Bounded sets: x = [1, 2] U [3, 4] c R

[939]

Hello. Please look over my answers!

Homework Statement

a) Prove that this set is not convex: x = [1, 2] U [3, 4] c R
b) Prove the intersection of two bounded sets is bounded

Homework Equations

for a) x = [1, 2] U [3, 4] c R

The Attempt at a Solution

a) A convex set is where you can draw a line at two chosen points inside a set and everything between them will be inside the set. But here, you can have a set of points at 1 and 3 and connect them,however not all of the line will be in the set and thus is not convex. More formally, a set C is convex if for two points k, p ∈ C, the point t(k) + (1-t)(p) ∈ C for all t∈ [0, 1]. However, here, if you let t = 0.4, the result is 2.2 which is NOT in the set and is thus not convex.

b) I am not exactly sure how to proove this. Every finite set is bounded. Thus, when two bounded sets intersect, the intersection also must be bounded because it has an upper and lower interval.

 

[Dick]

Hello. Please look over my answers!

Homework Statement

a) Prove that this set is not convex: x = [1, 2] U [3, 4] c R
b) Prove the intersection of two bounded sets is bounded

Homework Equations

for a) x = [1, 2] U [3, 4] c R

The Attempt at a Solution

a) A convex set is where you can draw a line at two chosen points inside a set and everything between them will be inside the set. But here, you can have a set of points at 1 and 3 and connect them,however not all of the line will be in the set and thus is not convex. More formally, a set C is convex if for two points k, p ∈ C, the point t(k) + (1-t)(p) ∈ C for all t∈ [0, 1]. However, here, if you let t = 0.4, the result is 2.2 which is NOT in the set and is thus not convex.

b) I am not exactly sure how to proove this. Every finite set is bounded. Thus, when two bounded sets intersect, the intersection also must be bounded because it has an upper and lower interval.

The first one is fine. I would have picked 2 and 3 and k=1/2, but tastes differ. It works. For the second one if A and B are bounded, then the intersection of A and B is contained in A. Isn't it? You don't even need both sets to be bounded. Just one.