Example of converging subnet, when there is no converging subsequence

I'm trying to wrap my head around the concept of nets/subnets, especially in the following example.

Let $X$ be the Banach space $\ell_{\infty}$ and $X^*$ its dual. We know by Banach-Alaoglu that the unit ball $B$ of $X^*$ is compact, but not metrizable (as $X$ is not separable). It is compact but not sequentially compact, meaning that every net has a convergent subnet, but not every sequence has a convergent subsequence.

Consider the sequence $\{ \delta_n\}_{n \geq 0}$ defined by $\delta_n(\ldots, a_1, a_2, \ldots ) = a_n$. I see that this has no convergent subsequence, but what is a convergent subnet of it?

More generally, examples/comments illuminating the distinction between subnets and subsequences are welcome.

Solution 1:

Let $\mathscr{U}$ be an ultrafilter on $\Bbb N$. For $x=\langle x_n:n\in\Bbb N\rangle\in X$ let $f_{\mathscr{U}}(x)=\mathscr{U}\text{-}\lim x$, where $\mathscr{U}\text{-}\lim x$ is the unique real number $y$ such that $\{n\in\Bbb N:x_n\in U\}\in\mathscr{U}$ for each open nbhd $U$ of $y$. (For basic information about limits along ultrafilters see this answer.) Then $f_{\mathscr{U}}\in X^*$, and indeed your $\delta_n$ is $f_{\mathscr{U}}$ for the principal ultrafilter over $n$.

Now let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. Let $D=\{\langle U,n\rangle\in\mathscr{U}\times\Bbb N:n\in U\}$, and define a pre-order $\preceq$ on $D$ by setting $\langle U,m\rangle\preceq\langle V,n\rangle$ iff $U\supseteq V$; this makes $\langle D,\preceq\rangle$ a directed set. Let $\nu$ be the net $\nu:D\to X^*:\langle U,n\rangle\mapsto\delta_n$; for $x=\langle x_n:n\in\Bbb N\rangle\in X$ we have

$$\lim_{\langle U,n\rangle\in D}\nu(\langle U,n\rangle)(x)=\lim_{\langle U,n\rangle\in D}\delta_n(x)=\lim_{\langle U,n\rangle\in D}x_n\;.$$

Now $y=\lim_{\langle U,n\rangle\in D}x_n$ iff for each open nbhd $G$ of $y$ there is a $\langle U_G,m_G\rangle\in D$ such that $x_n\in G$ whenever $\langle V,n\rangle\in D$ with $\langle U_G,m_G\rangle\preceq\langle V,n\rangle$, i.e., whenever $V\in\mathscr{U}$ and $n\in V\subseteq U_G$. In short, for each open nbhd $G$ of $y$ there is a $U_G\in\mathscr{U}$ such that $x_n\in G$ for all $n\in U_G$. Let $G$ be an open nbhd of $y$, and let $A=\{n\in\Bbb N:x_n\notin G\}$; clearly $A\cap U_G=\varnothing$, and $\mathscr{U}$ is an ultrafilter, so $\{n\in\Bbb N:x_n\in G\}=\Bbb N\setminus A\in\mathscr{U}$. In other words, $y=\lim_{\langle U,n\rangle\in D}x_n$ iff $y=f_{\mathscr{U}}(x)$, and the net $\nu$ converges to $f_{\mathscr{U}}\in X^*$.

It only remains to check that $\nu$ is a subnet of the sequence $\langle\delta_n:n\in\Bbb N\rangle$. There are three different definitions of subnet in use; see this question and answer for details. In the terminology used there, $\nu$ is a Kelley subnet (and hence also an AA-subnet) of $\langle\delta_n:n\in\Bbb N\rangle$. To see this, let

$$\varphi:D\to\Bbb N:\langle U,n\rangle\mapsto n\;.$$

Let $m\in\Bbb N$ be arbitrary, and let $T_m=\{n\in\Bbb N:n\ge m\}$. $\mathscr{U}$ is a free ultrafilter, so $T_m\in\mathscr{U}$, and it’s clear that $\varphi(\langle U,n\rangle)=n\ge m$ whenever $\langle T_m,m\rangle\preceq\langle U,n\rangle\in D$, i.e.,

$$\varphi[\{\langle U,n\rangle\in D:\langle T_m,m\rangle\preceq\langle U,n\rangle\}]\subseteq T_m\;.$$