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I was looking over previous comprehensive exams, and I saw this question about the Sorgenfrey line:

Can the Sorgenfrey line be represented as the union of three metrizable subspaces?

I haven't made any progress with this one, and I was wondering if anyone could help me out.

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In fact the Sorgenfrey line $S$ is not the union of countably many metrizable subspaces, because no uncountable subspace of $S$ is metrizable. (Obviously if $S$ is written as the union of countably many subspaces, at least one of the subspaces must be uncountable.)

Suppose that $X$ is an uncountable subspace of $S$. Let $\mathscr{I}$ be the set of open intervals in $\Bbb R$ with rational endpoints, and let $\mathscr{I}_0=\{I\in\mathscr{I}:I\cap X\ne\varnothing\}$. For each $I\in\mathscr{I}_0$ pick a point $x_I\in I\cap X$, and let $D_0=\{x_I:I\in\mathscr{I}_0\}$. $D_0$ is almost dense in $X$: if $x\in X$, $x<y\in\Bbb R$, and $(x,y)\cap X\ne\varnothing$, pick $z\in(x,y)\cap X$ and an $I\in\mathscr{I}_0$ such that $z\in I\subseteq(x,y)$, and observe that $x_I\in[x,y)\cap D_0$. However, if $x\in X\setminus D_0$, and there is a $y>x$ such that $[x,y)\cap X=\{x\}$, then $[x,y)\cap X$ is a non-empty open set in $X$ disjoint from $D_0$.

To fix this, let $L=\{x\in X:(x,y)\cap X=\varnothing\text{ for some }y>x\}$. For each $x\in L$ let $y_x>x$ be such that $(x,y_x)\cap X=\varnothing$. Then $\{(x,y_x):x\in L\}$ is a family of pairwise disjoint open intervals in $\Bbb R$, so it must be countable, and hence so must $L$. Let $D=D_0\cup L$; then $D$ is a countable dense subset of $X$, and $X$ is separable.

It follows that if $X$ were metric, it would be second countable. This, however, is impossible. To see this, let $\mathscr{B}$ be a base for $X$. Let $x\in X$ be such that $(\leftarrow,x)\cap X$ is uncountable. (There must be such an $x$. I’ll leave it to you to prove this; let me know if you get stuck.) For each $y\in(\leftarrow,x)\cap X$ there is a $B_y\in\mathscr{B}$ such that $y\in B_y\subseteq[y,x)$. Clearly the map $(\leftarrow,x)\cap X\to\mathscr{B}:y\mapsto B_y$ is injective, so $\mathscr{B}$ is uncountable.

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