Re: Homeomorphisms and Roman alphabet
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Mon, 20 Aug 2007 18:50:47 -0400
Gvnaena Pura wrote:
The real problem comes when I have to decide whether two letters are
homeomorphic. I exploit the fact that the continuous image of a
connected is connected. For instance, let's assume f:'A'->'B' is a
homeomorphism. Any restriction of f is still a homeomorphism, but I can
disconnect A by removing a point, while I need to remove at least two
points from B to disconnect it. Then f cannot exist.
That was easy. But are A and P homeomorphic? No, wait, maybe I can say
something like this: I can remove two points x,y from 'A' such that x
and y have two non intersecting open neighborhoods and, by
(bi)continuity, f(x) and f(y) should be apart, but there are no such
points in 'P' whose removal would keep 'P' connected.
Then, I can bend a letter into another but without altering the
"proximity" of its points. For instance I can't open an 'O' because I
would alter the "proximity" of two points, and, dually, I can't link two
point that were far apart.
I think that depending on how you view those letters. If you consider
the letter to be made of elastic solid (2 dimensional manifold), then
P and O are homeomorrphic (so you can "shink the leg into the body of
P"). If you consider them to be made of elastic stings tighted
together, then P and O are not homeomophic.
P and O are not homeomorphic. You can remove a single point from P and leave two separated sets. The same is not true of O.
--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan
.
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