By Weil W.

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Show that ext A = ∅, if and only if A does not contain any line. 2. Let K ⊂ Rn be compact and convex. (a) If n = 2, show that ext K is closed. (b) If n ≥ 3, show by an example that ext K need not be closed. 3. Let A ⊂ Rn be closed and convex. A subset M ⊂ A is called extreme (in A), if M is convex and if x, y ∈ A, (x, y) ∩ M = ∅ implies [x, y] ⊂ M . Show that: (a) Extreme sets M are closed. (b) Each support set of A is extreme. (c) If M, N ⊂ A are extreme, then M ∩ N is extreme. (d) If M is extreme in A and N ⊂ M is extreme in M , then N is extreme in A.

J−1) of K1 (j−1) , K2 , . . with (j) (j) Wj (K1 ) = Wj (K2 ) = · · · , for all j ∈ N (j ≥ 2). Since √ min d(x, y) ≤ (j) y∈Kl n 2j , (j) for all x ∈ Kk , we have √ (j) (j) d(Kk , Kl ) ≤ n 2j , for all k, l ∈ N, and all j. By the subsequence property we deduce √ n (j) (i) d(Kk , Kl ) ≤ i , for all k, l ∈ N, and all j ≥ i. 2 (k) In particular, if we choose the ’diagonal sequence’ Kk := Kk , k = 1, 2, . . , then √ n d(Kk , Kl ) ≤ l , for all k ≥ l. 2 Hence (Kk )k∈N is a Cauchy sequence in M. Let ˜ k := cl conv (Kk ∪ Kk+1 ∪ · · · ) K and ∞ ˜ k.

Proof. Suppose K = conv A and x ∈ ext K. 4, there is a representation x = α1 x1 + · · · + αk xk with k ∈ N, xi ∈ A, αi > 0, and k ≥ 2, αi = 1. In case k = 1, we have x = x1 ∈ A. If x = α1 x1 + (1 − α1 ) and α2 x2 + · · · + αk xk α2 + · · · + αk α2 x2 + · · · + αk xk ∈ K. α2 + · · · + αk Since x is extreme, we obtain x = x1 ∈ A. Thus, in both cases we have x ∈ A, therefore ext K ⊂ A. 34 CHAPTER 1. CONVEX SETS In the other direction, we need only show that K = conv ext K. We prove this by induction on n.