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Let us assume the space $X$ is not compact. Then there exists a covering with no finite subcovering, such that every set contains at least one point that no other does.

Select a countable number of such points assuming the axiom of countable choice. We have an infinite sequence $\langle x_{n}\rangle$.

There is no point in $X$ such that every open set containing it contains infinite points of $\langle x_{n}\rangle$, as for every point in $X$ there is at least one set which contains only one or no point of $\langle x_{n}\rangle$- namely the open set part of the infinite cover of $X$. Hence, there is no accumulation point in $X$, making $X$ not sequentially compact.

I know for a fact that there exist topological spaces which are sequentially compact but not compact. It would be great if someone pointed out the (perhaps glaring) flaw in the argument?

Thanks in advance!

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I'm sorry I mean there exist spaces that are sequentially compact but not compact. It has been corrected in the question too. –  Ayush Khaitan Jul 13 '13 at 7:27
    
@Ittay: Compactness does not imply sequential compactness. $\beta\omega$ is compact but not sequentially compact. –  Brian M. Scott Jul 13 '13 at 7:28
    
ohhhh I was reading this completely wrong. Sorry Ayush, and thank Brian for setting me straight. –  Ittay Weiss Jul 13 '13 at 7:30
    
The question is we know a sequentially compact space need not be compact. But my erroneous argument suggets that a sequentially compact space should be compact. I was looking for the 'hole' in the argument. Thanks –  Ayush Khaitan Jul 13 '13 at 7:31

1 Answer 1

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The flaw is in the clause such that every set contains at least one point that no other does. The ordinal space $\omega_1$ with the order topology is sequentially compact but not compact. The most obvious open cover with no finite subcover is the one consisting of the sets $V_\alpha=\{\xi:\xi<\alpha\}$ for $\alpha<\omega_1$: this is an uncountable nest of increasing open sets, so each point of the space is actually contained in uncountably many of the $V_\alpha$.

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Oh. My thought was if there is such a set in the cover such that every point in it is contained in some other set of the cover, then we can remove that set to still get a cover. Thanks for the answer Brian! –  Ayush Khaitan Jul 13 '13 at 7:36
    
@Ayush: In my example let $\mathscr{V}=\{V_\alpha:\alpha<\omega_1\}$. You can remove any countable subcollection from $\mathscr{V}$ and still have a cover. In fact, every uncountable subcollection of $\mathscr{V}$ is a cover. But if $\mathscr{U}$ is any subcover of $\mathscr{V}$, every point of the space is going to belong to uncountably many members of $\mathscr{U}$. You’re welcome! –  Brian M. Scott Jul 13 '13 at 7:40

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