnastics of the kind you were doing.
 
  • #8
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I looks like your second proof assumes A is open.
 
  • #9
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I understand the contradiction, but I can't seem to grasp why the interaction with the int being empty causes the interaction with boundary to be empty. :\
 
  • #10
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What is space X you work with? Metric? Topological? Something else? How are "open" and "closed" sets defined in it, what is a "neighborhood"? I am asking this because the proof may require one of those definitions, and since there are quite a few equivalent formulations, I would like to know which ones you are supposed to use.
 
  • #11
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X is a topological space. A set S as a subset X is open iff there exists a set for all p in S, there exists and open neighborhood completely contained in S.

I know this problem should be easier, I think I'm just at a point where I "can't see the forest because of all the trees." I'm able to do these other proofs fairly simply--I probably just need fresh eyes is all.
 
  • #12
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I "think" I have it now. I finally discussed the closure of X\Int A, thus it has SOME neighborhood not contained in it--thus the intersection with Int A is not empty.

As well, A-closure is closed, so U intersecting Ext A is not empty for SOME neighborhood.

Does that sound right?
 
  • #13
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I think the standard way to prove that statement is by introducing interior points, boundary points, points of closure and exterior points. Interior and exterior points are those contained in A and X\A, respectively, with some open neighborhoods; boundary points are those any neighborhood of which intersects with both A and X\A; points of closure are a union of interior and boundary points.

Then you establish that int A is the set of all the interior points (this is almost trivial); the closure of A is the set of all its points of closure (you do that by noticing that the set of the interior points of X\A is X \{A's points of closure}, then use the previous result); finally that the boundary of A is the set of all its boundary points, which follows from the previous results.

As a remark, all those things (int, ext, closure, boundary) can be defined equivalently in terms of the respective points, which I think is slightly more intuitive. But in that case you have to prove things the other way around :)
 
  • #14
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Firstly, thank you for all of your help!

I wanted to give one more try at this. I think I have it now!! (This time for real lol)

Spse p[itex]\in[/itex][itex]\delta[/itex]A
If p[itex]\in[/itex]A,
Then [itex]\forall[/itex] open U[itex]\subseteq[/itex]X s.t. p[itex]\in[/itex]U, U[itex]\cap[/itex]A≠∅.
Since A is closed, [itex]\forall[/itex] open U[itex]\subseteq[/itex]X s.t. p[itex]\in[/itex]U, U[itex]\cap[/itex]A is not a subset of A.
Thus, U[itex]\cap[/itex](A\X)≠∅.

And prove it the same for p not in A.

What do you think?
 
  • #15
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Also U intersect A for that first proof wasn't empty because p is an element.
 
  • #16
6,054
390


Firstly, thank you for all of your help!

I wanted to give one more try at this. I think I have it now!! (This time for real lol)

Spse p[itex]\in[/itex][itex]\delta[/itex]A
If p[itex]\in[/itex]A,
Then [itex]\forall[/itex] open U[itex]\subseteq[/itex]X s.t. p[itex]\in[/itex]U, U[itex]\cap[/itex]A≠∅.
Since A is closed, [itex]\forall[/itex] open U[itex]\subseteq[/itex]X s.t. p[itex]\in[/itex]U, U[itex]\cap[/itex]A is not a subset of A.
Thus, U[itex]\cap[/itex](A\X)≠∅.

And prove it the same for p not in A.

What do you think?
I do not follow. First the assumption is that p is in the boundary of A, then you assume it is in A? And why is A closed?
 

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