Exercise 30.9

Proof. $X$ is Lindelöf if and only if a collection of closed subsets of $X$ with empty intersection has a countable subcollection with empty intersection by taking complements in Theorem $30.3(a)$. Now suppose we have a collection $C$ of closed subsets of $A$ with empty intersection; it is then also a collection of closed subsets of $X$ with empty intersection by Theorem $17.3$ since $A$ is closed, and so has a countable subcollection with empty intersection since $X$ is Lindelöf. Thus, $A$ is also Lindelöf.

Now let $X={ℝ}_{\ell }^2$. We see ${\mathbb{Q}}^2$ is countable, and is dense in $X$ since if we take $x\in X$ and a neighborhood $U\ni x$, there exists a basis element $[a,b)\times [c,d)\subseteq U$ containing $x$, which intersects ${\mathbb{Q}}^2$ by the fact that $(a,b)\times (c,d)\cap {\mathbb{Q}}^2\ne \varnothing$ since $\mathbb{Q}$ is dense in $ℝ$. Thus, $X$ has a countable dense subset; we claim that $L=\{x\times (-x)\mid x\in {ℝ}_{\ell }\}$ is a closed subspace of $X$ that does not have a countable dense subset. $L$ is closed since if $(x_1,x_2)\in X\setminus L$, then letting $d=x_1+x_2$, the basis element $[x_1-d∕3,x_1+d∕3),[x_2-d∕3,x_2+d∕3)$ does not intersect $L$. But then, $L$ has the discrete topology since $\{(x,-x)\}=L\cap [x,b)\times [-x,d)$ is open in $L$. Thus, if $A\subseteq L$, $A=\overline{A}$ by discreteness, and so $\overline{A}=L$ is true if and only if $A=L$. Thus, $L$ has no countable dense subset. □