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In the previous parts, I was given that (X,τ) is compact. I have proved that every closed set of X is compact, and that every infinite subset of X has a limit point.

I was struggling with these question and found this thread

choose the correct following option?

This says that every sequence in a compact space need not have a convergent subsequence? I am very confused, as I don't think there is an error in the question, and I don't think they wouldn't phrase it like this if they were looking for a counterexample.

Is there something I have misunderstood or am missing?

Any help is great appreciated! Also any hints on how to start the proof would be a huge help!

srh
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1 Answers1

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There is nothing wrong with the statement. In a compact metric space every sequence has convergent subsequence but thus is not always true in compact topological spaces. Every sequence in a compact topological space has convergent subnet but not necessarily a convergent subsequence.

For additional information please search for 'sequential compactness' in Wikipedia.

Kavi Rama Murthy
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  • So I can't prove that every sequence in the given space has a convergent subsequence? – srh Dec 21 '19 at 12:18
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    No,you cannot. There is a distinction between compactness and sequential compactness of topological spaces. I suggest you take a look at sequential compactness in Wikipedia. @srh – Kavi Rama Murthy Dec 21 '19 at 12:19
  • That's really helpful thank you! – srh Dec 21 '19 at 12:27
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    @srh In the compact space ${0,1}^{\Bbb R}$, there are sequences without a convergent subsequence. Also in $\beta \Bbb N$, if you studied that space. – Henno Brandsma Dec 21 '19 at 12:38