# Topology Time

1. Mar 18, 2006

### JasonRox

So, here I am again.

This is to only have a proof reviewed. Like I said before, I will do this from time to time, so I know I'm staying on track.

Theorem - Cantor Theorem

Let ${C_1, C_2, ...}$ be a monotone decreasing countable family of non-void closed subsets of a T_1 space such that C_1 is countably compact. Then the intersection of the family above is non-void.

Proof

Now, create the subspace C_1 with respect to the T_1 space. Call this subspace C.

Since C_1 is countably compact the subspace C is countably compact, and it is also a T_1 space (herediatary).

Now, by Theorem 1.3 (proven), which says that...

Theorem 1.3

Let X be a countably compact T_1 space, and let F be a countable collection of open subsets of X. If F covers X, then some finite subfamily of F covers X (X is compact).

So, replacing X with C we get C is compact.

Now, also by Theorem 1.10 (proven), which says that...

Theorem 1.10

A space X is compact if and only if each family of closed subsets of X with the FIP (finite intersection property) has a non-null intersection.

So, now ${C_1, C_2, ...}$ form a family of closed subsets in C. Also, since they are non-void it is quickly observed that they have the FIP.

So, since C is compact this implies that the intersection of ${C_1, C_2, ...}$ is non-empty.

And we are done.

The early proofs made it very easy of course. This is what makes the text so good in my opinion. Proving small little things leads to an easy proof of something bigger.

2. Mar 18, 2006

### matt grime

C_1 is $C_1$, right?

how can you 'create' C_1 if is given, and I've no idea what it means to create with respect to something like this. In short, what the heck is C?

[quote[So, replacing X with C we get C is compact.[/quote]

it isn't clear what C is, and, no I don't believe you can claim C is compact because it is countably compact, that is only true for a metrizable space, apparently.

Countably compact and T_1 implies sequentially compact, if that is any use.

(source, planetmath, i've never 'heard' of countably compact, though it was clear what it should mean, and indeed does mean: every countable open cover has a finite subcover)

3. Mar 18, 2006

### Cincinnatus

What does "monotone decreasing" mean in this context?

Is it that C_2 is a proper subset of C_1;
C_3 is a proper subset of C_2 etc. ?

4. Mar 18, 2006

### matt grime

I would guess so, though it is a strange choice of words.

YOu want to show the intersection of the C_i is non-empty. It is unnecessary to do so, but let's do it by contradiction:

if the intersection is empty, the complement (inside C_1) is all of C_1, but the complement of the intesection is the union of the complements which are thus a countable cover of C_1, which has a finite refinement. you should be able to see now why this is a contradiction.

5. Mar 18, 2006

### JasonRox

That's only true if it is countably compact in a T_1 space.

6. Mar 18, 2006

### JasonRox

For matt,

C is the relative topology with respect to the T_1 space using C_1.

7. Mar 18, 2006

### mathwonk

why make such an elementary theorem look so hard?

8. Mar 18, 2006

### JasonRox

It doesn't look hard.

The proof can be done in two lines. I just added all those comments so you are on the same page. I can't refer to Theorem X if you don't got the text.

9. Mar 19, 2006

### matt grime

Surely what I wrote is the definition of countably compact: every countable subcover has a finite refinement, what do you think it means? It is certainly what planetmath thinks it means.

Theorem: let A be any infinite ordinal and C_a be a collection of nested sets indexed A with C_0 A-compact, then the intersection of all C_a is non-empty.

A-compact (my definition, though I suspect it already exists and means exactly this) means any open cover indexed by a cardinal A has a finite subcover.

This is exactly what the category theoretic notion of compactness implies that ought to mean.

Last edited: Mar 19, 2006
10. Mar 19, 2006

### matt grime

Cough, that still makes no sense. Why not just say C=C_1? Saying 'using' doesn't mean anything on its own, and there's no need to refer to the relative topology. You can't just say C is a topology without at least saying on what it is the topology. Indeed you explicitly say C is a subspace, not a topology.

Last edited: Mar 19, 2006
11. Mar 19, 2006

### matt grime

your proof is incorrect, and you don't need Theorem X which I believe when x=1.3 you have stated incorrectly, or you are not using the definition of these things as I've seen from other sources. In particular a T_1 countably compact space *IS NOT COMPACT* as you assert. It is sequentially compact.

Last edited: Mar 19, 2006
12. Mar 21, 2006

### Cincinnatus

Why isn't a T1 countably compact space also compact?

That certainly seems like it ought be true... (to me anyway, not that I've tried to prove it though)

13. Mar 21, 2006

### JasonRox

Let X be a T1 countably compact space then each countable open cover for X has a finite cover.

If it were compact, then it would be true for uncountable open covers to, but that's not always the case.

14. Mar 21, 2006

### matt grime

What on earth is the definition of a countably compact space if it is not that any countable cover has a finite refinement.

The only definition i found was that one i used, and if it is T1 then a countably compact space is sequentially compact.

15. Mar 21, 2006

### JasonRox

A subset X of a space is countably compact if and only if each infinite subset of X has a limit point in X.

16. Mar 21, 2006

### matt grime

that is seemingly not the definition given normally, though I believe it is the equivalent to the seemingly more 'obvious' one that I gave. If you look at yours, can you see any reason for using the word 'countable' at all for it?

http://en.wikipedia.org/wiki/Compact_space section 5