Consider a vector space $V$ and a family $\mathcal{P}$ of seminorms on $V$. For $p\in \mathcal{P}, z \in V$, define
$$p_z: V \to [0,\infty[: v \mapsto p(v-z)$$
Consider the collection of maps
$$\{p_z:z \in V, p \in \mathcal{P}\}$$
and let $\mathcal{T}$ be the initial topology w.r.t. this collection. I managed to show that $(V, \mathcal{T})$ is a topological vector space. Now, I'm focused on showing the following:
$$\{\{v \in V: \max_{i=1}^n p_i(v) < \epsilon\}: p_i \in \mathcal{P}, \epsilon > 0, n \geq 1\}$$
is a local basis for this topological space (thus for every open containing $0$, I can find a set in the above collection contained in it).
What confuses me is that this does not depend on the translations of seminorms in $\mathcal{P}$, and I tried to apply reverse triangle inequality to get rid of these translation terms but I was unsuccesful.
Thanks in advance for any help.
$\mathcal T$ is the coarsest topology making all functions $p_z$ continuous. A subbasis for $\mathcal T$ is given by the family of all $(p_z)^{-1}(U)$ with $p \in \mathcal P$, $z\in V$ and $U \subset [0,\infty)$ open. Obviously $p_0 = p$, thus certainly each set $$B(p_1,\ldots,p_n, \epsilon) = \{v \in V: \max_{i=1}^n p_i(v) < \epsilon\} = \bigcap_{i=1}^n p_i^{-1}([0,\epsilon)$$ is an open neighborhood of $0$.
Now let $O$ be any open neighborhood of $0$. Thus there exist $p_1,\ldots,p_n \in \mathcal P$, $z_1,\ldots,z_n \in V$ and open $U_1,\ldots,U_n \subset [0,\infty)$ such that $$0 \in \bigcap_{i=1}^n ((p_i)_{z_i})^{-1}(U_i) \subset O .$$ We have $(p_i)_{z_i}(0) = p_i(-z_i) \in U_i$. There exists $\epsilon > 0$ such that $(p_i(-z_i)+\epsilon,p_i(-z_i)+\epsilon) \cap [0,\infty) \subset U_i$ for $i =1,\ldots,n$. But then $(p_i)^{-1}([0,\epsilon) \subset ((p_i)_{z_i})^{-1}(U_i)$: In fact, $v \in (p_i)^{-1}([0,\epsilon)$ means $p_i(v) = p_i(-v) \in [0,\epsilon)$ and we conclude $$p_i(-z_i) - \epsilon < p_i(-z_i) - p_i(-v) = p_i(v-z_i - v) - p_i(-v) \le p_i(v-z_i) + p_i(-v) - p_i(-v)\\ = p_i(v-z_i) \le p_i(v) + p_i(-z_i) < p_i(-z_i) + \epsilon. $$ Thus $(p_i)_{z_i}(v) = p_i(v-z_i) \in (p_i(-z_i)+\epsilon,p_i(-z_i)+\epsilon) \cap [0,\infty) \subset U_i$, i.e. $v \in ((p_i)_{z_i})^{-1}(U_i)$.