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 $\BB_x = \set {\hointr x {x + \frac 1 n} | n \in \N_{>0} }$.

By definition of first-countability, it suffices to 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 Archimedean Principle there exists $n \in \N$ such that $n > \dfrac 1 \epsilon$ (that is, $\dfrac 1 n < \epsilon$).

So:
 * $x \in \hointr x {x + \dfrac 1 n} \subseteq \hointr x {x + \epsilon} \subseteq U$

Also see

 * Sorgenfrey Line is not Second-Countable
 * Sorgenfrey Line is Separable