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Some Basic Topological and Algebraic Properties of Convex Sets in What

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Some basic topological and algebraic properties of convex sets In what follows, C is a convex set in a vector space X. Topological interior. Let us start with the following easy observation. Observation 0.1. Let C be a convex set in a normed space X. If C contains an 0 open ball B (z0; r), then 0¡ ¢ B (1 ¡ t)z0 + ty; (1 ¡ t)r ½ C whenever y 2 C; t 2 [0; 1). 0 Proof. Fix y 2 C, t 2 [0; 1). Since C is convex, it contains the set (1 ¡ t)B (z0; r) + ty = B0((1 ¡ t)x + ty; (1 ¡ t)r). ¤ Corollary 0.2. Let C be a convex set in a normed space X. If x 2 int(C) and y 2 C, then [x; y) ½ int(C). In particular, int(C) is a convex set. Theorem 0.3. Let C be a convex set in a normed space X. (a) If either int(C) 6= ; or C is ¯nite-dimensional, then int(C) = int(C). (b) If int(C) 6= ;, then C = int(C). Proof. First part of (a). Suppose that int(C) 6= ;. Thus C contains an open ball 0 B (z0; r). One inclusion is obvious. Let x 2 int(C). There exists % > 0 such 0 0 that B (x; %) ½ C. If x = z0, we are done. If x 6= z0, ¯x y 2 B (x; %) such that x 2 (z0; y), that is, there exists t 2 (0; 1) with x = (1 ¡ t)z0 + ty. Since y 2 C, there 0 0 0 0 exists y 2 C be such that ky ¡y k < r. For x = (1¡t)z0 +ty , Observation 0.1 says that B0(x0; (1¡t)r) ½ C. Then x 2 int(C), since kx¡x0k = (1¡t)ky¡y0k < (1¡t)r. Second part of (a). Let C be ¯nite-dimensional. If a®(C) = X, then int(C) 6= ; by the \relative interior theorem"; hence we can apply the ¯rst part of (a). If a®(C) 6= X, the a±ne hull of C is closed (since it is a translate of a ¯nite-dimensional linear set) and has no interior points. Since int(C) ½ int(C) ½ int(a®(C)) = int(a®(C)) = ;, we have int(C) = int(C) = ;. (b) Obviously, C ⊃ int(C). If x 2 C, ¯x z0 2 int(C). Arbitrarily near to x we can ¯nd some y 2 C; but since [z0; y) ½ int(C) (Corollary 0.2), arbitrarily near to x we can ¯nd points of int(C). Thus x 2 int(C). ¤ Algebraic interior. Let C be a convex set in a vector space X, x 2 C. We say that x is an algebraically interior point of C if for every v 2 X there exists t > 0 such that x + tv 2 C. In this case, we shall write x 2 a-int(C). (For a nonconvex set E ½ X, an algebraically interior point is a point x 2 E such that for each v 2 X there exists t > 0 such that [x; x + tv] ½ E.) In other words, x 2 a-int(C) if and only if x 2 intL(C \ L) for each line L ½ X passing through x. The following proposition is an analogue of Corollary 0.2. Proposition 0.4. Let C be a convex set in a vector space X. If x 2 a-int(C) and y 2 C, then [x; y) 2 a-int(C). In particular, the set a-int(C) is convex. 1 2 Proof. Let 0 < ¸ < 1, z = (1 ¡ ¸)x + ¸y, v 2 X. Take t > 0 such that x + tv 2 C. Then z +(1¡¸)tv 2 C since z +(1¡¸)tv = (1¡¸)(x+tv)+¸y. Now, the assertion easily follows. ¤ If X is a normed space, then obviously int(C) ½ a-int(C). The following theo- rem states some su±cient conditions for equality of the two sets. Before stating it we recall the following fundamental theorem of R.L. Baire, which we state in the following form. Theorem 0.5 (Baire Category Theorem). Let T be either a complete metric space or a Hausdor® locally compact topologicalS space. If Fn (n 2 N) are closed sets without interior points, then also the union n2N Fn has no interior points. Exercise 0.6. Use the Baire Category Theorem to prove the following two facts. 1. Every compact countable set in a Hausdor® topological space has at least one isolated point. 2. The algebraic dimension of a Banach space is either ¯nite or uncountable. Theorem 0.7. Let C be a convex set in a normed linear space X. Then int(C) = a-int(C) provided at least one of the following conditions is satis¯ed. (a) int(C) 6= ;; (b) C is ¯nite-dimensional; (c) X is Banach and C is an Fσ-set. Proof. (a) Let x 2 a-int(C). Fix z0 2 int(C). If x 6= z0, there exists y 2 C such that x 2 (z0; y). Then x 2 int(C) by Corollary 0.2. (b) Let x 2 a-int(C). This implies that a®(C) = X and hence X is ¯nite- dimensional. By the \relative interior theorem", int(C) = ri(C) is nonempty. Apply the part (a). S (c) Let x 2 a-int(C). We can suppose that x = 0. Write C = k Fk where each Fk is a closed set. Then [ [ [ X = tC = nC = nFk : t>0 n n;k By the Baire Category Theorem, there exist n; k such that nFk has a nonempty interior. Thus int(Fk) 6= ; and hence int(C) 6= ;. Apply (a). ¤ Directions from a point. Let C be a convex set in a vector space X, x0 2 C. Let us consider two sets of directions: the set c(x0;C) of directions of the half-lines from x0 to the points of C, and the set i(x0;C) of directions of the lines L such that L \ C is a nondegenerate segment containing x0 in its relative interior. More precisely, c(x0;C) = fv 2 V : x0 + tv 2 C for some t > 0g ; i(x0;C) = fv 2 V : x0 § tv 2 C for some t > 0g : Let us state main properties of these two sets. 3 Proposition 0.8. Let X; C; x0 be as above. (a) x0 2 x0 + i(x0;C) ½ x0 + c(x0;C) ½ a®(C). (b) c(x0;C) = cone(C ¡ x0). (c) c(x0;C) is a convex cone,£ and i(¤x0;C) is a linear set. (d) i(x0;C) = c(x0;C) \ ¡c(x0;C) . (e) x0 + i(x0;C) is£ the largest a±ne set¤ A such that x0 2 intA(C \ A). (f) a®(C) = x0 + c(x0;C) ¡ c(x0;C) . Proof. (b) follows directly form the de¯nitions; hence c(x0;C) is a convex cone. Obviously, we have 0 2 i(x0;C) ½ c(x0;C). Since x0 + c(x0;C) is the set of all closed half-lines from x0 to the points of C (or fx0g if there are no such half-lines), it is contained in a®(C); hence (a) holds. The part (d) follows easily from de¯nitions. Since K \ (¡K) is a linear set whenever K is a convex cone, we have proved (c), too. Let us show (e). Clearly, x0 + i(x0;C) is an a±ne set. Let A be an a±ne set such Ã! that x0 2 int(C \ A). For every point x 2 A n x0 the set C \ x0x is a nondegenerate segment containing x0 in its relative interior. Since x0 + i(x0;C) is the union of all possible lines L with this property, we have A ½ x0 + i(x0;C). It remains to prove (f). By translation, we can suppose that x0 = 0. Thus, by (b), we have to show that, if 0 2 C, then span(C) = cone(C) ¡ cone(C). The inclusion P\⊃" is obvious. To show the other one, consider a (¯nite) linear combination x = i ®ici with ®i 6= 0, ci 2 C. Then X X x = ®ici ¡ (¡®i)ci 2 cone(C) ¡ cone(C) : ®i>0 ®i<0 ¤ Algebraic relative interior. Let us recall that a relative interior of a convex set C in a normed space is the set ri(C) = inta®(C)(C). Recall also the fact, proved in the chapter on ¯nite-dimensional spaces, that every ¯nite-dimensional convex set has a nonempty relative interior. The algebraic relative interior of C is de¯ned analogously in the following de¯nition. De¯nition 0.9. Let C be a convex set in a vector space X. The algebraic relative interior of C is the set a-ri(C) = a-inta®(C)(C). Observation 0.10. Let C be a convex set in a vector space X, x 2 C. Then x 2 a-ri(C) () x + i(x; C) = a®(C) () x + c(x; C) = a®(C) : Proof. Easy exercise. ¤ Observe that Theorem 0.7 immediately gives the following Corollary 0.11. Let C be a convex set in a normed space. Then a-ri(C) = ri(C) whenever at least one of the following conditions is satis¯ed: (a) ri(C) 6= ;; (b) dim(C) < 1; (c) X is a Banach space, a®(C) is closed, and C is an Fσ-set. 4 Inside points. Let C be a convex set in a normed space X. For x 2 C, let Lx be the family of all lines L ½ X, for which C \L is a nondegenerate segment containing x in its relative interior.
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