quasi
Posts:
11,867
Registered:
7/15/05


Re: Stone Cech
Posted:
Mar 22, 2013 5:44 AM


Butch Malahide wrote: >Butch Malahide wrote: >>Butch Malahide wrote: >>>quasi 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 obtained 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), >> >> Found two more. Never mind! > >And now I get 61. The hell with it.
Ullrich predicted it (hopeless squared).
For small n, say n < 10, it might be feasible to get the counts via a computer program, but my sense is that the development of such a program would be fairly challenging. If I get a chance, I may give it a try.
quasi

