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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? –  Vobo Oct 19 '12 at 7:53

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up vote 2 down vote accepted

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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I preferred the latter $\tau$. I got no problem of showing it a topology. Many thx. –  juniven Oct 19 '12 at 22:11

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