Let $X$ be a normed space. We can easily prove that if $F_1,F_2 \subset X^*$ then $\sigma(X,F_1)\subset \sigma(X,F_2)$. But if $F_2=span (F_1)$, $\sigma(X,F_1)= \sigma(X,F_2)?$ I try to find the answer for this question. I hope you can help me.
NOTE: the topology $\sigma(X,F)$ has the Basic include all sets $V(x;f_1,f_2,...f_k;\epsilon)=\bigcap_{i=1}^{k}\{y\in X: |f_i(y)-f_i(x)|<\epsilon\}$ with $k\in \mathbb N^*, x\in X; f_1,f_2,..,f_k\in F$ and $\epsilon>0$.
I assume that you mean $F_1\subseteq F_2$ (this is not mentioned explicitly).
To see the other inclusion, recall that $\sigma(X, F_2)$ is the smallest topology on $X$ that makes all functionals in $F_2$ continuous. But the topology $\sigma(X,F_1)$ also makes every functional in $F_2$ continuous, because an element in $F_2$ is a linear combination of elements in $F_1$. Hence, the other inclusion follows by minimality.