I've encountered this problem in Munkres General Topology, the problem ask if $X$ is a limit point compact set, then if $X$ is a subspace of a Hausdorff space $Z$, is $X$ necessarily closed?
I think I know the answer should be wrong, as there is a counter example here: Is a limit point compact subset of a Hausdorff space necessarily closed?
However, I'm not satisfied by it because it make use of the minimal uncountable set, which has no practical example. I've also worked up a proof which proves this implication is true. My idea is to show that every convergent sequence in $X$ converges to a point in $X$.
Proof: Let $\{x_1,x_2,\cdots\}$ be a convergent sequence in $X$ with limit $x\in Z$, if the sequence is a finite set, then $x=x_n$ for all sufficiently large $n$, which proves $x\in X$. Otherwise if the sequence is an infinite set, from the fact that $Z$ is Hausdorff, the sequence must have a unique limit $x$. By definition the sequence $\{x_1,x_2,\cdots\}$ has a limit point in $X$, as stated before this sequence can have only one limit point, so this limit point must be $x$, which is in $X$. So I've proved $X$ is closed because every convergent sequence in $X$ converges in $X$.
Can you point out which part of my proof is wrong? I really want to believe my proof is wrong, so I can put my effort into learning the minimal uncountable set. It will be better if you can provide a counterexample which doesn't make use of the minimal uncountable set.
Best Answer
Let $Z=[0,1]^{\Bbb R}$, all functions from $\Bbb R$ to $[0,1]$ in the product (aka pointwise) topology which is compact Hausdorff.
Let $X$ be its subspace of all functions $f$ that have at most countably many non-zero values, i.e. such that $C(f) = \{x \in \Bbb R\mid f(x) \neq 0\}$ is at most countable. This $X$ is dense in $Z$ (so in particular not closed) and limit point compact, even sequentially compact (which is stronger). This is an example which does not require ordinals, and is quite natural IMHO.