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June 10th, 2016, 05:09 AM   #1
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How to find out if the subset is closed?

We have a vector p = (0, 0, 2) in R^3 and we have the subset S = {xp where x >= 0} + T, where T is the convex hull of 5 vectors: (2,2,2), (4,2,2), (2,4,2), (4,4,6) and (2,2,10).

How do I show that the subset T is a closed and convex subset?
I'm not sure if I'm correct, but I'm using the following definition of a subset being convex:
A subset is called convex if it contains the line segment between any two of its points: (1-t)u + tv for every u and v in the subset.

I've tried to take two of those 5 vectors and see if there contains a line segment, but so far, it doesn't make any sense.
I hope that you can help me with this problem.
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June 10th, 2016, 02:43 PM   #2
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Quote:
Originally Posted by FightingMongooses View Post
We have a vector p = (0, 0, 2) in R^3 and we have the subset S = {xp where x >= 0} + T, where T is the convex hull of 5 vectors: (2,2,2), (4,2,2), (2,4,2), (4,4,6) and (2,2,10).

How do I show that the subset T is a closed and convex subset?
I'm not sure if I'm correct, but I'm using the following definition of a subset being convex:
A subset is called convex if it contains the line segment between any two of its points: (1-t)u + tv for every u and v in the subset.

I've tried to take two of those 5 vectors and see if there contains a line segment, but so far, it doesn't make any sense.
I hope that you can help me with this problem.
A convex hull is, by definition, convex.
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June 26th, 2016, 06:15 AM   #3
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The "convex hull" of a set of points (or vectors defining those points) is defined as "the smallest convex set containing those points" so, as mathman said, a "convex hull" is convex by definition. The simplest way to prove it is closed is probably to show that it contains all of its boundary points- and the boundary points of a convex hull of a (finite) set of points always consists of points on a straight line between two of the given points.
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