Normed Vector Space Let $V$ be a real normed vector space.
Suppose that $A$ is an open set from $V$. Show that the set $\frac12 A = \left\{ \frac12 x \, : \, x \in A \right\}$ is also open.
Let $V$ be a complex vector space. A norm on $V$ is a function $|| \cdot || : V \to R$ that satisfies the following three conditions:


*

*$||v|| \geq 0$, $\forall v \in V$, and $||v|| = 0 \Leftrightarrow v = 0$

*$||αv|| = |α| \cdot ||v||$, $\forall v \in V, α \in C$

*$||v + w|| \leq ||v|| + ||w||$, $\forall v, w \in V$


I'm not sure how to approach this problem, any pointers are appreciated! 

Let $x \in \dfrac12A$, then $2x \in A$. Since $A$ is open, $\exists\epsilon>0$ such that the open neighbourhood $B(2x,\epsilon)\subseteq A \iff B(x,\dfrac\epsilon 2) \subseteq \dfrac12A$, so $\dfrac12A$ is open, where the open ball $B(x,a)$ denotes the set $\{y \in V:\lVert y-x\rVert<a\}$ for any $a\in \Bbb R$
