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\label{thm:cylinder-basis}
The collection
$\mathcal{B} = \{C_k(x) : k \geq 1,\; x \in \tau^3\}$
is a basis for a topology on~$\tau^3$.
The resulting topology satisfies:
\begin{enumerate}
    \item[\textup{(1)}] \textbf{Covering.}
          $\tau^3 = \bigcup_{x \in \tau^3} C_1(x)$.
    \item[\textup{(2)}] \textbf{Intersection closure.}
          For any $C_k(x), C_l(y) \in \mathcal{B}$
          and any $z \in C_k(x) \cap C_l(y)$,
          there exists $C_m(z) \in \mathcal{B}$
          with $z \in C_m(z) \subseteq C_k(x) \cap C_l(y)$.
    \item[\textup{(3)}] \textbf{Totally disconnected.}
          The basis consists entirely of clopen sets.
    \item[\textup{(4)}] \textbf{Second countable.}
          The basis has a countable sub-basis:
          the collection $\{C_k(r) : k \geq 1,\;
          r \in \mathbb{Z}/P_k\}$ is countable.
    \item[\textup{(5)}] \textbf{Compatible with profinite structure.}
          The topology coincides with the inverse-limit topology
          on $\tau^3 = \varprojlim_k \mathbb{Z}/P_k$,
          where each $\mathbb{Z}/P_k$ carries the discrete topology.
\end{enumerate}

\emph{Clause~(1).}
For any $x \in \tau^3$,
$x \in C_1(x)$ by definition.
Hence $\bigcup_{x} C_1(x) = \tau^3$.

\emph{Clause~(2).}
Let $z \in C_k(x) \cap C_l(y)$
and set $m = \max(k, l)$.
By Proposition~\ref{prop:ch09-intersection},
$C_m(z) \subseteq C_k(z) = C_k(x)$
(since $z \in C_k(x)$ implies $C_k(z) = C_k(x)$)
and similarly $C_m(z) \subseteq C_l(z) = C_l(y)$.
Hence $C_m(z) \subseteq C_k(x) \cap C_l(y)$,
and $z \in C_m(z)$.
This verifies the basis intersection axiom.

\emph{Clause~(3).}
Each $C_k(x)$ is clopen by
Proposition~\ref{prop:ch09-complement}.

\emph{Clause~(4).}
At stage~$k$, there are exactly $P_k$
distinct cylinders $\{C_k(r)\}_{r \in \mathbb{Z}/P_k}$.
The total number of basis elements
$\bigcup_{k \geq 1} \{C_k(r) : r \in \mathbb{Z}/P_k\}$
is a countable union of finite sets,
hence countable.

\emph{Clause~(5).}
The inverse-limit topology
on $\varprojlim_k \mathbb{Z}/P_k$
has as a basis the preimages
$\pi_k^{-1}(U)$
for $U \subseteq \mathbb{Z}/P_k$ open.
Since $\mathbb{Z}/P_k$ carries the discrete topology,
every singleton $\{r\}$ is open,
and $\pi_k^{-1}(\{r\}) = C_k(r)$.
These singletons generate the discrete topology
on $\mathbb{Z}/P_k$,
so their preimages generate
the inverse-limit topology.
But these preimages are exactly
the elements of~$\mathcal{B}$.