$E$ be a normed vector space. Prove that $U+V$ is open in E

Question

Let $E$ be a normed vector space. Let $U$ be open in $E$ and $V\subset E$. Prove that $U+V$ is open in $E.$

My trial:

Since $U$ is open in $E$, then $\forall x\in U$, $\exists \;r>0$ such that $B(x,r)\subset U.$

From here, I don't know how to proceed! Please, can someone help out with proof with balls??

Answer 1

Notice that $$U + V = \bigcup_{v \in V} (U + \{ v \} ).$$ Since each element in the union is a translate of the open set $U$, it is open. Thus, $U + V$ is open, as it is the union of open sets.

Answer 2

For a fixed $v \in V$, the map $E \to E$ given by $x \mapsto x+v$ is continuous with a continuous inverse, given by $x \mapsto x-v$. Therefore, it is a homeomorphism and so maps open sets to open sets. Thus, $U+v$ is open. Finally, $U+V = \bigcup_{v \in V} U+v$ is open because it is the union of open sets.