A space \(X\) is said to be countably compact if every countable open covering of \(X\) contains a finite subcollection that covers \(X\). Show that for a \(T_1\) space \(X\), countable compactness is equivalent to limit point compactness. [Hint: If no finite subcollection of \(U_n\) covers \(X\), choose \(x_n \notin U_1 \cup \dots \cup U_n\), for each \(n\).]
Solution
Assume first that \(X\) is countably compact, and let \(A\subset X\) be infinite. Suppose \(A\) had no limit point. Choose a countably infinite subset \[ \{a_1,a_2,\dots\}\subset A. \] Since \(X\) is \(T_1\) and none of the \(a_n\) is a limit point of this subset, for each \(n\) there is an open set \(V_n\ni a_n\) with \[ V_n\cap\{a_1,a_2,\dots\}=\{a_n\}. \] The countable subset \(B=\{a_n:n\ge1\}\) has no limit point, hence is closed in a \(T_1\) space: if \(x\notin B\) lay in \(\overline B\), every neighborhood of \(x\) would meet \(B\), making \(x\) a limit point of \(B\). Therefore \[ X\setminus B,\ V_1,V_2,\dots \] is a countable open cover of \(X\). No finite subfamily covers \(B\). This contradicts countable compactness. Hence every infinite subset has a limit point, so \(X\) is limit point compact.
Conversely, suppose \(X\) is limit point compact and let \[ U_1,U_2,\dots \] be a countable open cover with no finite subcover. For each \(n\), choose \[ x_n\notin U_1\cup\cdots\cup U_n. \] The set \(A=\{x_n:n\ge1\}\) is infinite. Indeed, if it were finite, each of its points would belong to some \(U_j\); for sufficiently large \(N\), all points of \(A\) would then lie in \(U_1\cup\cdots\cup U_N\), contradicting \(x_N\notin U_1\cup\cdots\cup U_N\).
We show \(A\) has no limit point. Fix \(x\in X\), and choose \(m\) with \(x\in U_m\). For every \(n\ge m\), \[ x_n\notin U_m, \] so \(U_m\cap A\) is finite. Since \(X\) is \(T_1\), deleting the finitely many points of \((U_m\cap A)\setminus\{x\}\) leaves an open neighborhood of \(x\) meeting \(A\) in at most \(x\). Hence \(x\) is not a limit point of \(A\). This contradicts limit point compactness. Therefore every countable open cover has a finite subcover.
Thus, for \(T_1\) spaces, \[ \boxed{\text{countably compact}\iff\text{limit point compact}.} \]