Let $(X,\tau)$ be a topological vector space over $\Bbb{R}$ and $X^*$ be the topological dual of $X$. There are two ways to define the weak topology $\tau_w$ on $X$:
- $\tau_w$ is the initial topology on $X$ with respect to $X^*$;
- For each $f\in X^*$, $p_f(x):=|f(x)|$ gives a seminorm on $X$. $\tau_w$ is the topology on $X$ generated by the family of seminorms $\{p_f\mid {f\in X^*}\}$.
I'm trying to understand that why these two definitions are equivalent. Denote the first one defined as an initial topology as $\tau_1$ and the second one as $\tau_2$.
Let $$ E=\{f^{-1}(U)\mid f\in X^*,\ U\ \hbox{open in}\ \Bbb{R}\} $$ and $$ F=\{B_r^{f}(x)\mid f\in X^*, \ r>0,\ x\in X\}. $$ where $$ B_r^{f}(x)=\{y\in X \mid |f(y-x)|<r\}. $$ Then $\tau_1$ is generated by $E$ and $\tau_2$ is generated by $F$. It is not very hard to show that $F\subset E$ and thus $\tau_2\subset \tau_1$: $ y\in B_r^f(x) $ iff $ |f(y)-f(x)|<r $ iff $ y\in f^{-1}(B_\Bbb{R}(f(x),r)) $
I don't see why $E\subset F$ (not necessarily true?) or at least $E\subset \tau_2$. Could anyone help me to go on?
Thanks to @Daniel's comment:
Consider $W=f^{-1}(U)\in E$ and $x\in W$. Then $f(x)\in U$. Since $U$ is open in $\Bbb{R}$, there exists $r_x>0$ such that $B_\Bbb{R}(f(x),r_x)\subset U$ and thus $$ B_{r_x}^f(x)=f^{-1}(B_\Bbb{R}(f(x),r_x))\subset f^{-1}(U). $$ Since $$ W=\bigcup_{x\in W} B_{r_x}^f(x), $$ we have $W\in T_2$.