Convergent subsequences in compact spaces

[redbowlover]

My quick question is this: I know it's true that any sequence in a compact metric space has a convergent subsequence (ie metric spaces are sequentially compact). Also, any arbitrary compact topological space is limit point compact, ie every (infinite) sequence has a limit point.
So in general, are the compact spaces that are not sequentially compact?

This is part of a larger problem: If a real-valued function on a topological space X is proper, show the image of f is closed. My idea was to chose a limit point y of f(X) and a sequence f(x_n) in f(X) converging to Y. Cover this sequence by a closed interval I. The the preimage of I is compact and contains the sequence x_n. Now, x_n has a limit point, say x.

If I could show f(x)=y, I'd be done. But i get stuck without being able to use sequential compactness.


THanks ahead of time

 

[micromass]

My quick question is this: I know it's true that any sequence in a compact metric space has a convergent subsequence (ie metric spaces are sequentially compact). Also, any arbitrary compact topological space is limit point compact, ie every (infinite) sequence has a limit point.
So in general, are the compact spaces that are not sequentially compact?

Yes, sequentially compactness has nothing to do with compactness in general topological spaces. A counterexample is given by the topology of finite complements. Take \mathbb{N} equipped with the topology

\mathcal{T}=\{A\subseteq \mathbb{N}~\vert~\mathbb{N}\setminus A~\text{is finite}\}\cup \{\emptyset\}

Then the sequence

0,1,2,3,4,5,...

does not have a convergent subsequence. (in fact, the only convergent sequences in this topology are the one which eventually become constant).

This is part of a larger problem: If a real-valued function on a topological space X is proper, show the image of f is closed. My idea was to chose a limit point y of f(X) and a sequence f(x_n) in f(X) converging to Y. Cover this sequence by a closed interval I. The the preimage of I is compact and contains the sequence x_n. Now, x_n has a limit point, say x.

If I could show f(x)=y, I'd be done. But i get stuck without being able to use sequential compactness.


THanks ahead of time

What is proper with you? I imagine it's something like "the inverse images of compact sets are compact", but do you also demand things like surjective??

And are there conditions on X, like Hausdorff?

 

[micromass]

You can probably save your argument using filters or nets, instead of sequences

 

[redbowlover]

for me a proper map is a continuous map such that the preimage of a compact set is compact. Unfortunately there is no extra conditions on X.

and since I've never worked with nets before, a net-free solution would be best :-) I went with a sequence route because I know it works for proper maps from R^n to R...

 

[micromass]

OK, the key idea is the following:

If A_i is a collection of subsets of X such that its interiors cover X, and if each f\vert_{A_i} is closed, then f is closed.

Use properness to find such a collection of A_i's.

 

[redbowlover]

thanks...i'll think about this

 

[Hurkyl]

and since I've never worked with nets before, a net-free solution would be best :-)

This might be a good time to start -- one of the best ways to learn a concept is to see how it is used to extend the reach of other notions you understand.

 

[Bacle]

Wouldn't the use of nets be artificial here, tho? Metric spaces are 1st-countable,

so sequences work to detect continuity and convergence.

 

[micromass]

Wouldn't the use of nets be artificial here, tho? Metric spaces are 1st-countable,

so sequences work to detect continuity and convergence.

But we're not working in a first countable space here. Otherwise, I would agree with you!

 

[Bacle]

You're right, micromass; I read when s/he mentioned a metric space X, and I

thought that is what the ref. was about.

 

[dslowik]

Micromass Jul 15 3:11 pm:
Yes, sequentially compactness has nothing to do with compactness in general topological spaces. A counterexample is given by the topology of finite complements. Take N equipped with the topology

T={A⊆N | N∖A is finite}∪{∅}Then the sequence

0,1,2,3,4,5,...does not have a convergent subsequence. (in fact, the only convergent sequences in this topology are the one which eventually become constant).

***
But:
Isn't every point a limit point of this sequence (any open set containing an arbitrary point contains all points beyond some point of the above sequence)? This is example 18 (1) of Counterexamples in Topology by Steen & Seebach.

 

[micromass]

Micromass Jul 15 3:11 pm:
Yes, sequentially compactness has nothing to do with compactness in general topological spaces. A counterexample is given by the topology of finite complements. Take N equipped with the topology

T={A⊆N | N∖A is finite}∪{∅}


Then the sequence

0,1,2,3,4,5,...


does not have a convergent subsequence. (in fact, the only convergent sequences in this topology are the one which eventually become constant).

***
But:
Isn't every point a limit point of this sequence (any open set containing an arbitrary point contains all points beyond some point of the above sequence)? This is example 18 (1) of Counterexamples in Topology by Steen & Seebach.

Yes, so?
The finite complement topology is limit point compact and compact, but not sequentially compact. That's all I wanted to show here. If you want a space that's limit point compact, but not compact, you'll need some other example.