First of all I have asked this question one time before here:
But I was wondering if there might be another way to solve, using the following:
$(V,T)$ topological vector space over $\mathbb{F}$.
$U$ is an open neighbourhood of the origin, then there exists a balanced neighbourhood $W$ s.t $W \subset U$. Since $W$ is balanced then $\alpha W \subset W$ $\forall \alpha \in \mathbb{F} : |\alpha | \leq 1$. My question now is, can one prove that $\alpha W$ is an open neighbourhood aswell?
If you just want to show $ W \in \mathcal{T} \Rightarrow \alpha W \in \mathcal{T}$ for $\alpha \neq 0$ then there is not much to do, because you started with a topological vectorspace $(V,\mathcal{T})$.
You can easily show that $\phi_\alpha: V \to V,\; x \mapsto \alpha x$ is a homeomorphism. This results from an axiom of the topological vectorspace and the fact that $\phi_{\frac{1}{\alpha}}$ is the inverse of $\phi_\alpha$ https://en.wikipedia.org/wiki/Topological_vector_space#Definition
However this gives you a circular argument because you want to show that there is a balanced neighborhood $N$ of $U$ by using that you already have one ($W$).
