7
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 07 Aug 2023
7 points (88.9% liked)
math
276 readers
1 users here now
Interesting news and discussion centered around Mathematics
founded 2 years ago
MODERATORS
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!