Q:

How to show a collection is a local subbasis of the weak topology?

Let $X$ be a normed $\mathbb K$-linear space and for all $x \in X$ , $f \in X^*$, $\epsilon >0$ Let us define $$U_x(f,\epsilon)=\{y \in X : |f(y)-f(x)|<\epsilon\}\\ 
 =x+f^{-1}(B(0,\epsilon))$$
Then for each $x \in X,$ $\{U_x(f,\epsilon)|f \in X^*, \epsilon>0\}$ is a local subbasis of the weak topology of $x.$Here $X^*$ is the dual space of $X$.

Edit: (definition of weak topology) Let $X$ be a normed $\mathbb K$ -linear space . The weak topology of $X$ is the smallest topology $J_{\omega}$(which is a subset of the normed topology of X) such that $f:(X,J_{\omega})\to\mathbb K$ is continuous for all $f \in X^*.$
Clearly the sets $U_x(f,\epsilon)$ are open sets in the weak toplogy.
To show the above set is local subbasis of $x$ in the weak topology we have to take an open set $U$ from the weak topology $s.t.$ $x\in U$ and then we have to find a finite set $\{U_x(f_1,\epsilon), U_x(f_2,\epsilon),...,U_N(f_N,\epsilon)\}$ s.t. $x\in \displaystyle{ \cap_{i=1}^{N}U_x(f_i,\epsilon)\subseteq U\}}$.
But how can I find such types of finite sets? 
Please someone help.
Thank you..

A:

The weak topology has as a subbasis $\{U_x(f,\varepsilon):x\in X,f\in X^*,\varepsilon>0\}$.  
So given a weak open neighborhood $U$ of $x$, there exist $x_1,\ldots,x_n\in X$, $f_1,\ldots,f_n\in X^*$, and $\varepsilon_1,\ldots,\varepsilon_n>0$ such that
$$x\in\bigcap_{k=1}^nU_{x_k}(f_k,\varepsilon_k)\subset U.$$
What you have to show is that there exist $f'_1,\ldots,f'_m\in X'$ and $\varepsilon'_1,\ldots,\varepsilon'_m>0$ such that 
$$\bigcap_{k=1}^mU_x(f'_k,\varepsilon'_k)\subset\bigcap_{k=1}^nU_{x_k}(f_k,\varepsilon_k).$$
One last hint: $m$ will equal $n$, and $f_k'=f_k$ for $k=1,\ldots,n$.

