TheoremBase

Sketch: the intersection of the nonempty family of convex supersets of S is convex, contains S and lies in each of them; two such sets contain each other.

Proof

No choice is used; every object below is defined by a formula. Let C\mathcal{C} be the set of all C∈P(Rn)C\in\mathcal{P}(\mathbb{R}^{n}) that are convex and contain SS.

Let

H=⋂C,H=\bigcap\mathcal{C},

which is exactly the set displayed in (ii). Then HH is convex by Operations Preserving Convexity in Rn\mathbb{R}^n: Intersections, Translates, Scalar Multiples, Sums, Affine Images and Preimages, Closure and Interior (ii); it contains SS, because every member of C\mathcal{C} does; and it is contained in every convex C⊇SC\supseteq S, because such a CC is a member of C\mathcal{C}. So HH has the property required in (i).

For uniqueness, suppose that HH and H′H' both have this property. Each is a convex superset of SS, so the property of HH gives H⊆H′H\subseteq H' and the property of H′H' gives H′⊆HH'\subseteq H; hence H=H′H=H'. This proves (i), and since the unique such set is the intersection HH constructed above, it also proves (ii).

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