• 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.