this post was submitted on 07 Aug 2023
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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!