If $A$ is open, then $-A$ is also Open. Problem: Let $(X, ||\cdot||)$ be a normed vector space. Suppose $A \subseteq X $ is open.
And we have $-A=\{-a:a \in A\}$. Prove that $-A$ is open.
**Proof:**As $A$ is open,  $B(x,r) \subseteq A$. 
Let $x \in -A$.
Then suppose $y \in B(x,r)$
$\implies d(x,y)<r$ 
$\implies ||x-y||<r$ $\implies ||-x-(-y)||<r$. 
So, $-y \in B(-x,r). $ Thus $y\in -A$. 
I Guess this is incorrect. Also, I would like to see a full proof of it.
 A: Let $x \in -A$, then by the definition of $-A$ we have that $-x \in A$. Since $A$ is open there exists an $r >0$ such that $B(-x,r) \subset A$. Again by the definition of $-A$ we have that
$$-B(-x,r) \subset -A.$$
All that's left to show is that $-B(-x,r) = B(x,r)$, to do this we just note that
$$|-x - y| < r \iff |x - (-y)| < r.$$
Hence $B(x,r) \subset A$ and we are done.
A: Here's another approach, by general abstract nonsense:

*

*Let $X$ be a topological space and $f:X\to X$ an homeomorphism. If $A\subset X$ has the same "topological properties" as $f(A)$. For our purposes, $A$ is open in $X$ iff $f(A)$ is open in $X$.

*If $(X,\|\cdot\|)$ is a normed space (or more generally a topological vector space), then the map $f:X\to X$ defined by $x\mapsto -x$ is a homeomorphism.

*Letting $f$ be as before, then $f(A)=-A$.

Now put everything together.
A: I assume your normed vector space is over either the real field or the complex field. You have noted that $||\alpha w||$ = $|\alpha|\;||w||$, so that in particular
$||-w||$ = $||w||$.

Suppose $A \subseteq X $ is open.  And we have $-A=\{-a:a \in A\}$.
Prove that $-A$ is open.

As $A$ is open, we have that for any $x \in A,$ $B(x,r_x) \subseteq A$, where $r_x >0$.
Let $x \in -A$; we want to show that $x$ is interior. $-x \in A.$
Then suppose $y \in B(x,r_{-x})$
$\implies d(x,y)<r_{-x}$ 
$\implies ||x-y||<r_{-x}$ $\implies ||-x-(-y)||<r_{-x}$. 
So, $-y \in B(-x,r_{-x})$. Thus $-y \in A$, and hence $y\in -A$.
