Sorgenfrey Line is Hausdorff

From ProofWiki
Jump to navigation Jump to search


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

Then $T$ is Hausdorff.


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

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

From Real Numbers are Close Packed:

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


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

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.