Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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 23, 2013 1:43 PM

On Fri, 22 Mar 2013 23:52:17 -0700 (PDT), Butch Malahide
<fred.galvin@gmail.com> wrote:

>On Mar 22, 10:27 pm, Butch Malahide <fred.gal...@gmail.com> wrote:
>> On Mar 22, 10:52 am, David C. Ullrich <ullr...@math.okstate.edu>
>> wrote:
>>
>>
>>
>>
>>

>> > 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.

>>
>> You mean, non-isomorphic is a weaker condition than non-homeomorphic.
>>
>> The remedy for that is to stick to graphs with no vertices of degree
>> 2. (OK, you have to make an exception for the graph consisting of a
>> single vertex of degree 2, corresponding to the one-point
>> compactification of R.)

>
>Oops, I was thinking of connected graphs. In general, you want to work
>with graphs with the property that any vertex of degree 2 is the only
>vertex in its component.

I think that's right, maybe.

>OK, so you have graphs G and H that you want
>to test for homeomorphism. Remove vertices of degree 2 (and unify the
>two edges that were incident with that vertex) until you get graphs G'
>and H', homeomorphic to G and H respectively (G and H are subdivisions
>of G' and H') and having the aforesaid property; and then test G' and
>H' for isomorphism. So homeomorphism testing (of graphs) is no harder
>than isomorphism testing.

Right. Whether you got every detail right here (which is not so say
you didn't, just that I haven't thought about it carefully), it's
clear that this or something like this does give a way to enumerate
those compactifications by testing graphs for isomorphism. Thanks.

Still gonna be incredibly slow...

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