# Sequences and nets

According to the (Banach-)Alaoglu theorem, for any Banach space $X$ the closed unit ball of $X^*$ is compact in the weak* topology (the weakest/coarsest topology that makes all evaluation functionals $f\mapsto f(x)$ continuous).

For example, take $X=\ell_{\infty}$. The dual space $\ell_{\infty}^*$ contains an isometric copy of $\ell_1$ because $\ell_\infty^*=\ell_1^{**}$. The sequence $x_n=n^{-1}(e_1+\dots+e_n)$, where $e_n$ are the standard basis vectors, is contained in the unit sphere of $\ell_1$. Should $(x_n)$ have a weak*-convergent subsequence? Maybe it should, but it does not.

Indeed, take any subsequence $(x_{n_k})$. If necessary, choose a further subsequence so that $n_{k+1}\ge 3n_k$ for all $k$. Define

$\displaystyle y=\sum_{k}(-1)^k\sum_{n_{k-1}< j\le n_k} e_j$

where we set $n_0=0$. Two things to notice here: (1) $y\in \ell_{\infty}$; and (2) at least 2/3 of the coefficients of $e_j$, $1\le j\le n_k$, have the sign $(-1)^k$. Hence, $\langle x_{n_k},y\rangle\le -1/3$ when $k$ is odd and $\langle x_{n_k},y\rangle\ge 1/3$ when $k$ is even. This shows that $(x_{n_k})$ does not converge in the weak* topology.

The above does not contradict the Banach-Alaoglu theorem. Since $\ell_\infty$ is not separable, the weak* topology on the unit ball of its dual is not metrizable. The compactness can be stated in terms of nets instead of sequences: every bounded net in $X^*$ has a convergent subnet. In particular, the sequence $(x_n)$ has a convergent subnet (which is not a sequence). I personally find subnets a recipe for erroneous arguments. So I prefer to say: the infinite set $\{x_n\}$ has a cluster point $x$; namely, every neighborhood of $x$ contains some $x_n$. You can use the reverse inclusion of neighborhoods to define a subnet, but I’d rather not to. Everything we want to know about $x$ can be easily proved from the cluster point definition. For example,

• $\|x\|_{X^*}=1$
• $\langle x, 1_{\infty}\rangle =1$ where $1_{\infty}$ stands for the $\ell_\infty$ vector with all coordinates 1.
• $\langle x, y\rangle = \langle x, Sy\rangle$ where $S$ is the shift operator on $\ell_\infty$