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

In summary, the conversation discusses proving that a non-empty, convex, open subset of R^2 is homeomorphic to R^2. The suggested approach is to first prove that the intersection of a line in R^2 with the subset is homeomorphic to an open interval in R^1, using radial projections. The conversation also mentions using the completeness axiom for R and some basic definitions of homeomorphism and open/closed sets. A suggested bijection is provided for mapping the projected "lines" in the subset to all of R^2, but the continuity of this function is still being worked on. The conversation concludes with a question about whether there is a better approach to the proof.
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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.
 
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any ideas?
 

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

1. What is a convex subset?

A convex subset is a subset of a vector space that contains all the points on a line segment connecting any two points within the subset.

2. How is a convex subset different from a non-convex subset?

A convex subset is different from a non-convex subset in that a convex subset is always a connected region, whereas a non-convex subset may contain disconnected regions.

3. Can a convex subset be empty?

No, a convex subset cannot be empty. It must contain at least one point to satisfy the definition of being a subset of a vector space.

4. What are some examples of convex subsets?

Examples of convex subsets include circles, triangles, and squares in a two-dimensional space, and spheres, cones, and cubes in a three-dimensional space.

5. How is convexity useful in data analysis?

Convexity is useful in data analysis because it allows for simplification and optimization of complex data sets. Many algorithms and techniques in machine learning and optimization rely on the convexity of data to find global solutions efficiently.

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