Are limit points and interior points of a set contained in the set?

  • Thread starter Jamin2112
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  • #1
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Homework Statement



Just wondering if I'm understanding the definitions correctly. I honestly feel like an idiot for asking this.

Homework Equations



A point p is an interior point of E is there is a neighborhood N of p such that N contained in E.

A point p is a limit point if the set E if every neighborhood of p contains a point qp such that q is in element of E.

The Attempt at a Solution



It looks like an interior point of E must be contained in E, but a limit point of E is not necessarily contained in E. Am I right?
 

Answers and Replies

  • #2
Dick
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Yes, you are.
 
  • #3
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Yes, you are.
Sweet. Keep on standby for more questions.
 
  • #4
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Just tell me if these seems legit. I sense something wrong about it.

Let Eo denote the set of all interior points of a set E.
(a) Prove that Eo is always open.
(b) Prove that E is open if and only if Eo = E
(c) If G is a subset of E and G is open, prove that G is a subset of Eo


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  • #5
Dick
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Well, your first proof is not doing well. You say Nr(p) is not contained in E0. Just because E0 is contained in E that doesn't mean Nr(p) isn't contained in E. E is larger than E0. That's probably what's bothering you. The whole proof is needlessly indirect. Just use that E0 is a union of open sets.
 
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  • #6
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Well, your first proof is not doing well. You say Nr(p) is not contained in E0. Just because E0 is contained in E that doesn't mean Nr(p) isn't contained in E. E is larger than E0. That's probably what's bothering you. The whole proof is needlessly indirect. Just use that E0 is a union of open sets.
Is this an okay start, brah? Where should I got from here? I'm trying to prove that every point in Eo is an interior point by choosing an arbitrary element of Eo and showing that it cannot not be an interior point.

screen-capture-4-15.png
 
  • #7
Dick
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Is this an okay start, brah? Where should I got from here? I'm trying to prove that every point in Eo is an interior point by choosing an arbitrary element of Eo and showing that it cannot not be an interior point.

screen-capture-4-15.png
Read it. It's false. If p is not an interior point of E0 then every neighborhood contains at least one point in E0^C. It's certainly not necessarily true that it's CONTAINED in E0^C. Can't you think of a way to write E0 as a union of open sets?
 
  • #8
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Read it. It's false. If p is not an interior point of E0 then every neighborhood contains at least one point in E0^C. It's certainly not necessarily true that it's CONTAINED in E0^C. Can't you think of a way to write E0 as a union of open sets?
Sorry, brah. Let me give this another go.

If E is open, then clearly Eo is open.
If E is closed, then there exists an element p such that p is in E and (Eo)c.
Suppose Eo is closed.
-----> (Eo)c is open
-----> There is an r >0 s.t. Nr(p) is contained in (Eo)c


...... So close! I'm trying to derive a contradiction to the fact that Eo is all interior points of E (my only premise to contradict).
 
  • #9
Dick
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Sorry, brah. Let me give this another go.

If E is open, then clearly Eo is open.
If E is closed, then there exists an element p such that p is in E and (Eo)c.
Suppose Eo is closed.
-----> (Eo)c is open
-----> There is an r >0 s.t. Nr(p) is contained in (Eo)c


...... So close! I'm trying to derive a contradiction to the fact that Eo is all interior points of E (my only premise to contradict).
This isn't going well. And I don't think you are very close. If p is in E0 then there is an open ball Nr(p) that is contained in E, yes? Can you think of a reason why every other point p' in Nr(p) is also contained in E0?? Tell me that first. I don't think the 'contradiction' proof is working for you.
 
  • #10
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This isn't going well. And I don't think you are very close. If p is in E0 then there is an open ball Nr(p) that is contained in E, yes? Can you think of a reason why every other point p' in Nr(p) is also contained in E0?? Tell me that first. I don't think the 'contradiction' proof is working for you.
Just thinking about it visually, if I choose any point p' in Eo, then the neighborhood Nr-d(p,p')(p) will surely contain only points of E.
 
  • #11
Dick
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Just thinking about it visually, if I choose any point p' in Eo, then the neighborhood Nr-d(p,p')(p) will surely contain only points of E.
Sure, that works. So every point in E0 has a neighborhood contained in E0.
 
  • #12
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Sure, that works. So every point in E0 has a neighborhood contained in E0.
How do I show that? Triangle Inequality? I somehow need to show that a point q is in Eo if d(q,p') < r - d(p,p').
 
  • #13
Dick
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How do I show that? Triangle Inequality? I somehow need to show that a point q is in Eo if d(q,p') < r - d(p,p').
It's not really necessary to do that. And you probably don't want to assume you are in a metric space. If N is an OPEN neighborhood of p what do you know about every point p' in N?
 
  • #14
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It's not really necessary to do that. And you probably don't want to assume you are in a metric space. If N is an OPEN neighborhood of p what do you know about every point p' in N?
There is an open neighborhood centered at p' that is contained in the open neighborhood centered at p.
 
  • #15
Dick
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There is an open neighborhood centered at p' that is contained in the open neighborhood centered at p.
Ok, so can you use that to prove E0 is open?
 
  • #16
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I'm pretty sure we can assume E is a subset of a metric space, because all the definitions we're supposed to use (neighborhood, interior point, limit point, etc.) relate to metric spaces.
 
  • #17
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Ok, so can you use that to prove E0 is open?
Yeah. I've shown that an arbitrary point in N is an interior point of N. Thus N is open and Eo, the union of all interior points, is open.
 
  • #18
dextercioby
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You needn't use metric spaces. Proofs work for general topological spaces. The definitions you gave in post #1 are probably picked up from Kelley's book. Newer books define the interior as an open set.
 
  • #19
Dick
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Yeah. I've shown that an arbitrary point in N is an interior point of N. Thus N is open and Eo, the union of all interior points, is open.
That's not very well stated at all.
 

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