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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
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> <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:
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> >> > > > > Perhaps you could illustrate with the five different one to four point
> >> > > > > point compactifications of two open end line segements.
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> >> > > > (There are seven.)
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> >> > > Five without homeomorphisms.
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> >> > ||
> >> > |O
> >> > O O
> >> > |
> >> > 9
> >> > 8
> >> > O
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> >> 1 8
> >> 2 0 00 6
> >> 3 01 1
> >> 4 11
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> >> Ok, seven non-homeomophic finite Hausdorff compactications.
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> >How many will there be if you start with n segments instead of 2?
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> Surely there's no simple formula for that?
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> 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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