# A sequence of functionals without a weak-* convergent subsequence

I'm reading Royden Chap 15. The last remark in section 1 reads:

For $$X = l^{\infty}$$, $$B^*$$, the closed unity ball of $$X^*$$, is not weak-* sequentially compact. Indeed the sequence $$\{\psi_n\}\subset B^*$$ defined for each $$n$$ by $$\psi_n(\{x_k\})=x_n \qquad \forall \{x_k\}\in l^{\infty}$$ fails to have a weak-* convergent subsequence.

I am having trouble seeing how $$\{\psi_n\}$$ has no weak-* convergent subsequence. I'm having trouble even stating what I want to show. I think what I want to show is: $$\exists (c>0 \text{ and }\hat{x}\in X^{**}) \, \forall(n,m\in\mathbb{N})\text{ s.t. }c<|\hat{x}(\psi_n)-\hat{x}(\psi_m)|$$

Let's start by writing down what it would mean to have a weak$$^*$$-convergent subsequence of $$\{\psi_n\}_{n \geq 1}$$.
We would then have an increasing sequence $$n_k$$ and $$\psi \in X^*$$ such that for every $$x \in \ell^\infty$$, $$\psi_{n_k}(x) \to \psi(x)$$ as $$k \to \infty$$. Note that $$\psi_{n_k}(x) \to \psi(x)$$ if and only if $$x_{n_k} \to \psi(x)$$.
It is then not too hard to see that this cannot happen, since you can take $$x$$ to be the sequence $$x_n = \begin{cases} (-1)^k \qquad n = n_k \\ 0 \qquad \text{otherwise} \end{cases}$$ and then $$\psi_{n_k}(x) = (-1)^k$$ does not even converge.