Subnets and compactness in $\{ 0, \dots, 9 \}^{[0,1]}$ I'm going over my knowledge of nets in preparation for some more work in functional analysis, and I came across an area of knowledge which I thought I had understood, but now seems a little flaky.
The set $\{ 0, \dots, 9 \}^{[0,1]}$ is compact with the product topology, since it is the product of finite discrete (and therefore compact) sets. Therefore every net contains a convergent subnet. This is the classical example where sequential compactness fails, because if we take the sequence
$$ f_k(x) = \text{The $k$'th element of the binary expansion of x} $$
Then there is no convergent subsequence $f_{i_k}$, because we can always take the number
$$ y = \sum_{k = 0}^\infty \frac{(-1)^k + 1}{10^{i_k}} $$
and $f_{i_k}(y)$ alternates between 0 and 2, and hence cannot converge.
Now the theory of nets tells us that there must exist a subnet $\mathfrak{a} = f_{i_\alpha}$ which converges in this set, where $i$ is an order preserving map from the directed set which forms the domain of $\mathfrak{a}$ to $\mathbf{N}$. If we then take the elements $i_1 < i_2 < \dots$ which form the image of $i$ in $\mathbf{N}$ how does the construction of $y$ fail here? Can one construct a convergent subnet for this sequence explicitly?
 A: It depends on what you consider explicit. To simplify the explanation a little, I’ll look just at $\{0,1\}^{[0,1]}$, since it’s a closed subset of $\{0,1,\ldots,9\}^{[0,1]}$ that contains every possible cluster point of your sequence. I’ll also replace the index set by $\wp(\Bbb N)$, setting $X=\{0,1\}^{\wp(\Bbb N)}$, and for $k\in\Bbb N$ and $A\subseteq\Bbb N$ I’ll let
$$x_A^{(k)}=\begin{cases}
1,&\text{if }k\in A\\
0,&\text{if }k\notin A\;;
\end{cases}$$
this is just your example in slightly different clothing. For each $A\subseteq\Bbb N$ let $B_A^+=\{x\in X:x_A=1\}$ and $B_A^-=\{x\in X:x_A=0\}$; then
$$\{B_A^+:A\subseteq\Bbb N\}\cup\{B_A^-:A\subseteq\Bbb N\}$$
is a subbase for the product topology on $X$. For disjoint finite subsets $\mathscr{F}^+$ and $\mathscr{F}^-$ of $\wp(\Bbb N)$ let
$$B(\mathscr{F}^+,\mathscr{F}^-)=\bigcap_{A\in\mathscr{F}^+}B_A^+\cap\bigcap_{A\in\mathscr{F}^-}B_A^-\;;$$
the family of all such sets is the canonical base for the product topology on $X$.
Let $\mathscr{U}$ be a free (i.e., non-principal) ultrafilter on $\Bbb N$, and let $p\in X$ be the indicator function of $\mathscr{U}$; I’ll construct a subnet of the sequence $\sigma=\langle x^{(k)}:k\in\Bbb N\rangle$ converging to $p$.
Let $\mathscr{D}=\{\langle U,n\rangle\in\mathscr{U}\times\Bbb N:n\in U\}$, and for $\langle U,n\rangle,\langle V,m\rangle\in\mathscr{D}$ set $\langle U,n\rangle\preceq\langle V,m\rangle$ iff $U\supseteq V$ and $n\le m$; then $\langle\mathscr{D},\preceq\rangle$ is a directed set. Let
$$h:\mathscr{D}\to\Bbb N:\langle U,n\rangle\mapsto n\;;$$
$h$ is an order-preserving, cofinal map, so
$$\nu:\mathscr{D}\to X:\langle U,n\rangle\mapsto x^{(n)}$$
is a subnet of $\sigma$. Let $B(\mathscr{F}^+,\mathscr{F}^-)$ be any basic open nbhd of $p$. Then $A\in\mathscr{U}$ for each $A\in\mathscr{F}^+$, and, since $\mathscr{U}$ is an ultrafilter, $\Bbb N\setminus A\in\mathscr{U}$ for each $A\in\mathscr{F}^-$, so
$$\bigcap\mathscr{F}^+\setminus\bigcup\mathscr{F}^-=\bigcap\mathscr{F}^+\cap\bigcap_{A\in\mathscr{F}^-}(\Bbb N\setminus A)\in\mathscr{U}\;.$$
Let $U=\bigcap\mathscr{F}^+\setminus\bigcup\mathscr{F}^-$, and fix $n\in U$. Suppose that $\langle V,m\rangle\in\mathscr{D}$, and $\langle U,n\rangle\preceq\langle V,m\rangle$. Then $m\in U$, so
$$\nu_{\langle V,m\rangle}=x^{(m)}\in B(\mathscr{F}^+,\mathscr{F}^-)\;,$$
and therefore $\nu\to p$, as desired.
