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In the book Introduction to Functional Analysis written by A. E. Taylor there are the following theorems:

Theorem 1. Suppose that $X$ is a linear space and that $\mathscr{U}$ is a nonempty family of nonempty subsets of $X$ with the properties:

$(i)$ Each member of $\mathscr{U}$ is balanced, convex and absorbing;

$(ii)$ If $U\in \mathscr{U}$ there exists some $\alpha$ such that $0<\alpha\leq 1/2$ and $\alpha U\in\mathscr{U}$;

$(iii)$ If $U_1$ and $U_2$ are in $\mathscr{U}$ there exists $U_3\in\mathscr{U}$ such that $U_3\subset U_1\cap U_2$

$(iv)$ If $U\in \mathscr{U}$ there exists $V\in\mathscr{U}$ such that $x+V\subset U$.

Then there is a unique topology for $X$ such that $X$ is alocally convex topological vector space with $\mathscr{U}$ as a base at $0$.

Using this we might prove:

Theorem 2. Let $P$ be a nonempty family of seminorms defined on the linear space $X$. For each $p\in P$ let $V(p)$ be the set $\{x: p(x)<1\}$. Let $\mathscr{U}$ be the family of all finite intersections $$r_1V(p_1)\cap r_2V(p_2)\cap \ldots \cap r_n V(p_n),\ r_k>0,\ p_k\in P$$ Then $\mathscr{U}$ satisfies the conditions $(i)-(iv)$ of the previous theorem.

I have a doubt concerning the second theorem:

Question: It is not clear to me the definition of the set $$r_1V(p_1)\cap r_2V(p_2)\cap \ldots \cap r_n V(p_n),\ r_k>0,\ p_k\in P.$$ Does he mean $$U\in\mathscr{U}\Leftrightarrow \exists\ p_1, \ldots, p_k\in P\ \textrm{and}\ r_1, \ldots, r_k>0; U=\bigcap_{i=1}^k r_iV(p_i)?$$

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  • $\begingroup$ That's how I would interpret the statement. You're forming all finite intersections of $rV(p)$ for $r > 0$, $p \in P$. So $U\in\mathscr{U}$ is one of those intersections. $\endgroup$ – DisintegratingByParts Jan 9 '15 at 23:24
  • $\begingroup$ I agree with your interpretation, provided that by $P e r_{1}$ you mean $P$ and $r_{1}$. The way you had it typed looked far too deliberate for me to change it. $\endgroup$ – roo Jan 10 '15 at 23:14
  • $\begingroup$ Like the previous commenters: Yes, that is what he means. Something else: In condition $(iv)$, what is $x$? Should that be "If $U\in \mathscr{U}$ and $x\in U$ there exists $V\in \mathscr{U}$ such that $x+V\subset U$"? $\endgroup$ – Daniel Fischer Jan 10 '15 at 23:40
  • $\begingroup$ Thanks =) That "e" means "and" in portuguese, my mother tongue, I missed that when I was typing.. $\endgroup$ – PtF Jan 10 '15 at 23:44
  • $\begingroup$ I figured it was something like that since everything else was typed out so carefully. :) Good luck in your LCS adventures! $\endgroup$ – roo Jan 11 '15 at 3:30

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