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?
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.
The range of such a function isn't guaranteed to to not be a real line again, though, and mapping it into a single point is an definite counterexample.
Or do you mean some kind of ordering of the space of all such functions, maybe?
11 Comments
pdt@lemmy.sdf.org · 1 pts · 3y
it's been more than a decade since i last thought about such things but would IR^n with the dictionary order work
rouven@lemmy.sdf.org · 2 pts · 3y
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. ω
0\times 0. Is there an example that is nowhere locally homeomorphic to IR?CanadaPlus@lemmy.sdf.org · 1 pts · 3y
IR as in irrationals? Would that be complete? Actually, I'm realising property 3 is kind of confusing as written in the wiki.
pdt@lemmy.sdf.org · 1 pts · 3y
IR as in the real numbers in fake blackboard bold :)
CanadaPlus@lemmy.sdf.org · 0 pts · 3y
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.
pdt@lemmy.sdf.org · 1 pts · 3y
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.
CanadaPlus@lemmy.sdf.org · 0 pts · 3y
Hmm. Do you have a construction in mind?
pdt@lemmy.sdf.org · 2 pts · 3y
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.
CanadaPlus@lemmy.sdf.org · 2 pts · 3y
Ah, you're right, why didn't I think of that? Thanks for all the help!
sqw@lemmy.sdf.org · 0 pts · 3y
any continuous function that maps R onto something else, maybe?
CanadaPlus@lemmy.sdf.org · 1 pts · 3y
The range of such a function isn't guaranteed to to not be a real line again, though, and mapping it into a single point is an definite counterexample.
Or do you mean some kind of ordering of the space of all such functions, maybe?