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This is a theorem in Rudin's functional analysis:

Theorem. Suppose $\mathcal{P}$ is a separating family of seminorms on a real vector space $X$. Associate to each $p\in \mathcal{P}$ and to each $n\in \mathbb{N}$ the set $V(p,n)=\{x\in X: p(x)<\frac{1}{n}\}.$ Let $\mathcal{B}$ be the collection of all finite intersections of the sets $V(p,n)$. Then $\mathcal{B}$ is a convex balanced local base for a topology $\tau$ on X, which turns $X$ into locally convex space such that every $p\in \mathcal{P}$ is continuous.

Rudin declared that $A\subseteq X$ is open iff $A$ is a union of translates of members of $\mathcal{B}$. Does this mean that $ \tau = \{A \subseteq X: \forall x \in A, \exists y\in X \mbox{ and } B \in \mathcal{B} \mbox{ with }x \in y+B \subseteq A\}? $ If that is so, then how can we prove that $\tau$ is closed under finite intersection?

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    $\mathcal{B}$ forms a base for a system of neighbourhoods of 0. The set of translates of $\mathcal{B}$ form a base for a topology $\tau$ on $X$. Does this help?2012-10-19

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Rudin is right, you are too. Yes, your assumption on the definition of $\tau$ is correct, but in addition you can show $ \tau = \{ A \subseteq X : \forall x \in A \exists B \in \mathcal{B} \mbox{ with } x+B \subseteq A\} $ which is obviously a topology.

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    @Vobo can you check my question about it http://math.stackexchange.com/questions/1098160/doubt-concearning-the-definition-of-a-locally-convex-space-structure-through-sem?2015-01-09