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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

1 Answer 1

up vote 2 down vote accepted

Separability implies second countability in every metrisable space. Since the Sorgenfrey line is separable but not second countable, it is not metrisable.

By the Birkhoff-Kakutani theorem a topological group is metrisable if and only if it is both Hausdorff and first countable. Since the Sorgenfrey line is Hausdorff and first countable but not metrisable, it is not a topological group. Or to be more precise: there is no group operation on the Sorgenfrey line so that the multiplication and inverse operations are continuous.

However, it's not hard to see that the Sorgenfrey line with the usual addition as group operation is a paratopological group.

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