Sorgenfrey Line is Hausdorff

Theorem

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

Then $T$ is Hausdorff.

Proof

Take $x, y \in \R$ such that $x \ne y$.

Without loss of generality, assume that $x < y$.

From Real Numbers are Densely Ordered:

$\exists t \in \R: x < t < y$

Then:

$\hointr x t \cap \hointr t {y + 1} = \O$

We also have that $x \in \hointr x t$ by definition of half-open interval.

Also, as $t < y < y+1$ it is clear that $y \in \hointr t {y + 1}$.

By definition of the Sorgenfrey line, both are open in $T$.

Thus we have found two disjoint subsets of $\R$ which are open in $T$, such that one contains $x$ and the other contains $y$.

Hence the Sorgenfrey Line is Hausdorff by definition.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.): Part $\text {II}$: Counterexamples: $51$. Right Half-Open Interval Topology: $2$