TITLE: Norm and Weak Topologies agree?
QUESTION [7 upvotes]: so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_\alpha\rightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $f\in X^*$ we know by continuity that $f(x_\alpha)\rightarrow f(x)$, which means $x_\alpha\rightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.

REPLY [2 votes]: Another way to do this is to just go back to the definitions. Let $\epsilon>0,$ and $x_0\in X$. It suffices to  show that any open ball $B(x_0,\epsilon)$ is weakly open. We may use $\|\cdot\|_{\infty}$ as our norm.  And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=\sum^n_{i=1} x_ie_i$ for each $x\in X$ and for each $1\le i\le n,$  we have the functionals $e_i^*:X\to \mathbb C:x\mapsto x_i.$ 
Then, $B_{\epsilon }=\left \{ x:\left \| x\right \|<\epsilon  \right \}=\left \{ x:\forall\  1\le i\le n,\ |x_i|<\epsilon \right \},\ $ which is nothing more than the weakly open set $\left \{ x:\forall\  1\le i\le n,\ |e_i^*(x)|<\epsilon \right \}.$