%
\label{thm:hol-implies-cont}
Every $\omega$-germ transformer $f \colon \tau^3 \to \tau^3$
is continuous with respect to the cylinder topology
(equivalently, the ultrametric topology).

We give two formulations of the proof:
one via open preimages, one via the ultrametric.

\medskip

\noindent\textbf{Open-preimage formulation.}
By the Cylinder Basis Theorem
(II.T04, Chapter~\ref{ch:cylinder-domains}),
the stage-$k$ cylinders form a basis for the topology.
It suffices to show that the preimage of every basis set is open.

Let $C_k(z)$ be a cylinder in the codomain.
Consider the preimage $f^{-1}(C_k(z))$.
If $f^{-1}(C_k(z)) = \varnothing$, it is trivially open.
Otherwise, let $x \in f^{-1}(C_k(z))$,
so that $f(x) \in C_k(z)$,
i.e., $\pi_k(f(x)) = \pi_k(z)$.
By the cylinder compatibility lemma
(Lemma~\ref{lem:naturality-cylinder}),
$f(C_k(x)) \subseteq C_k(f(x)) = C_k(z)$,
since $\pi_k(f(x)) = \pi_k(z)$.
Hence $C_k(x) \subseteq f^{-1}(C_k(z))$.
Since $C_k(x)$ is open
(cylinders are clopen, Chapter~\ref{ch:cylinder-domains})
and contains~$x$,
every point of $f^{-1}(C_k(z))$ has an open neighborhood
contained in the preimage.
Therefore $f^{-1}(C_k(z))$ is open.

Since the preimage of every basis open set is open,
$f$ is continuous.

\medskip

\noindent\textbf{Ultrametric formulation.}
Let $d$ denote the ultrametric distance
(II.D13, Chapter~\ref{ch:ultrametric-depth}),
defined by $d(x, y) = 2^{-\delta(x,y)}$
where $\delta(x, y)$ is the first disagreement depth.
We claim that $f$ is $1$-Lipschitz:
\[
    \boxed{d\bigl(f(x),\, f(y)\bigr)
    \;\leq\;
    d(x,\, y)
    \qquad \text{for all } x, y \in \tau^3.}
\]
To see this, suppose $\delta(x, y) = k$,
meaning $x$ and~$y$ agree at stages $1, \ldots, k$
but disagree at stage~$k+1$.
In particular, $\pi_k(x) = \pi_k(y)$,
so $y \in C_k(x)$.
By Lemma~\ref{lem:naturality-cylinder},
$f(y) \in C_k(f(x))$,
which means $\pi_k(f(y)) = \pi_k(f(x))$,
i.e., $f(x)$ and~$f(y)$ agree at stage~$k$.
Therefore $\delta(f(x), f(y)) \geq k = \delta(x, y)$, and so
\[
    d\bigl(f(x), f(y)\bigr)
    \;=\; 2^{-\delta(f(x), f(y))}
    \;\leq\; 2^{-\delta(x, y)}
    \;=\; d(x, y).
\]
Every $1$-Lipschitz map is continuous.
In fact, $f$ is \emph{uniformly} continuous.