• CanadaPlus@lemmy.sdf.orgOP
      link
      fedilink
      arrow-up
      1
      arrow-down
      1
      ·
      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
        link
        fedilink
        arrow-up
        1
        ·
        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
          link
          fedilink
          arrow-up
          1
          arrow-down
          1
          ·
          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
            link
            fedilink
            arrow-up
            2
            ·
            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.