Let $V$ be a normed vector space and let $(p_{\alpha})_{\alpha \in A}$ be a set of seminorms on $V$ such that for all $x \in V$ such that $p_{\alpha}(x)=0$ for all $\alpha$ we have $x = 0$.
I know two different definitions of the topology induced by these seminorms:
Definition 1: The topology is the initial topology w.r.t. the maps $f_{(\alpha,v)}:V \to \mathbb{R}: x \mapsto p_{\alpha}(v-x)$ for $(\alpha,v) \in A \times V$.
Definition 2: The subbase of the topology consists of $\{x \in V\;|\;p_{\alpha}(v-x) <\epsilon\}$ for all $\alpha \in A$ and $v \in V$ and $\epsilon > 0$.
It is clear that the topology in definition 1 is finer than the one given in definition 2.
Is it true that these topologies are equivalent? Is every set $f_{(\alpha,v)}^{-1}(U)$ (where $U \subseteq \mathbb{R}$ is open) a union of open sets of topology in the second definition?
Yes, the two definitions are equivalent. A clean way to express this is: a seminorm on a topological vector space is continuous if and only if it is continuous at $0$. A proof can be found, for example, here, item 7. It boils down to the triangle inequality. Suppose $f$ is a seminorm, $U\subset\mathbb{R}$ is open, and $x\in f^{-1}(U)$, then there is $r>0$ such that the $r$-neighborhood of $f(x)$ is in $U$. By the continuity of seminorm at $0$ there is a neighborhood $W$ of $0$ such that $f<r$ in that neighborhood. But then the set $\{y:f(y-x)<r\}$ is open, contains $x$, and is contained in $U$.
Definition 2 ensures that we have a TVS on which the given seminorms are continuous at $0$. By the above, they are continuous.