Timeline for Must a linearly ordered, separable space be metrizable?
Current License: CC BY-SA 2.5
8 events
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| May 25, 2018 at 10:12 | comment | added | Henno Brandsma | As I also commented on the question: we need $X$ to have no jumps (points $x < y$ such that for no $z$, $x < y < z$, so places where order-denseness fails), or at most countably many.. No jumps happens when $X$ is connected, e.g. | |
| Mar 1, 2011 at 11:05 | history | edited | Jim Conant | CC BY-SA 2.5 |
added 187 characters in body; added 4 characters in body
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| Mar 1, 2011 at 11:04 | comment | added | Jim Conant | I think I was using a different definition of "dense." I was taking dense to mean that between any two elements there exists an element from my dense subset, but I guess it should mean that the closure is the whole space in this context. (Which is how you define separable.) This is one of those situations where overlapping terminology causes confusion. :) | |
| Mar 1, 2011 at 9:37 | comment | added | mathahada | I mean Apollo's example, not mine:) | |
| Mar 1, 2011 at 9:37 | comment | added | mathahada | In my example the space is not second countable. For every real number $x$ the open interval $U=(x-1,x')=(x-1,x]$ has maximal element $x$. A basis element contained in $U$ and containing $x$ must also have maximal element $x$ so there is a surjection from the set of all basis element to the reals | |
| Mar 1, 2011 at 1:16 | comment | added | Jim Conant | Indeed. And of course, that's easily arranged! | |
| Mar 1, 2011 at 0:51 | comment | added | Joel David Hamkins | You should also ensure the maximal and minimal elements of the order, if any exist, are in your countable dense set in order for the corresponding set of intervals to be a basis. | |
| Mar 1, 2011 at 0:47 | history | answered | Jim Conant | CC BY-SA 2.5 |