Arthur Fisher, would you please help me with a simple question? Why is the sorgenfrey line topology not linearly ordered? I'm not sure how I would show this.

@Jake1234 Just to clarify your question, you are asking why sorgenfrey-line does not have order topology derived from some linear order on $\mathbb R$?

Simply asking Google returns a few reasonable results: google.com/search?q=sorgenfrey+line+not+ordered

Henno Brandsma writes here at.yorku.ca/cgi-bin/…

A space X that is orderable and such that has a G_delta diagonal is metrisable.

(due to Lutzer: A metrization theorem for linearly orderable spaces, Proc Amer. math. Soc., 22, 1969, 557-558 (so short proof..))

And the Sorgenfrey line is not metrisable (no countable base but separable) and has a G_delta diagonal.

So it cannot be orderable.

Two remarks: 1) If you wanted Arthur Fischer to nice your message, it would have been better to ping him.

And I think that the question you asked @Jake1234 is interesting enough to be posted on the main.

I have searched a bit, I did not find this question already asked on the main.

Thank you, sadly I actually found this approach by Henno Brandsma in Engelking, after looking through all the pages that contained the word "sorgenfrey".

@MartinSleziak I might make a question out of it next time I'm thinking this through problem again. My Prof. mentioned this being true, and so I thought it's something I should be able to figure out from what we've taken so far this semester, but since we haven't looked at metrisable spaces yet, I think I'll let it go for the time being. Also thanks for the link to that FAQ, I will use it next time.

@Jake1234 Yes, it's there: Problem 5.2.22(l) (Lutzer [1969]) Deduce from (k) that there is no linear order on the Sorgenfrey line which induces the topology of that space.

Thanks for mentioning Engelking - it is good that we also have some reference for this.