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Topic: Stone Cech
Replies: 49   Last Post: Mar 28, 2013 12:15 PM

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Stone Cech
Posted: Mar 22, 2013 11:52 AM

On Fri, 22 Mar 2013 08:43:53 -0600, David C. Ullrich
<ullrich@math.okstate.edu> wrote:

>On Fri, 22 Mar 2013 05:34:37 -0500, quasi <quasi@null.set> wrote:
>

>>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.

>>>>>>>>>>>>>>

>>>>>>>>>>>>>
>>>>>>>>>>>>>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).

>
>Hey, you guys are making progress!
>Now all we need is an f such that
>f(2) = 7, f(3) = 21 or whatever it was,
>and f(4) = 61. I doubt there's more than
>one such f...
>

>>
>>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.

>
>My prediiction is that anything remotely resembling brute
>force will take forever to run, even in a compiled language.
>Just enumerating all the possible equivalence relations
>on 2n points is going to take a long time; now a
>brute force check whether _one_ pair of equivalence
>relations defines homeomorphic spaces will also take
>a long time, and there will be N(N-1)/2 such checks,
>where N is huge even for small n.

I just realized I have no idea whatever how to write
a correct algorithm here, even without worrying

An equivalence relation on the 2n endpoints
determines a compactification. It also determines
a graph, where the vertices are the equivalence
classes of endpoints and the edges are the
original segments. Earlier, thinking that I was
thinking about deciding when two of these
compactifications are homeomorphic, I was
actually thinking about how to determine
whether two of the graphs are isomorphic.
That's a strictly weaker condition... I have
no idea how I'd check whether two equivalences
gave homeomorphic compactifications.

>
>But don't let me stop you. I'd start in Python, which
>would be much too slow - the point being that Python
>is so easy to _write_ that it would be the quickest
>way to get the algorithm straight. I'd test the
>Python code for n = 2 and 3 maybe; for n = 3
>it would take forever. Once I had a correct
>algorithm I'd rewrite it in a compiled language.
>take a trip somewhere while waiting for the
>

>>
>>quasi

Date Subject Author
3/14/13 William Elliot
3/14/13 fom
3/15/13 fom
3/16/13 William Elliot
3/15/13 David C. Ullrich
3/17/13 William Elliot
3/17/13 David C. Ullrich
3/17/13 fom
3/18/13 David C. Ullrich
3/18/13 fom
3/18/13 David Hartley
3/19/13 William Elliot
3/19/13 David Hartley
3/19/13 William Elliot
3/20/13 Butch Malahide
3/20/13 David C. Ullrich
3/20/13 Butch Malahide
3/20/13 Butch Malahide
3/21/13 quasi
3/21/13 quasi
3/21/13 quasi
3/21/13 quasi
3/21/13 Butch Malahide
3/21/13 quasi
3/22/13 Butch Malahide
3/22/13 Butch Malahide
3/22/13 Butch Malahide
3/22/13 quasi
3/22/13 David C. Ullrich
3/22/13 David C. Ullrich
3/22/13 Butch Malahide
3/23/13 Butch Malahide
3/23/13 David C. Ullrich
3/23/13 David C. Ullrich
3/23/13 Frederick Williams
3/23/13 David C. Ullrich
3/23/13 Frederick Williams
3/22/13 Butch Malahide
3/23/13 David C. Ullrich
3/22/13 Butch Malahide
3/23/13 quasi
3/23/13 Butch Malahide
3/23/13 Butch Malahide
3/24/13 quasi
3/24/13 Frederick Williams
3/24/13 quasi
3/25/13 Frederick Williams
3/28/13 Frederick Williams
3/25/13 quasi
3/19/13 David C. Ullrich