91Ó°ÊÓ

Let \(A\) and \(B\) be convex, disjoint, closed, nonempty sets in \(\mathbb{R}^{N}\). Show that \(A \cup B\) can never be convex.

Let \(x=\left(a_{1}, a_{2}, \ldots, a_{n}, \ldots\right)\) and \(y=\left(b_{1}, b_{2}, \ldots, b_{n}, \ldots\right)\) where \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) are bounded sequences of real numbers. Define $$ d(x, y)=\sup _{n}\left|a_{n}-b_{n}\right| $$ Show that the space \(S\) consisting of the totality of such bounded sequences with the distance defined above is a metric space. Is the metric space complete?

Let \(S_{1}, S_{2}\) be metric spaces and suppose that \(f: S_{1} \rightarrow S_{2}\) is an isometry. Suppose that \(S_{2}\) is complete. Is the space \(f\left(S_{1}\right)\) complete? Justify your statement.

A subset \(K\) of \(\mathbb{R}^{N}\) is called a cone if, whenever \(x \in K\), then \(\lambda x \in K\) for every \(\lambda \geqslant 0\) Show that a cone \(K\) is convex if and only if \(x+y \in K\) whenever \(x, y \in K\). Give an example of a nonconvex cone. Show that if \(K_{1}\) and \(K_{2}\) are cones in \(\mathbb{R}^{N}\), then \(K_{1} \cap K_{2}\) and \(K_{1} \cup K_{2}\) are cones. If \(K_{1}\) and \(K_{2}\) are convex, under what conditions, does it follows that \(K_{1} \cap K_{2}\) and \(K_{1} \cup K_{2}\) are convex cones?

In \(\mathbb{R}^{N}\), define the cylinder \(C=\left\\{\left(x_{1}, x_{2}, \ldots, x_{N}\right): x_{1}^{2}+x_{2}^{2}+\cdots+x_{N-1}^{2} \leqslant 1,0 \leqslant\right.\) \(\left.x_{N} \leqslant 1\right\\}\). Show that \(C\) is a convex set.