Sorgenfrey Line is Topology

Theorem
The Sorgenfrey Line is a topological space.

Proof
We have to check that $\mathcal B = \left\{{\left[{a \,.\,.\, b}\right): a, b \in \R}\right\}$ fulfills the axioms of being a basis for a topology.

From Equivalent Definitions of Synthetic Basis we only have to check that:
 * $(1): \quad \bigcup \mathcal B = \R$
 * $(2): \quad \forall B_1, B_2 \in \mathcal B: \exists V \in \mathcal B: V \subseteq B_1 \cap B_2$

We have that:
 * $\forall n \in \Z: \left[{n \,.\,.\, n+1}\right) \in \mathcal B$
 * $\R = \displaystyle \bigcup_{n \in \Z} \left[{n \,.\,.\, n+1}\right) \subseteq \bigcup \mathcal B$

Hence $\R = \bigcup \mathcal B$ and condition $(1)$ is fulfilled.

Now take $ B_1, B_2 \in \mathcal B$ where:
 * $B_1 = \left[{a_1 \,.\,.\, b_1}\right)$
 * $B_2 = \left[{a_2 \,.\,.\, b_2}\right)$

Let $B_3$ be constructed as:
 * $B_3 := \left[{\max \left\{{a_1, a_2}\right\} \,.\,.\, \min \left\{{b_1, b_2}\right\}}\right) \in \mathcal B$

From the method of construction, it is clear that $B_3 = B_1 \cap B_2$.

Thus taking $V = B_3$, condition $(2)$ is fulfilled.