Show whether S-omega is SC, CC, LPC, or compact

[synergy]

This is a topology question.
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

 

[synergy]

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.

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

 

[wais]

To show whether S-omega is SC, CC, LPC, or compact, we need to consider the definitions of each of these properties and see which ones apply to S-omega.

Starting with sequentially compact, we can try to find a counterexample that shows that S-omega is not sequentially compact. A counterexample would be a sequence that does not have a convergent subsequence in S-omega. For example, the sequence (1, 2, 3, ...) does not have a convergent subsequence in S-omega since S-omega does not contain the point omega+1. Therefore, we can conclude that S-omega is not sequentially compact.

Next, for countably compact, we need to show that every COUNTABLE open cover contains a finite subcover. Let's consider a countable open cover of S-omega, which would consist of countably many open sets. Since S-omega is the smallest uncountable set, this means that any open set in the cover must contain infinitely many points of S-omega. Therefore, we cannot choose a finite subcover from this countable open cover, as there will always be points left uncovered. Thus, S-omega is not countably compact.

For limit point compactness, we need to show that every infinite subset of S-omega has a limit point. This is true for S-omega, as any infinite subset must contain the point omega, which is a limit point in S-omega. Therefore, S-omega is limit point compact.

Lastly, for compactness, we need to show that every open cover contains a finite subcover. Let's consider an open cover of S-omega. Since S-omega is the smallest uncountable set, this means that any open set in the cover must contain infinitely many points of S-omega. However, we can choose a finite subcover from this open cover by taking one open set that contains the point omega and infinitely many other open sets that cover the remaining points. Therefore, S-omega is compact.

Moving on to [0,1]^|R, we can use similar reasoning to show that it is not sequentially compact. Consider the sequence (0, 1/2, 0, 1/4, 0, 1/8, ...). This sequence does not have a convergent subsequence in [0,1]^|R, as it oscillates between 0 and 1