Sum of open/closed/compact sets in $\mathbb{R}^n$ open/closed/compact

Solution 1:

Something really useful to learn for these type of exercises is the following piece of information.

Let $y\in \mathbb{R}^{n}$ be fixed and define $\varphi_{y}:\mathbb{R}^{n}\to \mathbb{R}^{n}$ by setting $\varphi_{y}(x)=x+y$. Now $\varphi_{-y}\circ \varphi_{y}(x)=(x+y)-y=x$ and similarly $\varphi_{y}\circ \varphi_{-y}(x)=x$. Hence $\varphi_{y}$ is a bijection and $\varphi_{y}^{-1}=\varphi_{-y}$. Both $\varphi_{y}$ and $\varphi_{-y}$ are continuous functions so $\varphi_{y}$ is a homeomorphism.

Using the above property, many exercises of this sort become much easier.

In particular, for any open set $X\subseteq \mathbb{R}^{n}$ the set $X+y$ is an open set for all $y\in\mathbb{R}^{n}$ (since $\varphi_{y}$ is a homeomorphism). In particular, \begin{align*} X+Y=\bigcup_{y\in Y}(X+y) \end{align*} is an open set as the union of open sets.
Since arbitrary unions of closed sets aren't closed, we can't apply the above reasoning. But there is a standard counter example for this part. Take $X=\{-n+\frac{1}{n}:n\in\mathbb{N}\}$ and $Y=\mathbb{N}$. Both $X$ and $Y$ are closed but $X+Y=\left\{(m-n)+\frac1n\:;\:m,n\in\mathbb{N}\right\}$ has a subset $A:=\{\frac{1}{n}:n\in\mathbb{N}\}$, which has a limit point at zero but $0\notin X+Y$. Hence $X+Y$ is not closed.
Since $X$ and $Y$ are compact then $X\times Y\subseteq\mathbb{R}^{2n}$ is compact, and use the continuity of the function $(x,y)\mapsto x+y$ and the fact that continuous images of compact sets are compact, to conclude that $X+Y$ is compact.