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This Wikipedia article gives two equivalent definitions of locally convex space (l.c.s). I don't see clearly the equivalence and I'd like to make it crystal clear.

  • Definition 1 Let $(V,\tau)$ be a TVS. It is called a l.c.s. if the origin has a local base of convex balanced absorbing sets.
  • Definitions 2 Let $(V,\tau)$ be a TVS. It is called a l.c.s. if $\tau$ is generated by a family of seminorms on $V$.

Suppose $(V,\tau)$ satisfies Definition 1. Let $\mathcal{B}$ be a local base at $0$ such that for every $C\in\mathcal{B}$, $C$ is convex, balanced and absorbing. A known result shows that the Minkowski functional $\mu_C$ is a seminorm on $V$ for $C\in\mathcal{B}$.

Question 1: Why can we conclude that $\tau$ is generated by $\{\mu_C\}_{C\in\mathcal{B}}$ so that Definition 1 implies Definition 2?

Suppose $(V,\tau)$ satisfies Definition 2 and $\tau$ is generated by a family of seminorms $\{p_\alpha\}_{\alpha\in A}$. For every finite subset $F\subset A$ and $r>0$, define $$ S_{F,r}=\bigcap_{\alpha\in F}\{x\in V:p_\alpha(x)<r\}. $$

Question 2: Why $\{S_{F,r}\}$ form a base of convex balanced absorbing sets at the origin?


Let $\tau'$ be the topology generated by $\{\mu_C\}$ in Question 1. I don't know at all why $\tau=\tau'$. For Question 2, I can show that the sets $S_{F,r}$ form a convex and balanced base at $0$. Why is each $S_{F,r}$ absorbing and are the two definitions of absorbing sets in this question the same here?

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Let's begin with the last point. For a balanced set, the two definitions of "absorbing" coincide. If $A$ is balanced, and $x \in tA$ for some $t \neq 0$, then $x \in sA$ for all $s$ with $\lvert s\rvert \geqslant \lvert t\rvert$. Because $\lvert s\rvert \geqslant \lvert t\rvert > 0$ means $\bigl\lvert \frac{t}{s}\bigr\rvert \leqslant 1$, and hence $s^{-1}x = \frac{t}{s} t^{-1} x \in A$ follows from $t^{-1}x \in A$, which is the same as $x \in tA$.

Coming to question 1, we note that the $C \in \mathcal{B}$ are all neighbourhoods of $0$, so the seminorms $\mu_C$ are continuous (with respect to $\tau$). Hence the topology induced by $\{ \mu_C : C \in \mathcal{B}\}$ is coarser than $\tau$, i.e. $\tau' \subset \tau$. But, since $\mathcal{B}$ is a neighbourhood basis of $0$ (for $\tau$), given any $\tau$-neighbourhood $U$ of $0$, there is a $C\in \mathcal{B}$ with $C \subset U$. Since $C$ is by definition a $\tau'$-neighbourhood of $0$, it follows that $U$ is a $\tau'$-neighbourhood of $0$. Since vector space topologies are determined by the neighbourhood filter of $0$ it follows that $\tau'$ is finer than $\tau$. Being both coarser and finer than $\tau$ means that $\tau' = \tau$, so the topology is indeed induced by the family $\{\mu_C : C \in \mathcal{B}\}$ of seminorms.

Regarding question 2, note that each $S_{F,r}$ is a neighbourhood of $0$. And every neighbourhood of $0$ is absorbing by the continuity of scalar multiplication. For every $x\in V$, the map $s_x \colon t \mapsto tx$ is continuous, and we have $s_x(0) = 0$, so by continuity there is a $\delta > 0$ such that $\lvert t\rvert < \delta \implies tx \in S_{F,r}$. But that means $x \in u\cdot S_{F,r}$ for all $u$ with $\lvert u\rvert > \delta^{-1}$.

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  • $\begingroup$ Why $C$ being heighbourhood of $0$ implies that $\mu_C$ is continuous? $\endgroup$ – Jack Dec 18 '15 at 23:04
  • $\begingroup$ Why is "$C$ by definition a $\tau'$-neighbourhood of $0$"? (I can only see that $C$ is by definition a $\tau$-neighbourhood of $0$. ) $\endgroup$ – Jack Dec 18 '15 at 23:07
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    $\begingroup$ A seminorm is continuous if and only if its unit ball is a neighbourhood of $0$. The one direction is clear, if $p$ is a continuous seminorm, then $p^{-1}([0,1)) = p^{-1}((-\infty,1))$ is open and contains $0$, hence is a neighbourhood of $0$. But that set is the (open) unit ball of $p$. Conversely, if the unit ball is a neighbourhood of $0$, then $p$ is continuous at $0$, since $p^{-1}([0,r)) = r\cdot p^{-1}([0,1))$, and the family of neighbourhoods of $0$ is stable under scaling. And the triangle inequality gives $\lvert p(x) - p(y)\rvert \leqslant p(x-y)$, which shows the continuity at $x$. $\endgroup$ – Daniel Fischer Dec 18 '15 at 23:09
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    $\begingroup$ Note that $x \in O \iff (\exists 0 < t < 1)(x \in tC)$. But since $C$ is balanced, we have $tC\subset C$ for $0 \leqslant t \leqslant 1$. That shows $O \subset C$. On the other hand, if $x\in C$, then clearly $\inf \{\lambda > 0 : x \in \lambda C\} \leqslant 1$, since $1$ is an element of the set whose infimum we take. Thus $C \subset R$. $\endgroup$ – Daniel Fischer Dec 18 '15 at 23:32
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    $\begingroup$ We start with the condition that $C$ is a neighbourhood of $0$. Since $C\subset R$, it follows that $R$ is a neighbourhood of $0$. Then $\frac{1}{2}R$ is also a neighbourhood of $0$. But $\frac{1}{2}R = \bigl\{ x\in V : \mu_C(x) \leqslant \frac{1}{2}\bigr\} \subset O$, hence $O$ is a neighbourhood of $0$. $\endgroup$ – Daniel Fischer Dec 18 '15 at 23:34

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