# Fewer open sets in the weak topology

Consider $E^*$ the dual of a Banach space $E$.

Every open set in the strong (or norm) topology can be written as: $A_{\epsilon\tau}= \{ f \in E^* : \|f - \tau \| < \epsilon \}$, for some shift $\tau$, and $\epsilon > 0$. Here, the norm is the "operator norm," that is, for $f\in E^*$, $\|f\| = \sup \{ |f(x)| : \|x\| \leq 1, x \in E \}$

Every open set in the weak-* topology can be written as the union of finite intersections of its sub-basis: $B_{\epsilon\tau k} = \{ f\in E^* : \|f-t\|_k < \epsilon \}$, for some shift $\tau, \epsilon >0$ and $k \in E$. Here, the seminorm $\| \cdot \|_k$ is defined as $\| f\|_k = |f(k)|$

The reason why the latter topology is called the "weak"-* topology is because it has fewer open sets, so it is "easier" to converge, and hence called "weak". On the other hand, the norm topology has more open sets, hence it is "harder" to converge, and is called "strong." The connection between open sets and convergence is that a sequence $f_n \rightarrow f$ if and only if for every open set $A$, there exists an $N$ such that for $n>N$, $f_n \in A$.

Please correct me if any of the above is incorrect. My question:

Why are there "more" open sets in the strong topology, and "less" in the weak-* topology? I am struggling to see this. I would have liked to prove: for every open set in the weak-* topology, there exists a corresponding open set in the norm topology, but not vice versa. I am not sure how to go about proving this, or if it is even the right way to think about it.

• @user399601 then how is it possible there are "more" open sets defined by the strong topology? Or is that not true? Commented Apr 9, 2017 at 16:40
• I'm sorry, I got that backwards. Every weak-* open set is strong open. Anyway you don't need a complicated correspondence Commented Apr 9, 2017 at 16:41
• @user399601 Okay. But it is not clear to me how to show that every weak-* open set is strong open. And then I would like to find a strong open set that is not weak-* open. Commented Apr 9, 2017 at 17:20
• Assuming $E$ is infinite-dimensional, you can show that every weak$^*$ open neighborhood of 0 contains an infinite-dimensional subspace, which is not true for the norm topology. So the open unit ball (which is norm bounded) is not weak$^*$ open. Commented Apr 9, 2017 at 17:29