Minkowski sum of two sets in NLS

Question

if $(X, \left\Vert \right\Vert)$ is an NLS, and U is open in X, then how to justify that $A+ U= \bigcup_{a\in A}(a+U)$ for any set $A \subsetneq X$

Answer 1

If $x \in A+U = \{a+u: a \in A, u \in U\}$ we know that there is some $a_1 \in A$ and $u_1 \in U$ such that $x=a_1+u_1$. But this also says that $x \in a_1+U$ and this is one of the sets of the union $\bigcup_{a \in A} (a+U)$, so $x$ lies in that union too. The reverse is equally simple.