11
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.