```Date: Mar 22, 2013 5:44 AM
Author: quasi
Subject: Re: Stone Cech

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 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?>>>>>>>>>>>>>>>>>>>>>> ...>>>>>>>>>>>>>>>>>>>> ...>>>>>>>>>>>>>>>>>>>>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 3-component spaces: OOO, OO|, O||, |||.>>>>>>>>>>7 2-component 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 countsvia 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
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