# Demonstration that the strong and weak topologies are equivalent in finite dimensions, and may be different in infinite dimensions?

I am trying to get a better understanding of the weak topology. I have read that the strong and weak topologies are equivalent in finite dimensions, and may be different in infinite dimensions. By the strong topology, I mean the topology due to the norm, and by the weak topology I mean the weakest topology such that every bounded linear functional is continuous.

So does anyone know of explicit examples that shows how these topologies are equivalent in finite dimensions, but may be different in infinite dimensions?

• My A needs a total re-write but I have a headache so for now my A is deleted. – DanielWainfleet Jul 19 '17 at 19:06

So, $\mathbb{R}^n$ with the sup norm is what we look at (it's isomorphic to any finite dimensional normed space). Any open set in the weak topology is open in norm topology automatically. $B(x,r) = \bigcap_{i=1}^n \pi_i^{-1}(B(x_i,r))$, where the $\pi_i$ are bounded linear functionals. Under the weak topology $\pi_i^{-1}(B(x_i,r))$ will also be open, so that $B(x,r)$ is as well. So the topologies are equivalent.
In infinite dimensions, they weak topology is always strictly weaker than the norm topology. The easy way to tell this is to look at $\ell_2$, where unit ball is sequentially compact under the weak topology (I say sequentially because the weak topology is not metrizable), while it fails to be under the norm topology.