# Norm topology and weak topology induce the same Borel sigma algebra on a Hilbert space

Let $$H$$ be a separable Hilbert space. Then the weak topology and the norm topology induce the same Borel sigma-algebra on $$H$$. I suspect there is something wrong with the following argument, but I'm not sure what it is:

Since the weak topology is weaker than the norm topology and $$H$$ is separable it suffices to show the Borel sigma algebra induced by the weak topology contains all closed balls. We have an isometric isomorphism from $$H$$ to its dual space given by $$x \to \left$$, from which we see that the weak and weak star topologies coincide. Then by the Banach-Alaoglu theorem any closed ball $$B = \{x\in H : \Vert x - y \Vert \leq r\}$$ is compact in the weak topology, which implies $$B$$ is closed in the weak topology since it's Hausdorff.

• What you wrote is correct, I would recommend carrying out the step: all closed balls in sigma algebra $\implies$ all open balls in sigma algebra Commented Aug 21, 2019 at 19:12
• I wouldn't say the proof is wrong. It's just like showing up to a fist fight with a tank :) Slightly overkill. Commented Aug 21, 2019 at 19:21

If $$X$$ is a locally convex space (e.g. a Hilbert space) and $$C\subset X$$ is convex, then $$C$$ is closed in the original topology iff it is closed in the weak topology.