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On Sun, 30 Apr 2006, Ulysse J. Keller wrote:
> <marsh@xxxxxxxxxxxxxxxxxx> wrote:
>
> >(...) In general I doubt that
> >every topology contains a minimal Hausdorff topology. Namely as
> >you see, there's a problem applying Zorn's lemma. (...)
> >
> What you doubt is indeed false. I find in the book suggested by
> G. A. Edgar (Handbook of the history of general topology, Vol. 2)
> in the chapter on minimal Hausdorff spaces the fact that the usual
> topology on the set of rationals isn't finer than any minimal
> Hausdorff topology (content of result 2.2 on p.678 without
> using the extra expressions defined there) - this is stronger
> than the fact mentioned and proven in this thread that |Q is
> an example of what the OP asked (reduced to considering
> identity map, i.e. its top. isn't finer than any compact H. topology)
>
> This implies that there exists a totally ordered set S of Hausdorff
> topologies on |Q coarser than the usual one, such that the
> intersection of S isn't Hausdorff. May-be even a decreasing sequence ?
>
Certainly a chain of Hausdorff topologies coarser than Q
with non-Hausdorff intersection.
> BTW the same book chapter also contains an example of
> a countable non-compact minimal Hausdorff space. Therefore
> the set |Q itself has a non-compact minimal Hausdorff topology
> which of course is not comparable (for inclusion partial order)
> with the usual one. I haven't seen yet the book "Counterexamples
> in Topology", so I can't say whether this example is essentially
> the same as #100 in that, but it somehow replaces it for me ;-)
>
It gives the order topology to
A = (omega_0 + 1) follow by omega_0 reversed
and adds to AxN two points with blah-blah nhoods.
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