Let $(E,\|\cdot\|_E)$ be a separable normed vector space over a field $\mathbb{K}$ and $E'$ its dual.
Theorem: Each bounded sequence in $E'$ has a weak-* convergent subsequence.
Question: With $(E,\|\cdot\|_E)$ not being separable, provide an example of a bounded sequence in $E'$ without any weak-* convergent subsequence.
I thought about an example on the space of continuous functions over a non compact set $(C((0,1)),\|\cdot\|_\infty)$, where $\|\cdot\|_\infty$ is the supremum's norm.
But any other would be highly appreciated.
Thanks.
Consider the sequence $\{f_n\}\subset(\ell^\infty)^*$ where each $f_n$ is the standard coordinate functional. I.e $f_n(e_j)=1$ if $n=j$ and $0$ otherwise. Clearly $\|f_n\|=1$ for all $n\in \mathbb N$, so it is a bounded sequence. Now we know that for $\{f_n\}$ to have a weak* convergent subsequence we would require some subsequence $\{f_{n_k}\}$ such that $f_{n_k}(x)\to f(x)$ for some $f\in (\ell^\infty)^*$ for all $x\in \ell^\infty$. However, for any subsequence $\{f_{n_k}\}$ we can construct the element $x_{n_k}\in \ell^\infty$ by $f_n(x_{n_k})=0$ if $n\neq n_k$ for some $k\in \mathbb N$, and $f_{n_k}(x_{n_k})=(-1)^k$. Clearly $(f_{n_k}(x_{n_k}))\subset \mathbb R$ is not convergent, so $\{f_{n_k}\}$ cannot be convergent. As this is true for any subsequence we conclude that $\{f_n\}$ has no convergent subsequence.