A question on a Luzin space A Luzin space is a crowded (no isolated points) Hausdorff space in which every nowhere dense set is countable. Now there is a Luzin space with its cardinality is $\omega_1$. Is it a Lindelöf space? Thanks for any help!
 A: You actually get somewhat more: Every Luzin space is hereditarily Lindelöf.
Suppose that $\mathcal{U}$ is a family of open subsets of a Luzin space $L$.  If no countable subfamily has the same union as $\mathcal{U}$ we can pick sequences $\langle x_\xi \rangle_{\xi < \omega}$ and $\langle U_\xi \rangle_{\xi < \omega_1}$ such that


*

*$U_\xi \in \mathcal{U}$; and

*$x_\xi \in U_\xi \setminus \bigcup_{\eta < \xi} U_\eta$.


Clearly $A = \{ x_\xi : \xi < \omega_1 \}$ is an uncountable subset of $L$.  Consider $V = \mathrm{Int} ( \overline{A} )$.
By induction on $\xi$ we show that $x_\xi \notin V$.  Suppose that $x_\eta \notin V$ for all $\eta < \xi$.  If $x_\xi \in V$ it follows that $U_\xi \cap V$ is a nonempty open set.  Note that $( U_\xi \cap V ) \cap A = \{ x_\xi \}$ (since by hypothesis $V \cap \{ x_\eta : \eta < \xi \} = \emptyset$ and by construction $U_\xi \cap A \subseteq \{ x_\eta : \eta \leq \xi \}$).  It follows that 
$$ x_\xi \in U_\xi \cap V = ( U_\xi \cap V ) \cap \overline{A} \subseteq \overline{ ( U_\xi \cap V ) \cap \overline{A} } = \overline{ U_\xi \cap V \cap A } = \overline{ \{ x_\xi \} } = \{ x_\xi \},$$ contradicting that $L$ has no isolated points!  Therefore $x_\xi \notin V$.
As $V \cap A = \emptyset$ it follows that $V \cap \overline{A} = \emptyset$, meaning that $V = \mathrm{Int} ( \overline{A} ) = \emptyset$, contradicting that no uncountable subset of $L$ is nowhere dense!
A: Here’s a slightly shorter argument showing that a Luzin space is hereditarily Lindelöf.
Let $X$ be a Luzin space, and suppose that $D\subseteq X$ is discrete. $X$ has no isolated points, so $D$ is nowhere dense in $X$ and therefore countable. Thus, $s(X)=\omega$, and $X$ is hereditarily ccc. (In words, $X$ has countable spread, and the spread of a space is its hereditary cellularity.)
Now let $Y\subseteq X$ be uncountable, and let $\mathscr{U}$ be an open (in $X$) cover of $Y$. $\bigcup\mathscr{U}$ is ccc, so there is a countable $\mathscr{U})\subseteq\mathscr{U}$ such that $\operatorname{cl}\bigcup\mathscr{U}_0\supseteq\bigcup\mathscr{U}$. The boundary of $\bigcup\mathscr{U}_0$ is nowhere dense in $X$, so it must be countable, and therefore $Y\setminus\bigcup\mathscr{U}_0\subseteq\left(\bigcup\mathscr{U}\right)\setminus\bigcup\mathscr{U}_0\subseteq\operatorname{bdry}\bigcup\mathscr{U}_0$ is countable. Thus, the countable subfamily $\mathscr{U}_0$ of $\mathscr{U}$ covers all but countably many points of $Y$, so there is a countable $\mathscr{V}\subseteq\mathscr{U}$ covering $Y$, and $Y$ is Lindelöf. 
