Countable compactness vs. limit point compactness

[radou]

Homework Statement

If X is a T1 space, countable compactness is equivalent to limit point compactness.

The Attempt at a Solution

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Let X be limit point compact, and assume X is not countable compact. So, there exists a countable open cover for X such that no finite subcover covers X. So there is an element x1 not belonging to U1, x2 not belonging to U1 U U2, etc. in general xn doesn't belong to U1 U ... U Un. Now, the infinite set S = {x1, x2, ...} has a limit point x. Since X is T1, every neighborhood of x intersects S in infinitely many points. Let Uj be an open set from the countable open cover containing x (and infinitely many points of S). Choose a finite number of sets which cover the finite number of remaining elements of S, let's say m of them. Now we have a finite subcover which covers X, contradicting the fact that X is not countable compact.

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Now, this is the direction I'm having trouble with, however I attack the problem, I don't seem to get anywhere. Any suggestions?

 

[Office_Shredder]

Claim: If S has no limit points, then the complement of S is open.

Use this along with some choice neighborhoods of elements of S to construct a countable cover with no finite subcover. You don't even need T₁ to do this direction

 

[radou]

I already considered that, and I found this thread which confuses me on that specific point:

https://www.physicsforums.com/showthread.php?t=180371

 

[micromass]

If S has no limit points. Then S is closed.
Furthermore, for every x in S there exists an open set $$G_x$$ such that $$G_x\cap S=\{x\}$$

Then $$\{G_x~\vert~x\in S\}\cup \{X\setminus S\}$$ is the open cover your looking for.


Can you tell what about that link you found bothering? Their definition of limit point is not the same as our definition of limit point (allthough in T1 spaces it is the same).

 

[radou]

If S has no limit points. Then S is closed.
Furthermore, for every x in S there exists an open set $$G_x$$ such that $$G_x\cap S=\{x\}$$

Then $$\{G_x~\vert~x\in S\}\cup \{X\setminus S\}$$ is the open cover your looking for.


Can you tell what about that link you found bothering? Their definition of limit point is not the same as our definition of limit point (allthough in T1 spaces it is the same).

Out definition of a limit point is:

"x is a limit point of the set A if every neighborhood of x intersects A in some point other than x itself "

right?

 

[micromass]

Right.

While in the other tread a limit point was a point x such that every neighbourhood of x contains infinitely many points of A. In standard terminology this is called a $$\omega$$-accumulation point.

 

[radou]

Right.

While in the other tread a limit point was a point x such that every neighbourhood of x contains infinitely many points of A. In standard terminology this is called a $$\omega$$-accumulation point.

OK. By the way, as mentioned above, if X is a T1 space, then every limit point is an accumulation point, i.e. x is a limit point of A if and only if every neighborhood of X contains infinitely many points of A.

If S has no limit points. Then S is closed.
Furthermore, for every x in S there exists an open set $$G_x$$ such that $$G_x\cap S=\{x\}$$

Then $$\{G_x~\vert~x\in S\}\cup \{X\setminus S\}$$ is the open cover your looking for.

OK, so the strategy of our proof for this direction is to prove the contrapositive right?

So, if X is not limit point compact, X is not countable compact. Somehow I expect to find a countable open cover for X with no finite subcover, but I don't see how what you wrote fits into it? Perhaps I misunderstood it.

 

[micromass]

OK. By the way, as mentioned above, if X is a T1 space, then every limit point is an accumulation point, i.e. x is a limit point of A if and only if every neighborhood of X contains infinitely many points of A.

Yes. In fact a space is T1 if and only if every limit point is an $$\omega$$-accumulation point.

OK, so the strategy of our proof for this direction is to prove the contrapositive right?

So, if X is not limit point compact, X is not countable compact. Somehow I expect to find a countable open cover for X with no finite subcover, but I don't see how what you wrote fits into it? Perhaps I misunderstood it.

Yes, so the proof is contrapositive. It goes like this:

If X is not limit point compact, then there exists a set S without limit point.
If S has no limit points. Then S is closed.
Furthermore, for every x in S there exists an open set$$G_x$$ such that $$G_x\cap S=\{x\}$$.

Then $$\{G_x~\vert~x\in S\}\cup \{X\setminus S\}$$ is a countable cover without finite subcover. Thus X is not countably compact.


Note that we did not use the T1 property here. Thus countably compact $$\Rightarrow$$ limit point compact is always true.

 

[radou]

Hm, but how do we know that S is countable? (which implies that the family Gx is countable)

 

[micromass]

Ah yes. That's a little gap in the proof. But it's easily solved. Let S be a set without limit points, then any countable subset of S also has no limit points. So we can assume S to be countable.

 

[radou]

Ah yes. That's a little gap in the proof. But it's easily solved. Let S be a set without limit points, then any countable subset of S also has no limit points. So we can assume S to be countable.

So, basically, instead of S, in our proof we take a countable subset S' of S and complete the proof with it, right?

 

[micromass]

Right!

 

[radou]

OK, thanks a lot for your help!