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

>>>>>>>>>>>>>>(There are seven.)

>>>>>>>>>>>>>Ok, seven non-homeomophic finite Hausdorff

>>>>>>>>>>>>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.

>>>>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.