The constructive proof of countably compactness implies limit point compactness In topological space, countably compactness implies limit point compactness. The proof is by contradiction, can anyone provide a constructive proof,i.e. the process of finding such limit point.
 A: I suspect that this is not possible. In the absence of the axiom of choice we can let $X$ be a strictly amorphous set and give $X$ the discrete topology; $X$ itself is an infinite set without a limit point, and $X$ is not limit-point compact, but I’ll show that $X$ is countably compact.
Now let $\mathscr{U}$ be a countable open cover of $X$. If $\mathscr{U}$ has no finite subcover, then each member of $\mathscr{U}$ is finite, $\mathscr{U}$ is countably infinite, and we can faithfully index $\mathscr{U}$ as $\mathscr{U}=\{U_n:n\in\omega\}$. For each $n\in\omega$ let $V_n=U_n\setminus\bigcup_{k<n}U_k$, let $N=\{n\in\omega:V_n\ne\varnothing\}$, and let $\mathscr{V}=\{V_n:n\in N\}$; $\mathscr{V}$ is a partition of $X$ into finite sets, and $X$ is strictly amorphous, so infinitely many of the sets $V_n$ are singletons. Let $M=\{n\in N:|V_n|=1\}$, and for each $m\in M$ let $V_m=\{x_m\}$. $M$ is an infinite subset of $\omega$, so there is a bijection $f:\omega\to M$, and the map 
$$h:\omega\to X:n\mapsto x_{f(n)}$$
is an injection, contradicting the amorphousness of $X$. Thus, $\mathscr{U}$ must have a finite subcover, and $X$ is countably compact.
This means that the theorem that every countably compact space is limit-point compact requires at least some of the strength of the axiom of choice. (Countable choice should be enough.) Depending on just what you count as constructive, that seems likely to be a serious obstacle.
