If $S,U\subset \mathbb R^n,\,$ with $U$ open, then $S+U$ is open.

Question

Attempt:

Take a $x\in U+S$ such that $x=u+s$ for some $u\in U,s\in S$. Since $U$ is open every element in $U$ there exists a ball containing them which lies in $U$ s.t. $\forall u\in U,\exists B(u,r)$ for some $r>0$. Now let's look at the translation of $u$ to $u+s$ and $B(u,r)$ to $B(u+s,r)$ so we have a ball containing $u+s$ at the center and lies in $U+S$ (Why?)

Answer 1

If $U$ is open, then it is a union of open bslls, i.e., $U=\bigcup_{i\in I} B(x_i,r_i)$, and if $s\in S$, then $$s+U=\bigcup_{i\in I} B(s+x_i,r_i),$$ and hence $s+U$ is also open. Consequently $$ S+U=\bigcup_{s\in S}(s+U) $$ and hence, $S+U$ is a union of open sets, and thus open.

Answer 2

Note that for each $s \in S$, the function $f_s(x)=s+x$ is a homeomorphism of $\mathbb{R}^n$: continuous, bijective and its inverse is also a (continuous) translation. It follows that $f_s$ is an open map and so $f_s[U] = \{s+x: x \in U\} = s+U$ is open when $U$ is open. Finally, $S + U = \bigcup \{s+U: s \in S\}$ is a union of open sets, hence open too.