Can anyone give me a example of space which is limit point compact but not compact?(Except the popular sets $S_\Omega$ and $Z \times Y$ )
I assume everyone knows that two sets.
I really having hard time to find out one. Actually I need it to give a trivial counter example of the statement "If $Y$ is a Hausdorff space and $X$ which is a subspace of that space is limit point compact space . Then $X$ is closed." If $Z \times Y$ was Hausdroff it would have been of use to me.
Thank You in Advance.
Another classic example: let $Y=[0,1]^I$ where $I$ is an uncountable subset (if you like it minimal take $I = \aleph_1$, in the product topology.
$Y$ is compact and Hausdorff (and it's not yet our example):
let $X= \{f \in [0,1]^I : |\{i: f(i) \neq 0\}| \le \aleph_0\}$, the subset of its points (functions, really) that are almost always $0$, except for at most countably many values, of course in the subspace topology. This is called the $\Sigma$-product w.r.t. $0$.
Then $X$ is Hausdorff and countably compact and even sequentially compact, but $X$ is a dense subset of $Y$, and so not closed in $Y$ and not compact.
The first is clear from the observation that if $f_n$ is a sequence from $X$, each $f_n$ "lives" on an at most countable subset $s(f_n)$ of $I$ and so there is a countable subset $I_0$ (the union of all $s(f_n)$) of $I$ such that for all $i \notin I_0$ we have $f_n(i) =0$ for all $n$. Restricted to $I_0$ we have that all $f_n$ really live on $[0,1]^{I_0}$ which is compact metrisable so we can find a convergent subsequence there, which is still a convergent subsequence on $Y$ and thus $X$. This shows sequential compactness and thus countable compactness too.
That $X$ is dense in the full product is clear from the definition of the product topology and a moment's thought.