Is it true that if $B$ is compact then $\operatorname{Cl}(A+B)=\operatorname{Cl}(A)+\operatorname{Cl}(B)$? Suppose that $A,B$ are in $\mathbb{R}^n$, is it true that if $B$ is compact then $\operatorname{Cl}(A+B)=\operatorname{Cl}(A)+\operatorname{Cl}(B)$?
I am trying to prove that if a sequence ($a_n + b_n$) in $A+B$ converges to a+b, then both sequences $(a_n)$ and $(b_n)$ converge. I tried subsequences to conclude that $a_{n_k}$ and $b_{n_k}$ do converge, but i'm afraid i'm still missing something out.
edit: I mean $A+B=\{a+b: a \in A\text{ and }b \in B\}$, and $\operatorname{Cl}(X)$ the closure of $X$.
 A: Since $B$ is compact, it’s closed, and we’re really trying to show that $\operatorname{cl}(A+B)=\operatorname{cl}A + B$. It’s clear that $\operatorname{cl}A+B\subseteq\operatorname{cl}(A+B)$, so the problem is to show that $\operatorname{cl}(A+B)\subseteq\operatorname{cl}A+B$. Suppose that $p\in\operatorname{cl}(A+B)$. Then, as you say, there are sequences $\langle a_n:n\in\mathbb{N}\rangle$ in $A$ and $\langle b_n:n\in\mathbb{N}\rangle$ in $B$ such that $\langle a_n+b_n:n\in\mathbb{N}\rangle$ converges to $p$.
Because $B$ is compact, the sequence $\langle b_n:n\in\mathbb{N}\rangle$ must have a convergent subsequence, say $\langle b_{n_k}:k\in\mathbb{N}\rangle$, and the limit, $b$, of this subsequence must be in $B$. For $k\in\mathbb{N}$ let $x_k=a_{n_k}+b$. Then
$$\lim_{k\to\infty}\|(a_{n_k}+b_{n_k})-x_k\|=\lim_{k\to\infty}\|b_{n_k}-b\|=0\;;$$
can you take it from there to show that $\langle x_k:k\in\mathbb{N}\rangle$ converges to $p$ and therefore that $\langle a_{n_k}:k\in\mathbb{N}\rangle$ converges to $p-b$, which of course must then be in $\operatorname{cl}A$?
