Let $E$ a normed vector space of finite dimension. Then the strong topology and the weak topology are the same. To prove it, we take $x_0\in E$ and $U$ an open set (for the strong topology) that contain $x_0$. Then we have to prove that it's an open set for the weak topology, i.e. that there is $V$ weakly open s.t. $V\subset U$. Do do, in my book it's written that we have to find a finite sequence $(f_i)_{i\in I}$ of $E'$ and an $\varepsilon>0$ s.t. $$V=\{x\in E : \forall i\in I|,\left<f_i, x-x_0\right>|<\varepsilon\}\subset U.$$
My questions : Why we need only a finite sequence $(f_i)_{i\in I}$ to describe $V$ ? Shouldn't it be $$V=\{x\in E : \forall f\in E', |\left<f_i, x-x_0\right>|<\varepsilon\}.$$
No, $I$ is finite since the subbasis for the weak topology is $$\mathcal S:=\{f^{-1}(U)\mid f\in E', U\text{ open in }\mathbb F\}.$$ Therefore, the basis is given by finite intersection of element of $\mathcal S$, i.e. an element of the basis is $$\bigcap_{\substack{i\in I \\ |I|<\infty }}f_{i}^{-1}(U_i),$$ where $U_i$ are open an $f_i\in E'$.