Closed sum of sets

Solution 1:

Here's a proof that works in any topological vector space.

If $z \notin X + Y$, then for every $x \in X$ we have $z - x \notin Y$; since $Y$ is closed and addition is continuous there are open sets $B_x$ containing $x$ and $U_x$ containing $z$ such that $(U_x - B_x) \cap Y = \emptyset$. By compactness of $X$, a finite collection $B_{x_1}, \ldots, B_{x_n}$ covers $X$. Thus $U = U_{x_1} \cap \ldots \cap U_{x_n}$ is an open set containing $z$ which does not intersect $X + Y$.

Solution 2:

(I assume the setting for this question is a normed vector space $V$)

Suppose $z_n$ is a sequence in $X+Y$, and assume it converges to some point $z$ in $V$. We want to show that $z$ is in $X+Y$. By definition of the Minkowski sum, there must exist $x_n \in X, y_n \in Y$ such that $z_n=x_n+y_n$. Since $X$ is compact, there is a subsequence $x_{n_i}$ such that $x_{n_i} \rightarrow x$ as $i \rightarrow \infty$, where $x$ is in $X$. Thus $y_{n_i}=z_{n_i}-x_{n_i} \rightarrow z-x$. Since $Y$ is closed, we have that $z-x$ is in $Y$, and so $z=x+(z-x)$ is in $X+Y$ as desired.

It's worth noting that it's not enough to simply have $X$ and $Y$ both closed. For example, suppose $V=\mathbb{R}$, $x=\mathbb{Z}$ and $y=\sqrt{2}\mathbb{Z}$. Then $X$ and $Y$ are both closed but $X+Y$ is a countable dense set.