# Convex compact sets

• Apr 8th 2010, 06:24 AM
Richard
Convex compact sets
Given a Euclidean n-space, take any compact convex set $\displaystyle S$ in that space, with a distinguished point $\displaystyle p$ in its interior. Let $\displaystyle \lambda$ be any positive real, and let $\displaystyle S^{\prime}$ be the result of scaling $\displaystyle S$ as centered on $\displaystyle p$ by the factor $\displaystyle \lambda$. I would have thought that $\displaystyle S^{\prime}$ must be compact and convex as well. However, I didn't manage prove this. Is there any easy proof for this?
• Apr 8th 2010, 06:44 AM
Failure
Quote:

Originally Posted by Richard
Given a Euclidean n-space, take any compact convex set $\displaystyle S$ in that space, with a distinguished point $\displaystyle p$ in its interior. Let $\displaystyle \lambda$ be any positive real, and let $\displaystyle S^{\prime}$ be the result of scaling $\displaystyle S$ as centered on $\displaystyle p$ by the factor $\displaystyle \lambda$. I would have thought that $\displaystyle S^{\prime}$ must be compact and convex as well. However, I didn't manage prove this. Is there any easy proof for this?

I don't think that it is necessary for P to be an interior point of S to prove it.
First, compactness: if S' has an open cover, then the inversely scaled cover is an open cover of S, hence has a finite subcover for S, and the scaled version of this is a finite subcover of the original open cover of S'.
Second, convexity: the inverse scaling of two points of S', $\displaystyle x'_1,x'_2$ say, are two points $\displaystyle x_1,x_2$ of S. If S is convex, then the entire line segment between these inversely scaled points must be in S, and therefore the scaled version of that line segment must be in S'. So to really show this in detail, you need to show that the scaled segment between $\displaystyle x_1$ and $\displaystyle x_2$ is identical to the line segment between $\displaystyle x'_1$ and $\displaystyle x'_2$.