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

[Jamin2112]

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 q ≠ p 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?

 

[Dick]

Yes, you are.

 

[Jamin2112]

Yes, you are.

Sweet. Keep on standby for more questions.

 

[Jamin2112]

Just tell me if these seems legit. I sense something wrong about it.

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

 

[Dick]

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.

 

[Jamin2112]

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 E^o is an interior point by choosing an arbitrary element of E^o and showing that it cannot not be an interior point.

 

[Dick]

Is this an okay start, brah? Where should I got from here? I'm trying to prove that every point in E^o is an interior point by choosing an arbitrary element of E^o and showing that it cannot not be an interior point.

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?

 

[Jamin2112]

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


... So close! I'm trying to derive a contradiction to the fact that E^o is all interior points of E (my only premise to contradict).

 

[Dick]

Sorry, brah. Let me give this another go.

If E is open, then clearly E^o is open.
If E is closed, then there exists an element p such that p is in E and (E^o)^c.
Suppose E^o is closed.
-----> (E^o)^c is open
-----> There is an r >0 s.t. N_r(p) is contained in (E^o)^c... So close! I'm trying to derive a contradiction to the fact that E^o 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.

 

[Jamin2112]

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 E^o, then the neighborhood N_r-d(p,p')(p) will surely contain only points of E.

 

[Dick]

Just thinking about it visually, if I choose any point p' in E^o, then the neighborhood N_r-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.

 

[Jamin2112]

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 E^o if d(q,p') < r - d(p,p').

 

[Dick]

How do I show that? Triangle Inequality? I somehow need to show that a point q is in E^o 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?

 

[Jamin2112]

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.

 

[Dick]

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?

 

[Jamin2112]

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.

 

[Jamin2112]

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 E^o, the union of all interior points, is open.

 

[dextercioby]

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.

 

[Dick]

Yeah. I've shown that an arbitrary point in N is an interior point of N. Thus N is open and E^o, the union of all interior points, is open.

That's not very well stated at all.