Separating family of seminorms covers a vector space Let $\mathcal{P}$ be a separating family of seminorms on a vector space $X$. Show that if $x\in X-\{0\}$ then $\exists p\in\mathcal{P}$ such that $p(x)\leq1$.
Context: This is from theorem 1.37 in Rudin's FA, part of an attempt to prove that the collection of $V(p,n)\equiv\{x\in X|p(x)<\frac{1}{n}\}$'s is a subbasis for a topology on $X$. In order to show that I must show that the union of all these sets give $X$. However, that's where I get stuck: as $\mathcal{P}$ is separating, it is clear that for $x\neq0$ there is some $p\in\mathcal{P}$ such that $p(x)\neq0$. If $p(x)\leq1$ then we are finished as we can just pick any $n\in\mathbb{N}-\{0\}$. However, what to do if all $p\in\mathcal{P}$ have $p(x)>1$?

I think this is a subbasis for the basis of open sets at 0, not for the whole space. Note that in the case that X is a normed space, then $\{\|\cdot\|\}$ is a (trivial) separating family, but in general you do not get the whole topology from just the open balls around the origin.
