Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S^n$ be the $n$-dimensional unit sphere in $\mathbb R^{n+1}$.

Let $X \subset S^n$ such that for any $x, y \in X$ the angle between $x$ and $y$ is smaller than $180°$.

Then $S^n - X$ does contain a semisphere, i.e., something isometric to $S^n \cap \{ x_{n+1} \leq 0 \}$. Can you give a proof with elementary tools and without directly using the Hahn–Banach separation theorem?

share|improve this question
1  
You want to include some convexity assumption on $X$, I suppose. –  user53153 Jan 25 '13 at 19:01
    
The original statement was blatantly wrong. I have editted the question into what I actually meant. –  shuhalo Jan 27 '13 at 14:49
    
It's still blatantly wrong, though. // Your statement reminds of a striking theorem by F.A. Valentine: if $X$ is a subset of $S^n$ and $f:X\to S^n$ is a $1$-Lipschitz map that is not an isometry, then $f(X)$ is contained in a closed hemisphere. Original source: "A Lipschitz condition preserving extension for a vector function", Amer. J. Math. 67 (1945), 83–93. Modern reference: Brudnyi and Brudnyi, "Methods of geometric analysis in extension and trace problems", Volume 1. –  user53153 Jan 28 '13 at 4:05
add comment

1 Answer 1

up vote 2 down vote accepted

This is wrong: Let $X$ consist of the four vertices of a regular tetrahedron inscribed in $S^2$. Then the distance between any two points of $X$ is $<\pi$; but there is no hemisphere contained in $S^2\setminus X$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.