Sequentially compact subset of Hausdorff space closed? We know compact subset of a Hausdorff space is closed. Does the same hold for  sequentially compact subset as well ?
Let X be a topological space with topology  $\tau$ which is Hausdorff. $A\subset X$ is sequentially compact that is every sequence in A has a subsequence converging to some point a in $A$. Then can we say anything about close or open Ness of $A$. I know if $X$ is first countable as well then A would be closed but what happens if $X$ is not first countable? 
Or is there a counter example to show that A can be closed or open or neither ?
Thanks.
 A: If $X = [0, \omega_1]$ and $A = [0, \omega_1)$, then $A$ is sequentially compact but not closed in $X$.  (And $A$ happens to be open in $X$.)
If $X = [0, \omega_1]$ and $A = [\omega, \omega_1)$, then $A$ is sequentially compact but it is neither closed nor open in $X$.
A: Another classical example (besides $\omega_1$ in its compactification) is the $\Sigma$-product: 
Let $\prod_{i \in I} [0,1]_i$ be the product of copies of $[0,1]$, where $|I|\ge \aleph_1$, so an uncountable power of $[0,1]$ (also denoted by $[0,1]^I$), in the product topology. By the Tychonoff theorem this is compact and it's also Hausdorff as any product of Hausdorff spaces.
Define $$\Sigma_I [0,1]_i = \{f \in [0,1]^I: |I\setminus f^{-1}[\{0\}|\le \aleph_0\}$$ the set of all members of $[0,1]^I$ with countable support: there are at most countably many coordinates $i$ where $f(i) \neq 0$.
This set is dense in the product $[0,1]^I$: if $U$ is a basic open set of $[0,1]$, then there are finitely many $J \subseteq I$ and for each $j \in J$ a non-empty open set $U_j$ of $[0,1]$ such that $$U = \{f \in [0,1]^I: \forall j \in J: f(j) \in U_j\}$$ a set determined by finitely many open restrictions. Pick $x_j \in U_j$, for every $j \in J$ and define $f(j) = x_j$ and continue defining $f$ by $f(i) = 0$ for $i \notin J$. Then $f \in \Sigma_I [0,1]_i \cap U$ (as the finitely many demands are fulfilled and the support is even at most finite), so the $\Sigma$-product intersects all basic open sets, and so is dense.
This set is also sequentially compact: Let $(f_n) \in \Sigma_{i \in I} [0,1]_i$ be any sequence. Define $I_n  = f^{-1}[(0,1]]$, which is at most countable, for each $n$. Let $I' = \cup_n I_n$ which is also countable. Then for all $f_n$, $f_n(i) = 0$ for $i \notin I'$. Now $[0,1]^{I'}$ is metrisable and compact, and so $f_n | I'$ has a convergent subsequence in $[0,1]^{I'}$. As outside $I'$ all of them are $0$ this means this is still a convergent subsequence in $[0,1]^I$ so in $\Sigma_{i \in I} [0,1]_i$ as well. So  $\Sigma_{i \in I} [0,1]_i$ is sequentially compact and dense so not closed (nor open, if you think about it).
