Sorgenfrey Line is Perfectly Normal

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Theorem

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.

Then $T$ is perfectly normal.

Proof

From the definition of perfectly normal space, it is necessary to prove that $T$ is a $T_1$ space and that any closed set is $G_\delta$.

From $T_2$ Space is $T_1$ Space and Sorgenfrey Line is Hausdorff:

the Sorgenfrey line is a $T_1$ space.

From Complement of $F_\sigma$ Set is $G_\delta$ Set it is sufficient to prove that an open set of $T$ is $F_\sigma$.

Let $W$ be any open set in $T$.

Let $O \subseteq W$ be the interior of $W$ with respect to the metric space:

$\R = \struct {\R, d}$

where $d$ is the usual metric on $\R$.

From the definition of $T$, for each $x \in W \setminus O$, we can choose $h_x \in W$ such that $\hointr x {h_x} \subseteq W$.

Suppose $\hointr x {h_x} \cap \hointr y {h_y} \ne \O$ for some distinct points $x, y \in W \setminus O$.

Then either $x < y < h_x$ or $y < x < h_y$.

If $x < y < h_x$, then $y \in \openint x {h_x} \subseteq O$, contradicting $y \in W \setminus O$.

Similarly, if $y < x < h_y$, then $x \in \openint y {h_y}) \subseteq O$, contradicting $x \in W \setminus O$.

Thus $\family {\hointr x {h_x} }_{x \mathop \in W \setminus O}$ is a pairwise disjoint indexed family of open sets of $T$.

From Sorgenfrey Line is Separable and Separable Space satisfies Countable Chain Condition:

$\set {\hointr x {h_x} : x \in W \setminus O}$ is countable

and thus:

$W \setminus O$ is countable.

From Metric Space is Perfectly T4:

$O$ is an $F_\sigma$ set in $\R$.

Thus from Sorgenfrey Line is Expansion of Real Line:

$O$ is an $F_\sigma$ set in the Sorgenfrey line.

Since $W \setminus O$ is a countable union of singletons and $T$ is a $T_1$ space:

$W \setminus O$ is an $F_\sigma$ set in $T$.

Since $W = O \cup \left({W \setminus O}\right)$ and $F_\sigma$ sets are closed under unions:

$W$ is an $F_\sigma$ set in $T$.

$\blacksquare$