Jason (jcreed) wrote,
Jason
jcreed

-

Mmmmm. Polyhedra.
Drew Olbrich (http://www.traipse.com/) has this hack
that figures out what a nonagon should look like.
I decided to implement this on a whim.
It 'finds' the tetrahedron, cube, dodecahedron, but finds
a d10-like shape for n=8, a triangular prism for n=5,
a pentagonal prism for n=7. But it finds a non-d10-like shape
for n=10 (square, two rows of 4 pentagons, square) finds
*something* for n=11, gets the same result for n=9 as
Olbrich, but gets n=20 horribly wrong. Maybe I need to
tweak parameters to get it to converge better.
Generally nifty, though.

Hm, there's an easy proof that in any reasonable (i.e. T1)
space, if there exists a finitely additive integral,
then there exists an additive one. For suppose we have C
and c \in C. Then IC = I(C u x) = IC u Ic hence
Ix \subseteq IC. So UIc \subseteq IC, consq.
DUIx \subseteq DIC = C. But obviously
DUIx \supseteq UDIc = Uc = C.
So DUIc = C, and we can define I*C = UIc,
and I* is an additive integral.

Another property that might be worth investigating:
an integral I is *decomposable* if, for every closed
C and c \in C, there exists a subset S \subseteq IC
such that DS = {c}. An additive integral is clearly
decomposable, but I'm not sure of the relationship
between finitely additivity and decomposibility.
Subscribe
  • Post a new comment

    Error

    Anonymous comments are disabled in this journal

    default userpic

    Your reply will be screened

    Your IP address will be recorded 

  • 0 comments