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:

1. $||v|| \geq 0$, $\forall v \in V$, and $||v|| = 0 \Leftrightarrow v = 0$
2. $||αv|| = |α| \cdot ||v||$, $\forall v \in V, α \in C$
3. $||v + w|| \leq ||v|| + ||w||$, $\forall v, w \in V$

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

• Firstly, do you know what is the definition of an open set? – E.Lim Dec 5 '15 at 7:43
• We were given several definitions in class but the one I usually go with is "A set $S$ is open if for any $x \in S$, there exists a neighborhood of $x$ included in $S$. " – Iff Dec 5 '15 at 7:48
• So given $x$ in $\frac{1}{2}A$, you want to find a neighbourhood of it that lies in $\frac{1}{2}A$. Can you find one by relating $x$ to an element in $A$? – E.Lim Dec 5 '15 at 7:51
• Pick $x$ to be $\frac14$? – Iff Dec 5 '15 at 8:10
• We don't know if $\frac{1}{4}$ is in $\frac{1}{2}A$ or even in $V$. Also, are you familiar with the open balls definition of an open set? It may be more apt for this question. – E.Lim Dec 5 '15 at 8:15

1 Answer

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 $a\in \Bbb R$