Skip to main content
9 events
when toggle format what by license comment
Sep 13, 2015 at 20:02 comment added Taras Banakh It suffices to find a paracompact Hausdorff space $X$ whose $k$-coreflexion $k(X)$ is not regular. For sequential coreflexions such a compact Hausdorff space was constructed by Franklin and Rajagopalan in 1970 (see repository.cmu.edu/cgi/…).
May 29, 2011 at 14:22 history bounty ended David Carchedi
May 23, 2011 at 1:31 comment added David Carchedi Thanks, I'm aware of this result, but I'm not sure how to use it. In fact, this is and if and only if, i.e. it characterizes compactly generated spaces. Moreover, for compactly generated Hausdorff spaces, they are the obvious quotient of the disjoint union of all their compact subsets, and if $X$ is not compactly generated, this quotient is $k\left(X\right).$ This means when $X$ is paracompact Hausdorff, $k\left(X\right)$ is a quotient of a space which is is both locally compact and paracompact Hausdorff. I'm not sure where to go from here.
May 22, 2011 at 19:13 comment added David White Every compactly generated space is a quotient of a locally compact Hausdorff space. That may help, but not in the naive way. You definitely can't conclude $k(X)$ is paracompact just because it's a quotient of a paracompact space.
May 22, 2011 at 14:01 history bounty started David Carchedi
May 19, 2011 at 5:40 history edited David Carchedi CC BY-SA 3.0
deleted 9 characters in body
May 18, 2011 at 23:47 comment added David Carchedi Yes, it is indeed Hausdorff; I know that $k$ is a functor $$k:Haus \to CGH,$$ the question is whether or not it is paracompact.
May 18, 2011 at 22:33 comment added wildildildlife It seems that it's certainly Hausdorff, as the topology of $k(X)$ is finer (if $U$ is open in $X$ then $U\cap K$ is open in $K$ for all compacta $K$, by definition of the subspace topology.) So the two separating sets that worked for $X$ still work for $k(X)$.
May 18, 2011 at 15:07 history asked David Carchedi CC BY-SA 3.0