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!
Best Answer
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 $V_\alpha$.