# A question on Weak and Norm topology

How to prove that If $$F$$ is closed with respect to the weak topology then $$F$$ is closed with respect to the norm topology...

Weak topology means let $$V$$ Banach space the weak topology on $$V*$$ is smallest topology in which each function is continuous

norm topology is the topology generated by $$\mathscr{B}=\{B(x,\epsilon):x\in X, \epsilon>0 \}$$

and also tell me what are balls that generated weak topology....

thank you somuch

• In general the weak topology may fail to be metrizable. – DanielWainfleet Feb 21 '19 at 23:24

It is called the weak topology because it is weaker than the strong topology. Let $$T_w$$ be the set of weakly open sets and let $$T_s$$ be the set of strongly open sets. Let $$U$$ be the set of all topologies on $$V^*$$ such that each $$f\in V^*$$ is continuous. Then $$T_w=\cap U,$$ and $$T_s\in U,$$ so $$T_w\subset T_s.$$
So: $$F$$ is weakly closed $$\implies V^*$$ \ $$F \in T_w \implies V^*$$ \ $$F \in T_s \implies F$$ is strongly closed.