# If A $\subset S^n$ has spherical diameter < $\pi$, then $S^n - A$ contains a closed semisphere.

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?

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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

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 $S^2\setminus X$.