Prove $\overline{x+A} = x+\bar{A}$ and $\alpha\bar{A} = \overline{\alpha A}$ If $A$ is a subset of $(V,\parallel.\parallel)$, then let $\bar{A}$ denote its closure. Show that if $x\in V$ and $\alpha \in \mathbb{R}$, then $\overline{x+A} = x+\bar{A}$ and $\alpha\bar{A} = \overline{\alpha A}$.
Honestly, I'm pretty stumped as to where to start, any help would be appreciated.

Start with the definitions. I'm not sure how a set's closure was defined for you, but let me define it and we'll go from there.
The point $q \in A$ is a closure point of $A$ if for every $\epsilon>0$, there is a point $p\in A$ such that $\|p-q\|<\epsilon$. (Note $p=q$ is fine.)
The closure of a set is the set of all its closure points.
So is $\overline{x+A} = x+\bar{A}$? The closure of $\{x+A\}$ must contain all its closure points. Let's suppose otherwise that there is a closure point of $\{x+A\}$, call it $y$, that is not contained in $x+\bar{A}$. 
So for every $\epsilon>0$, there is a point $p \in \{x+A\}$ such that $\|y-p\|<\epsilon$. Since $p \in \{x+A\}$, we can write it as $x+q$ where $q \in A$; so $\|y-x-q\|<\epsilon$. But that makes $y-x$ a closure point of $A$, and since $\bar{A}$ contains all the closure points of $A$, then $y-x \in \bar{A}$, which is a contradiction.
Proceed in the same way for proving $\alpha \bar{A} = \overline{\alpha A}$. Let me know if you get stuck!
