Sorgenfrey Line is First-Countable
Theorem
Let $\R$ be the set of real numbers.
Let $\BB = \set {\hointr a b: a, b \in \R}$.
Let $\tau$ be the topology generated by $\BB$, that is, the Sorgenfrey line.
Then $\tau$ is first-countable.
Proof
Let $\BB_x = \set {\hointr x {x + \dfrac 1 n} : n \in \N_{>0} }$.
We will show that:
- $(1): \quad \BB_x$ is countable
- $(2): \quad \BB_x$ is a local basis at $x$
$(1)$ follows from the fact that $\BB_x$ is a bijection from the set of natural numbers.
$(2)$ is demonstrated as follows:
By definition of local basis, it suffices to show that $\forall U \in \tau: x \in U: \exists B \in \BB_x: B \subseteq U$.
Pick any $U$ in $\tau$.
By definition of $\tau$, there exists $\hointr x {x + \epsilon} \subseteq U$ for some $\epsilon \in \R_{>0}$.
By the Axiom of Archimedes there exists $n \in \N$ such that $n > \dfrac 1 \epsilon$ (that is, $\dfrac 1 n < \epsilon$).
So:
- $x \in \hointr x {x + \dfrac 1 n} \subseteq \hointr x {x + \epsilon} \subseteq U$
Therefore $\BB_x$ is a local basis at $x$.
We have seen that the Sorgenfrey line has a countable local basis at every point.
By definition of first-countability, $\tau$ is first-countable.
$\blacksquare$