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Basic Topology

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  • #1
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1. Prove that a set is closed if and only if it contains all of its cluster points.



2. Can I use part of the Lemma here that states: Every interior point of A is a Cluster point.
Also what exactly is the definition of a closed set other than a set is closed if its compliment is open.

3. Just by using the definitions i feel like I should be able to prove this. It seems so simple yet I dont know where to start exactly. I just need some direction.......and maybe better definitions.
 

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  • #2
Fredrik
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For metric spaces, "E is closed" can be defined as "the limit of every convergent sequence in E is in E". I like this definition, because it ensures that "closed" means "closed under limits". Unfortunately this definition isn't equivalent to "Ec is open" when you're dealing with topological spaces that aren't metric spaces. I guess "Ec is open" has turned out to be a more useful definition, since it's the one that's used in all the books.

Suppose that there's a cluster point of E in Ec. Can Ec be open?
 
  • #3
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If there is one cluster point in Ec I think it could be open but not all cluster points. Ummm, but you said a cluster point in E also in Ec, how could it be in both.

I have a hard time understanding all of this to begin with sorry if my reasoning and questions do not make since.

The definition I have for cluster point is: Suppose A [tex]\subseteq[/tex] [tex]\Re[/tex] and x [tex]\in[/tex] [tex]\Re[/tex] then x is a cluster point of A if every neighborhood contains a point of A other than x.
So thats why i dont think a cluster point could be in E and Ec
 
  • #4
Fredrik
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Ummm, but you said a cluster point in E also in Ec,
No, I said a cluster point of E in Ec. I don't have time to answer more than that right now.
 
  • #5
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I dont understand than. Oh well thanks for trying
 
  • #6
Fredrik
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I don't understand. What is unclear about the sentence "Suppose that there's a cluster point of E in Ec."?

Edit: OK, I think I see the problem. You think the definition of "cluster point" means that a cluster point of a set is a member of that set. It doesn't. 0 is a cluster point of {1/n|n positive integer}, and every real number is a cluster point of the set of rational numbers.
 
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  • #7
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How can it be of and not in.........the wording to me just doesn't make sense, this whole subject does not make sense to me. its not logical to me.
 
  • #8
Fredrik
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Topology is pretty difficult. What makes it worse is that when you study functional analysis, you're expected to know these things perfectly.

This exercise wouldn't make sense if "of" implied "in", since every set would contain all of its cluster points.
 
  • #9
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ok how you explained cluster point to me helps a little bit. Yeah topology is very difficult and i dont think i like it.......
 
  • #10
Fredrik
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Did you understand that "There's a cluster point of E in Ec " is the negation of "E contains all its cluster points"? If you can prove that this assumption implies "Ec is not open", you have proved that if Ec is open, E contains all its cluster points. (Recall that [itex]A\Rightarrow B[/itex] is equivalent to [itex]\lnot B\Rightarrow \lnot A[/itex]).

(I won't be near my computer for a few hours now).
 
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  • #11
Fredrik
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Too bad the time limit for edits has been lowered, so that I can't remove the emphasis on the word "is", which makes no sense. My point was that the implication

E is closed [itex]\Rightarrow[/itex] E contains its cluster points,​

which is one of the two implications you're supposed to prove, is equivalent to

Ec is open [itex]\Rightarrow[/itex] E contains its cluster points,​

which is equivalent to

There's a cluster point of E in Ec [itex]\Rightarrow[/itex] Ec is not open.​

Have you proved this yet? Did you figure out how to do the other half of the problem?
 
  • #12
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1. Prove that a set is closed if and only if it contains all of its cluster points.


Since everyone uses a slightly different definition of closed, can you post yours?
 
  • #13
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So if I prove that

Ec is open [itex]\Rightarrow[/itex] E contains its cluster points,​

then I proved one way of this proof since it is a iff

Have you proved this yet? Did you figure out how to do the other half of the problem?[/QUOTE]

No i still haven't I have to figure it out by tomorrow.....I had to take a break and study for two more finals I had
 
  • #14
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@earlh the only definition I have is A is closed if A compliment is open
 
  • #15
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@earlh the only definition I have is A is closed if A compliment is open
And what's your definition of an open set?
 
  • #16
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A is closed if A compliment is open
it's a theorem, not definition
 
  • #17
Fredrik
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it's a theorem, not definition
It's a valid definition. It's a good one because it works in the context of topological spaces too. (In that context, it's the standard way to define "closed"). Other valid definitions of "E is closed", in the context of metric spaces, include "E contains all of its cluster points" and "for every x in E there's a sequence in E with limit x".

Cluster points are often called limit points or accumulation points instead.
 
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  • #18
Fredrik
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So if I prove that

Ec is open [itex]\Rightarrow[/itex] E contains its cluster points,​

then I proved one way of this proof since it is a iff
Right. And I recommend that you do that by proving

There's a cluster point of E in Ec [itex]\Rightarrow[/itex] Ec is not open.​

We can talk about how to prove the converse after you've done this.
 
  • #19
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it's a theorem, not definition

[tex]A[/tex] is closed if [tex]A^{C}[/tex] is open is a perfectly reasonable definition, at least in topology, where the definition of open is the elements of the topology set. It's the standard definition, and it's used by Munkres which is the only text I have here.
 

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