# How is a sequentially compact space not compact?

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?

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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

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 $V_\alpha$.
@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