Limit point compactness implies countable compactness for a $T_1$ -space. Let $X$ be a limit point compact $T_1$ -space and $\{U_n\}$ be a countable open covering of $X$ such that there is no finite subcovering of $X$ . Let $V_n=U_1\cup . . .\cup U_n$ . Note that for every $n$ , $V_n$ does not cover $X$ , but for every $x\in X$ there is minimal $n_x$ such that $x\in V_{n_x}$ . Let $x_0\in X$ . For each $n\ge 1$ let $x_n\in X-V_{n_{x_{n-1}}}$ . This defines an infinite subset of $X$ that must have a limit point $a$ . But then the neighborhood $V_{n_a}$ of $a$ contains only finite number of elements in the sequence. And for each of them different from $a$ we can find a neighborhood of $a$ that does not contain it (here we use the $T_1$ -property — consider Example 1 to see why it does not work in general). The finite intersection of all these neighborhoods with $V_{n_a}$ is a neighborhood of $a$ that does not contain any point of the sequence different from $a$ . This contradicts the fact that $a$ is a limit point of the sequence.Countable compactness implies limit point compactness. Suppose $X$ is a countably compact space and $Y$ is an infinite subset of $X$ . There is a countably infinite subset $Z\subseteq Y$ and every limit point of $Z$ is a limit point of $Y$ as well. If no point in $Z$ is a limit point of $Z$ then every point in $Z$ has a neighborhood that does not contain any other points in $Z$ , and the countable collection of such neighborhoods covers $Z$ . Since each set in the collection covers one point of $Z$ only and $Z$ is infinite, there is no finite subcollection covering $Z$ . Therefore, $Z$ is not closed (a closed subset of a countably compact space is countably compact: add the complement of the subset to any its covering) and there is a limit point of $Z$ .