
Re: Stone Cech
Posted:
Mar 22, 2013 4:15 AM


On Mar 21, 11:38 pm, Butch Malahide <fred.gal...@gmail.com> wrote: > On Mar 21, 11:08 pm, Butch Malahide <fred.gal...@gmail.com> wrote: > > > > > > > On Mar 21, 2:03 pm, quasi <qu...@null.set> wrote: > > > > 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: > > > > >> >>>>> >> >>Perhaps you could illustrate with the five > > > >> >>>>> >> >>different one to four point point > > > >> >>>>> >> >>compactifications of two open end line > > > >> >>>>> >> >>segements. > > > > >> >>>>> >> >(There are seven.) > > > > >> >>>>> >>Ok, seven nonhomeomophic 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? > > > > >> >>>>> ... > > > > >> >>>> ... > > > > >> >>>>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.) > > > > >> >>>For n = 3, my hand count yields 19 distinct > > > >> >>>compactifications, up to homeomorphism. > > > > >> >>>Perhaps I missed some cases. > > > > >> >>I found 1 more case. > > > > >> >>My count is now 20. > > > > >> >I found still 1 more case. > > > > >> >So 21 it is! > > > > >> >But after that, there are no more  I'm certain. > > > > >> Oops  the last one I found was bogus. > > > > >> So my count is back to 20. > > > > >Hmm. I counted them again, and I still get 21. > > > > >4 3component spaces: OOO, OO, O, . > > > > >7 2component spaces: OO, O, O6, O8, , 6, 8. > > > > >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. > > > > Thanks. > > > > It appears I missed the plain "Y", but other than that, > > > everything matches. > > > > So yes, 21 distinct types. > > > For n = 4 I get 56 types. If I counted right (very iffy), it may or > > Found two more. Never mind!
And now I get 61. The hell with it.

