# Local basis for topological vector space with topology induced by a family seminorms.

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.

• The translations are not necessary for local bases at $0$. For other points yes. Feb 21 '20 at 22:05
• @HennoBrandsma Would you be so kind to demonstrate why?
– user745578
Feb 21 '20 at 22:11
• It's rather obvious: all your basic open sets are neighbourhoods of $0$, as $p(0) =0$ for any seminorm. If you want a base at $v_0$ use finitely many $p_{v_0,i}$ instead; the translates of the $0$-neighbourhoods. $\mathcal{T}$ is also just the initial topology wrt $\mathcal{P}$. Feb 21 '20 at 22:14
• I need to show that given an open set $O$ containing $0$, I can find seminorms $p_1, \dots, p_n \in \mathcal{P}$ and $\epsilon > 0$ such that $\{\max_i p_i < \epsilon\} \subseteq O$. This is not clear to me. Since finite intersections of the subbasis giving the initial topology can contain translates, this gives me difficulties... How to construct this $\epsilon$ and these seminorms.
– user745578
Feb 21 '20 at 22:17

$$\mathcal T$$ 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. $$v \in ((p_i)_{z_i})^{-1}(U_i)$$.