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Re: Stone Cech
Posted:
Mar 22, 2013 12:38 AM
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On Mar 21, 11:08 pm, Butch Malahide <fred.gal...@gmail.com> wrote:
> On Mar 21, 2:03 pm, quasi <qu...@null.set> wrote:
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> > Butch Malahide wrote:
> > >quasi wrote:
> > >> quasi wrote:
> > >> >quasi wrote:
> > >> >>quasi wrote:
> > >> >>>Butch Malahide wrote:
> > >> >>>>David C. Ullrich wrote:
> > >> >>>>>Butch Malahide wrote
> > >> >>>>> >William Elliot wrote:
> > >> >>>>> >>David Hartley wrote:
> > >> >>>>> >> >William Elliot 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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> > >> >>>>> >>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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> > >> >>>>> ...
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> > >> >>>> ...
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> > >> >>>>I wasn't necessarily expecting a *complete* answer, such
> > >> >>>>as an explicit generating function. Maybe someone could
> > >> >>>>give a partial answer, such as an asymptotic formula, or
> > >> >>>>nontrivial upper and lower bounds, or a reference to a
> > >> >>>>table of small values, or the ID number in the
> > >> >>>>Encyclopedia of Integer Sequences, or just the value for
> > >> >>>>n = 3. (I got 21 from a hurried hand count.)
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> > >> >>>For n = 3, my hand count yields 19 distinct
> > >> >>>compactifications, up to homeomorphism.
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> > >> >>>Perhaps I missed some cases.
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> > >> >>I found 1 more case.
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> > >> >>My count is now 20.
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> > >> >I found still 1 more case.
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> > >> >So 21 it is!
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> > >> >But after that, there are no more -- I'm certain.
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> > >> Oops -- the last one I found was bogus.
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> > >> So my count is back to 20.
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> > >Hmm. I counted them again, and I still get 21.
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> > >4 3-component spaces: OOO, OO|, O||, |||.
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> > >7 2-component spaces: OO, O|, O6, O8, ||, |6, |8.
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> > >10 connected spaces: O, |, 6, 8, Y, theta, dumbbell, and the
> > >spaces btained by taking a Y and gluing one, two, or all
> > >three of the endpoints to the central node.
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> > Thanks.
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> > It appears I missed the plain "Y", but other than that,
> > everything matches.
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> > So yes, 21 distinct types.
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> For n = 4 I get 56 types. If I counted right (very iffy), it may or
Found two more. Never mind!
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