Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 12F (Analysis II)

True. Boundedness follows from $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$. To prove closedness of the Minkowski sum, let $z_j=x_j+y_j\in X+Y$ converge to $z$. By the Bolzano-Weierstrass theorem, the bounded sequence $x_j$ has a subsequence converging to $x$. Closedness gives $x\in X$. Along that subsequence, $y_j=z_j-x_j\to z-x$, which belongs to $Y$ by closedness. Thus $z=x+(z-x)\in X+Y$. This proves that the sum is bounded and closed.

True. The same subsequence argument applies to the sum of a compact set and a closed set. For a convergent sequence $z_j=x_j+y_j$, boundedness of $X$ provides a convergent subsequence of $x_j$, with limit $x\in X$. Then $y_j\to z-x\in Y$ along that subsequence. Consequently $z\in X+Y$. No boundedness of $Y$ is needed.

False. In $\mathbb{R}$, letEach is closed: its distinct points escape to infinity, so it has no finite accumulation point. Their Minkowski sum contains $1/n$, tending to zero. However zero is not in the sum. If $m-n+1/n=0$ with positive integers $m,n$, then $n-m=1/n$. For $n>1$ the right side is strictly between zero and one, while the left side is an integer; for $n=1$ the equation would require $m=0$. Thus the sum of two closed sets need not be closed. The same counterexample embeds along a coordinate axis in ${\mathbb{R}}^n$.

True, and the closedness assumption on $Y$ is unnecessary. The Minkowski sum has the representationEach translate of the open set $X$ is open, and an arbitrary union of open sets is open. If $Y$ is empty the union is empty, which is also open.