# Counterexample showing that weakly* compact sets might not be norm bounded in a normed spaces.

There is the following result for Banach spaces:

Let $X$ be a Banach space. Then a weakly* compact subset of the dual $X^*$ is bounded.

To show that this result does not hold for normed spaces, consider the space $(c_{00} (\mathbb{N}), \| \cdot \|_1)$. For each $m \in \mathbb{N}$, define $$f_m : c_{00} (\mathbb{N}) \to \mathbb{F}, \quad (a_n)_{n \in \mathbb{N}} \mapsto m \cdot a_m,$$ and let $A := \{ f_m \; | \; m \in \mathbb{N} \}$. I have shown that the set $A$ is not bounded, but I am not able to show that it is weakly* compact yet.

Any help is appreciated.

• Take a sequence of elements of $A$ of which infinitely many are distinct and an element of $c_{00}$. Then the subsequence consisting of the distinct elements, when evaluated on that element, goes to zero. A general sequence either has this property or has a constant subsequence. – Ian Aug 11 '16 at 10:06
• Thanks. Does the same hold for nets? The weak* topology on $X^*$ is not metrisable. – user342207 Aug 11 '16 at 10:17

This particular collection is not weak-$\ast$ compact (although $c_{00}$ does seem to be the perfect place to construct a counterexample): For each $n\in \mathbb{N}$, let $e_n$ denote the $n^{th}$ standard basis vector, and set $$U_n = \{f \in X^{\ast} : |f(e_n)| > n-1\}$$ where $X = (c_{00}, \|\cdot\|_1)$. Then $U_n$ is weak-$\ast$ open and $f_n \in U_n$, so $\{U_n : n\in \mathbb{N}\}$ is an open cover for $A$. However, $f_n\notin U_k$ for any $k\neq n$, so it does not have a finite subcover.