• pdt@lemmy.sdf.org
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    1 year ago

    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
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      1 year ago

      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?

    • CanadaPlus@lemmy.sdf.orgOP
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      1 year ago

      IR as in irrationals? Would that be complete? Actually, I’m realising property 3 is kind of confusing as written in the wiki.

        • CanadaPlus@lemmy.sdf.orgOP
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          1 year ago

          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
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            1 year ago

            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.orgOP
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              1 year ago

              IIANM you can construct an uncountable set of pairwise disjoint open intervals in this topology

              Hmm. Do you have a construction in mind?

              • pdt@lemmy.sdf.org
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                1 year ago

                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.