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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.

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  • $\begingroup$ 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. $\endgroup$ – Ian Aug 11 '16 at 10:06
  • $\begingroup$ Thanks. Does the same hold for nets? The weak* topology on $X^*$ is not metrisable. $\endgroup$ – user342207 Aug 11 '16 at 10:17
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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.

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  • $\begingroup$ Thanks. Are you aware of an counterexample that will work? $\endgroup$ – user342207 Aug 12 '16 at 8:11

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