Sorgenfrey line is not orderable

216 Views Asked by At

How can I prove that the Sorgenfrey line is not a linearly ordered topological space, i.e., that the Sorgenfrey topology on $\mathbb R$ is not the order topology for some linear ordering?

I tried to prove this by using compactness or unconnectness...

1

There are 1 best solutions below

3
On BEST ANSWER

The usual argument is as follows: Lutzer showed in 1969 (in this paper) that an orderable space $X$ is metrisable iff it has a $G_\delta$ diagonal (i.e. the set $\Delta_X = \{(x,x) : x \in X\}$ is a countable intersection of open sets in $X \times X$ in the product topology).

And the Sorgenfrey line has a $G_\delta$ diagonal but is not metrisable, so cannot be orderable.