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 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