1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ $\mathcal{B}$ 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? $\endgroup$
    – Vobo
    Oct 19 '12 at 7:53
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I preferred the latter $\tau$. I got no problem of showing it a topology. Many thx. $\endgroup$
    – Juniven
    Oct 19 '12 at 22:11
  • $\begingroup$ @Vobo can you check my question about it math.stackexchange.com/questions/1098160/…? $\endgroup$
    – PtF
    Jan 9 '15 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.