Sorgenfrey Line is Topology
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Theorem
The Sorgenfrey Line is a topological space.
Proof
We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a basis for a topology.
By definition of synthetic basis we only have to check that:
- $(1): \quad \bigcup \BB = \R$
- $(2): \quad \forall B_1, B_2 \in \BB: \exists V \in \BB: V \subseteq B_1 \cap B_2$
We have that:
- $\forall n \in \Z: \hointr n {n + 1} \in \BB$
- $\R = \ds \bigcup_{n \mathop \in \Z} \hointr n {n + 1} \subseteq \bigcup \BB$
Hence $\R = \bigcup \BB$ and condition $(1)$ is fulfilled.
Now take $ B_1, B_2 \in \BB$ where:
- $B_1 = \hointr {a_1} {b_1}$
- $B_2 = \hointr {a_2} {b_2}$
Let $B_3$ be constructed as:
- $B_3 := \hointr {\max \set {a_1, a_2} } {\min \set {b_1, b_2} } \in \BB$
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous): Part $\text {II}$: Counterexamples: $51$. Right Half-Open Interval Topology