# Does the Sorgenfrey Line have a group operation compatible with its order topology?

The title is the question, but let me explain. Let $\mathbb{L}$ denote the Sorgenfrey line. I and a friend were trying to develop some of the properties of the sorgenfrey line. (if it's metrizable, paracompact, or whatevs.) And we've stumbled upon the following problem:

Can one define a metric and a group operation in $\mathbb{L}$ that yields the usual order-induced topology and that makes it a topological group? What about semitopological group or something weaker?

So far we've introduced the metric as follows:

$\mathbb{L}$ = $(0,1)$ x $\mathbb{R}$ and $x = (t,r), \ y = (t',r')$ $\ \in$ $\mathbb{L}$
$D(x,y) = 1$ if $t \neq t'$

$D(x,y) =$ $\frac{\parallel x - y \parallel}{1 + \parallel x - y \parallel}$ if $t = t'$

We were trying to find countinous group operations unsuccessfully.

Many thanks.

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The Sorgenfrey line is not metrizable, since it's separable but not second-countable. –  Qiaochu Yuan Oct 14 '11 at 1:28
I see. Still any continuous group operations? –  Henrique Tyrrell Oct 14 '11 at 1:34