Q:

Minkowski functional of a open set

Given $X$ is a topological  vector space and $V$ is a convex, balanced neighbourhood of $0$ in X. Then  for $x \in V$, the Minkowski functional $\mu_{V}(x)$ = inf $\{t>0: t^{-1}x\in V\}$ $<$ $1$.
The reason given for strict inequality is "$V$ is open". But I dont get it. I can observe the following. Since $x\in V$, $\mu_{V}(x)$ $\leq 1$ and since $V$ is balanced $tV$ $\subseteq V$ for all $0<t<1$. Let $x\in V$. As $V$ is open, there exists a neighbourhood $U$ of $x$ such that $x\in U \subset V$. 
I dont know how to proceed with this. Please help!

A:

Using continuity of the map $M\colon X\times \mathbf R\to x$ defined by $M(x,\lambda):=\lambda x$ (continuity for the product topology), we have that the map $t\mapsto tx$ is continuous, hence the set
$$S:=\{t>0,t^{-1}x\in V\}$$
is an open  subset of $(0,+\infty)$.
Since $1$ belongs to $S$, the infimum of $S$ is necessarily strictly smaller than $1$.

