Ah. IR^n is separable, though. By Cantor’s mentioned theorem (which is irritatingly not cited) it must be order-isomorphic to IR if it meets the 3 conditions and is separable.
There has to be a simple example, though, right? Suslin added the fourth condition. I thought of the long line, but that seemed tricky for a couple of reasons.
I didn’t mean IR^n with its usual topology. I meant IR^n with the order topology for the dictionary order. IIANM you can construct an uncountable set of pairwise disjoint open intervals in this topology so it can’t have a countable dense subset. But as I said it’s been years since I touched a topology book.
I think you could just take an open interval in the order topology and then create a collection by turning the first dimension into a parameter. IIANM for each value of the parameter you’d get an open set, they’d be pairwise disjoint, and there’d be uncountably many of them.
Yeah that should work I think. Maybe more interesting would be whether there exists an example which is not locally homeomorphic to IR (I think you’re example still fulfills that). But I believe that is solved by using something like the long line and looking at e.g. ω0\times 0. Is there an example that is nowhere locally homeomorphic to IR?
it’s been more than a decade since i last thought about such things but would IR^n with the dictionary order work
IR as in irrationals? Would that be complete? Actually, I’m realising property 3 is kind of confusing as written in the wiki.
IR as in the real numbers in fake blackboard bold :)
Ah. IR^n is separable, though. By Cantor’s mentioned theorem (which is irritatingly not cited) it must be order-isomorphic to IR if it meets the 3 conditions and is separable.
There has to be a simple example, though, right? Suslin added the fourth condition. I thought of the long line, but that seemed tricky for a couple of reasons.
I didn’t mean IR^n with its usual topology. I meant IR^n with the order topology for the dictionary order. IIANM you can construct an uncountable set of pairwise disjoint open intervals in this topology so it can’t have a countable dense subset. But as I said it’s been years since I touched a topology book.
Hmm. Do you have a construction in mind?
I think you could just take an open interval in the order topology and then create a collection by turning the first dimension into a parameter. IIANM for each value of the parameter you’d get an open set, they’d be pairwise disjoint, and there’d be uncountably many of them.
Ah, you’re right, why didn’t I think of that? Thanks for all the help!
Yeah that should work I think. Maybe more interesting would be whether there exists an example which is not locally homeomorphic to IR (I think you’re example still fulfills that). But I believe that is solved by using something like the long line and looking at e.g. ω0\times 0. Is there an example that is nowhere locally homeomorphic to IR?