How can I prove that X is COUNTABLY compact?

  • Context: Graduate 
  • Thread starter Thread starter synergy
  • Start date Start date
  • Tags Tags
    Topology
Click For Summary

Discussion Overview

The discussion revolves around proving that a topological space X is countably compact, specifically exploring the equivalence between countable compactness and the property that every nested sequence of closed nonempty subsets has a nonempty intersection. Participants also seek to analyze specific sets, including S-omega and the product space [0,1]^|R, in terms of their compactness properties, including sequential compactness, limit point compactness, and countable compactness.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant asks for suggestions on proving that a topological space X is countably compact if and only if every nested sequence of closed nonempty subsets has a nonempty intersection.
  • Another participant suggests starting by considering different proof strategies and working from definitions.
  • There is a discussion about the properties of S-omega, with one participant questioning the topology used and suggesting it may be the smallest ordinal number.
  • Participants discuss whether [0,1]^|R is sequentially compact, with one suggesting interpreting it as the set of functions from R to [0,1].
  • One participant mentions that X x Y is limit point compact but not countably compact, providing reasoning based on the existence of an open cover without a finite subcover.
  • Another participant questions the limit point compactness of X x Y and suggests that the hints provided by the professor may be incorrect.
  • One participant reports progress on proving that [0,1]^|R is sequentially compact, referencing Tychonoff's theorem.
  • There is clarification about S-omega being the long line, with a participant describing its structure and topology.

Areas of Agreement / Disagreement

Participants express uncertainty about the properties of S-omega and whether the hints provided by the professor are accurate. There is no consensus on the implications of limit point compactness for X x Y, and multiple viewpoints on the interpretations of the topologies involved are present.

Contextual Notes

Participants note the complexity of definitions and theorems involved in the discussion, indicating a reliance on specific topological properties that may not be universally agreed upon. The discussion includes various assumptions about the topologies being used, which remain unresolved.

synergy
Messages
62
Reaction score
0
:bugeye:
Does anybody have any suggestions on how to prove that a topological space X is COUNTABLY compact (i.e. every COUNTABLE open cover has a finite subcover), IF AND ONLY IF, EVERY NESTED SEQUENCE of closed nonempty subsets of X has a nonempty intersection?

I also need hints on how to show whether S-omega is SC, CC, LPC, or compact, where SC is sequentially compact(every sequence has a convergent subsequence), CC is countably compact (every COUNTABLE open cover contains a finite subcover, LPC is limit point compact (every infinite subset has a limit point), and compact is EVERY open cover contains a finite subcover). S-omega is the smallest uncountable set (properties: cut it off anywhere before the point omega and you have a countable set, the point omega has no IMMEDIATE predecessor).

I also need to show whether [0,1]^|R, that is, [0,1]x[0,1]x... an uncountable number of times, whether that is sequentially compact, i.e. whether EVERY sequence has a convergent subsequence.

I also need to show that X x Y is limit point compact but not countably compact, where X=the natural numbers, the topology on X is the power set, Y={0,1} and its topology is ( {0,1} , the empty set ). Limit pt. compact means every infinite subset has a limit point, and countably compact means every countable open cover contains a finite subcover.

If somebody can help with hints, I'd be much obliged. I've tried drawing pictures, and some of the other parts of the questions I've already gotten, but this has been a hard class in general for me because of the sheer volume of definitions and theorems. The professor helps very much when I go to his office, some problems he practically does for us, but I'm hardly ever free when he has office hours. I usually work all day until 9 or 10:00 Thursday and Friday night to get it turned in on Friday (he has a "slide it under my door" policy). I can't work Monday or Tuesday on it because I have a job that takes a lot of my time, plus I have to work on my other classes SOMETIME, and that sometime is monday and tuesday. I'm feeling a little more behind than usual this week, and I'll have to share his one hour of office time with a lot of other students tomorrow. Please post any ideas you may have, any hint is better than none.

Thanks in advance for any help. All hints are greatly appreciated.
Aaron
 
Physics news on Phys.org
Ack, lots of questions at once! One step at a time!

Does anybody have any suggestions on how to prove that a topological space X is COUNTABLY compact (i.e. every COUNTABLE open cover has a finite subcover), IF AND ONLY IF, EVERY NESTED SEQUENCE of closed nonempty subsets of X has a nonempty intersection?

Well, I usually start this sort of thing by considering each of the different ways of proving parts of this statement (use one to prove the other, assume one is false the other is true, etc) and working from the definitions.


I also need hints on how to show whether S-omega is SC, CC, LPC, or compact, where SC is sequentially compact(every sequence has a convergent subsequence), CC is countably compact (every COUNTABLE open cover contains a finite subcover, LPC is limit point compact (every infinite subset has a limit point), and compact is EVERY open cover contains a finite subcover). S-omega is the smallest uncountable set (properties: cut it off anywhere before the point omega and you have a countable set, the point omega has no IMMEDIATE predecessor).

S-omega... is that supposed to be [itex]S^{\omega}[/itex] or [itex]S_{\omega}[/itex]? It sounds like you're trying to say it's the smallest ordinal number. What topology are you using? The order topology doesn't seem particularly interesting; wouldn't that work out to the discrete topology?


I also need to show whether [0,1]^|R, that is, [0,1]x[0,1]x... an uncountable number of times, whether that is sequentially compact, i.e. whether EVERY sequence has a convergent subsequence.

(Is the topology on R irrelevant?) Recall that [0,1]^R is the set of functions from R to [0,1] -- maybe if you can interpret the topology in this way, it would help.


I also need to show that X x Y is limit point compact but not countably compact, where X=the natural numbers, the topology on X is the power set, Y={0,1} and its topology is ( {0,1} , the empty set ). Limit pt. compact means every infinite subset has a limit point, and countably compact means every countable open cover contains a finite subcover.

There's an obvious open cover that has no finite subcover. :smile: You sure it's supposed to be limit point compact? What's a limit point of (0, 0), (1, 0), (2, 0), (3, 0), ...?
 
Hurkyl said:
There's an obvious open cover that has no finite subcover. :smile: You sure it's supposed to be limit point compact? What's a limit point of (0, 0), (1, 0), (2, 0), (3, 0), ...?

Yes, the hints the professor posted clearly say it is limit point compact, and I thought of your counter-example also, so maybe the hints are wrong? I'll be seeing him for a minute later, I'll ask.

Thanks for your wise suggestions. Oh, and it's S-sub-capitalOmega. I need to check the topology again, I think it's similar to the reals. I believe the professor said when he first introduced it that it's sometimes called the "long line" if you want to look it up on the internet.

Thanks again,
Aaron
 
progress report

In the case of [0,1]^|R, the product of compact spaces is compact by Tychonoff's theorem, and in particular is sequentially compact, so I've got that one. Stupid me , Tychonoff's theorem was in the hints the professor posted.

Hurkyl, I already had this when I wrote the above post, guess I just forgot.
Let Z = X x Y. Let L = {z_1, z_2,...} be an infinite sequence of points in Z. We can write z_i=(x_i, y_i) for x_i in X and y_i in Y. Without loss of generality, we may assume that y_1=0. Let z=(x_1, 1). Then ANY closed set containing z will also contain z_1, and hence z is a limit point of L. (Basically, we use the fact that Y is not a T_1 space here.)

So I've got those, does anyone have any ideas on the rest?
Aaron
 
Ah, I know the long line. It's two long rays placed end to end!

The long ray is described by the set Sx[0,1) where S is the first uncountable ordinal number, and the topology is the order topology. Bleh, I forget which way the order is supposed to work when multiplying order types... the way I've written it, I want (0,0.5) < (1, 0.0). It's supposed to be pictured as S copies of [0,1) lined up end to end.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
889
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K