Proposition 3.6 - " Functional Analysis , Sobolev Spaces and Partial Differential Equations ",Brezis "When $E$ is finite-dimensional, the weak topology $\sigma ( E,E') $ and the usual topology are the same."
We just need to show that every strongly open set ( I think it's the same of open set in the usual topology, that is the topology generated by a norm ) is weakly open. Let $x_{0} \in E$ and let $U$ be a neighborhood of $x_{0}$ in the strong topology. We have to find a neighborhood $V$ of $x_{0}$ in the weak topology $\sigma ( E,E') $ such that $V\subset U$. Let's assume we have already proved this thing, why would the topologies be the same? 
I think I have some problem with the definition of basis of neighborhoods. Could anyone help me?

A set $M$ is open if for every $x_0\in M$ it contains a neighborhood $U$ of $x_0$:
$$x_0\in U\subseteq M.$$
Brezis shows that if $E$ is finite dimensional and $U$ is a neighborhood in the strong topology it contains a neighborhood $V$ in the weak topology:
$$x_0\in V\subseteq U\subseteq M$$
thus showing that every strongly open set is also weakly open in the finite dimensional case. Since the weak topology is always coarser than the strong topology (also in the infinite dimensional case ) it follows that the set of weakly open sets and strongly open sets is the same , i.e. the topology is the same (in the finite dimensional case!).
