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Re: Stone Cech
Posted:
Mar 20, 2013 3:20 PM
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On Mar 20, 8:53 am, David C. Ullrich <ullr...@math.okstate.edu> wrote: > On Tue, 19 Mar 2013 21:25:44 -0700 (PDT), Butch Malahide > > > > > > <fred.gal...@gmail.com> wrote: > >On Mar 19, 8:49 pm, William Elliot <ma...@panix.com> wrote: > >> On Tue, 19 Mar 2013, David Hartley wrote: > >> > <ma...@panix.com> writes > >> > > On Mon, 18 Mar 2013, David Hartley wrote: > > >> > > > > Perhaps you could illustrate with the five different one to four point > >> > > > > point compactifications of two open end line segements. > > >> > > > (There are seven.) > > >> > > Five without homeomorphisms. > > >> > || > >> > |O > >> > O O > >> > | > >> > 9 > >> > 8 > >> > O > > >> 1 8 > >> 2 0 00 6 > >> 3 01 1 > >> 4 11 > > >> Ok, seven non-homeomophic finite Hausdorff compactications. > > >How many will there be if you start with n segments instead of 2? > > Surely there's no simple formula for that? > > The "obvious" approach seems like starting with the number > ofpartitions of 2n, which is already hopeless, and then > somehow factoring out the ones that give homeomorphic > compactifications... hopeless squared. Is there a better > way, or are you just making trouble again or what?- Hide quoted text -
Just making trouble again.
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