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Topology questions - Help ?

  1. Nov 4, 2004 #1
    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 alot 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 alot 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.
  2. jcsd
  3. Nov 4, 2004 #2


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    Ack, lots of questions at once! One step at a time!

    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.

    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?

    (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.

    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), ...?
  4. Nov 5, 2004 #3
    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,
  5. Nov 5, 2004 #4
    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?
  6. Nov 5, 2004 #5


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    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.
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