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I was reading about the convex set and its various properties. I came up with a theorem but don't know how to prove it.

Theorem: A convex set has empty interior if and only if there is a hyperplane containing the set.

Someone please give an idea how to prove it.

Greg Martin
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tourism
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    Is your convex set $S$ in $\Bbb R^n$? If so, here's an outline. Pick distinct points $p_0,p_1\in S$. Since $S$ isn't contained in the line through $p_0,p_1$, pick a point $p_2\in S$ that's not in that line. Since $S$ isn't contained in the plane through $p_0,p_1,p_2$, pick a point $p_3\in S$ that's not in that plane. And so on until you have $p_0,\dots,p_n$. Then $S$ contains the simplex with those points as vertices, which has empty interior. – Greg Martin Nov 05 '15 at 06:27
  • could you please explain the last line more clearly.... What is the mean that S contain simplex of these points which has empty interior – tourism Nov 05 '15 at 06:38
  • consider the affine hull of that convex set and the maximal hyperplane that contains it – Neutral Element Nov 05 '15 at 18:20

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