Given a Euclidean n-space, take any compact convex set $S$ in that space, with a distinguished point $p$ in its interior. Let $\lambda$ be any positive real, and let $S^{\prime}$ be the result of scaling $S$ as centered on $p$ by the factor $\lambda$. I would have thought that $S^{\prime}$ must be compact and convex as well. However, I didn't manage prove this. Is there any easy proof for this?
Given a Euclidean n-space, take any compact convex set $S$ in that space, with a distinguished point $p$ in its interior. Let $\lambda$ be any positive real, and let $S^{\prime}$ be the result of scaling $S$ as centered on $p$ by the factor $\lambda$. I would have thought that $S^{\prime}$ must be compact and convex as well. However, I didn't manage prove this. Is there any easy proof for this?
Second, convexity: the inverse scaling of two points of S', $x'_1,x'_2$ say, are two points $x_1,x_2$ 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 $x_1$ and $x_2$ is identical to the line segment between $x'_1$ and $x'_2$.