On Tue, 19 Mar 2013 21:25:44 -0700 (PDT), Butch Malahide <email@example.com> wrote:
>On Mar 19, 8:49 pm, William Elliot <ma...@panix.com> wrote: >> On Tue, 19 Mar 2013, David Hartley wrote: >> > <ma...@panix.com> writes >> > > On Mon, 18 Mar 2013, David Hartley wrote: >> >> > > > > Perhaps you could illustrate with the five different one to four point >> > > > > point compactifications of two open end line segements. >> >> > > > (There are seven.) >> >> > > Five without homeomorphisms. >> >> > || >> > |O >> > O O >> > | >> > 9 >> > 8 >> > O >> >> 1 8 >> 2 0 00 6 >> 3 01 1 >> 4 11 >> >> 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?
The "obvious" approach seems like starting with the number ofpartitions of 2n, which is already hopeless, and then somehow factoring out the ones that give homeomorphic compactifications... hopeless squared. Is there a better way, or are you just making trouble again or what?