What is the Homeomorphism Proof for Non-Empty Convex Open Subsets of R^2?

SpaceTag
Messages
12
Reaction score
0

Homework Statement



Let U be a non-empty, convex, open subset of R^2. Prove that U is homeomorphic to R^2.
Hint: First prove that the intersection of a line in R^2 with U (if non-empty) is homeomorphic to an open interval in R^1. Then use radial projections.

Homework Equations



We just have the basic definition of homeomorphism and some standard results about open/closed sets to work with. And of course the completeness axiom for R.

The Attempt at a Solution



Ok well, I have proven that the intersection of a line in R^2 with U (if non-empty) is homeomorphic to an open interval of R^1. I see the idea of the proof is to translate U homeomorphically so that the origin is at an interior point of U, and then radiate lines outward from the origin in all directions (think: polar coordinates). Since U is convex, the lines from the origin to the "boundary" of U are fully contained in U and comprise all of U; Since U is open, such lines to the "boundary" can be mapped to the FULL lines extending forever outward in the corresponding direction in R^2.

So I can make a bijection from such projected "lines" in U to all of R^2; in fact for the case where U is bounded I've made an explicit bijection which I suspect is both-ways continuous but I'm having trouble proving it.

The notation I'm using is

R(theta) = supremum of the distances of points of the "line" in U which points in the direction of theta. (ie, R(theta) is the distance from the origin to the "boundary" of U along the direction of theta)

I believe the following bijection sends a "line" in U in the direction of theta to the full line in R^2 in the direction of theta:

f(r) = r/ [R(Theta)- r]

where r is interpreted as distances along the direction of theta in U.

If I could prove that R(theta) is a continuous function of theta then I think I would know how to proceed. It seems obviously true because a jump discontinuity in the boundary of a convex space would seem to produce a contradiction to convexity. But that's just for the bounded case!

In any case, I am stumped, and I'm thinking there must be a better way to do this. Is there a better approach? Thanks guys.
 
Last edited:
Physics news on Phys.org
any ideas?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top