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(a) By definition a set is convex if for any points P and Q in the set, the segment `bar(PQ)` is also in the set.
Consider the set `L=L_1 nn L_2` where `L_1,L_2` are convex.
Take `P,Q in L` . By definition `P,Q in L_1` and since `L_1` is convex `bar(PQ) in L_1` . Simarly `bar(PQ) in L_2` so `bar(PQ) in L_1 nn L_2` which implies that L is convex.
** If the intersection consists of 1 point L is convex since P and Q are the same point so the "segment" is in L. If the intersection is empty then L is convex since the empty set is convex. **
(b) Consider the set `L_1` in Euclidean 2-space -- it is the line segment with endpoints (0,0) and (2,2). Let `L_2` be a line segment in Euclidean 2-space with endpoints (2,2) and (4,0).
Both `L_1,L_2` are convex. Also `(1,1) in L_1 uu L_2,(3,1) in L_1 uu L_2` . But only the endpoints of the segment joining (1,1) and (3,1) are in the union, so the union is not convex.
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