A collection \(\mathcal{A}\) of subsets of \(X\) has the countable intersection property if every countable intersection of elements of \(\mathcal{A}\) is nonempty. Show that \(X\) is a Lindelöf space if and only if for every collection \(\mathcal{A}\) of subsets of \(X\) having the countable intersection property,
\[ \bigcap_{A \in \mathcal{A}} \overline{A} \]
is nonempty.
Solution
Assume first that \(X\) is Lindelöf, and let \(\mathcal A\) have the countable intersection property. Suppose \[ \bigcap_{A\in\mathcal A}\overline A=\varnothing. \] Then the open sets \[ X\setminus\overline A,\qquad A\in\mathcal A, \] cover \(X\). Lindelöfness gives a countable subcover \[ X=\bigcup_{n=1}^\infty (X\setminus\overline{A_n}). \] Taking complements, \[ \bigcap_{n=1}^\infty\overline{A_n}=\varnothing, \] so certainly \(\bigcap_{n=1}^\infty A_n=\varnothing\), contradicting the countable intersection property. Hence the intersection of all closures is nonempty.
Conversely, assume the stated closure-intersection property, and let \[ \mathcal U=\{U_i:i\in I\} \] be an open cover of \(X\). Suppose no countable subfamily covers \(X\). Put \[ A_i=X\setminus U_i. \] For every countable set of indices \(i_1,i_2,\dots\), the union \(U_{i_1}\cup U_{i_2}\cup\cdots\) is not all of \(X\), so \[ A_{i_1}\cap A_{i_2}\cap\cdots\ne\varnothing. \] Thus \(\{A_i:i\in I\}\) has the countable intersection property. Each \(A_i\) is closed, so the assumed property gives \[ \varnothing\ne\bigcap_{i\in I}\overline{A_i} =\bigcap_{i\in I}A_i =X\setminus\bigcup_{i\in I}U_i, \] contradicting that \(\mathcal U\) covers \(X\). Therefore every open cover has a countable subcover, and \(X\) is Lindelöf.