%
\label{thm:finite-spectral-support}
Let $f = (f_k)_{k \geq 1}$
be a $\tau$-holomorphic function on~$\tau^3$.
Then for each stage $k \geq 1$,
the \textbf{spectral support}
\[
    S_k(f)
    \;:=\;
    \bigl\{\,
    (p, v, \sigma) \in \Lambda_k
    \;\big|\;
    \varphi_{p,v}^{(\sigma)} \neq 0
    \,\bigr\}
\]
is a \textbf{finite} set.
Moreover, the full spectral support
\[
    \boxed{%
    S(f)
    \;:=\;
    \bigcup_{k \geq 1} S_k(f)
    \;\subset\;
    \Lambda_\tau}
\]
satisfies the following:
at each finite stage~$k$,
only finitely many basis elements are active.
No $\tau$-holomorphic function requires
an uncountable collection of spectral coefficients.

The proof has two components.

\smallskip
\noindent\textbf{Stage-wise finiteness.}
At each stage~$k$,
the domain of~$f_k$ is the finite set
$\mathbb{Z}/P_k\mathbb{Z}$.
A function on a finite set
with values in a finite-dimensional algebra
is determined by finitely many values.
The indexing set $\Lambda_k$ is finite
(Remark~\ref{rem:ch35-finiteness}),
so $S_k(f) \subseteq \Lambda_k$ is finite.

\smallskip
\noindent\textbf{No uncountable basis needed.}
Book~I's unique infinity theorem (I.T36, Book~I)
establishes that the $\tau$-framework
admits a single infinity $\omega$
as the profinite limit of finite stages.
The spectral support $S(f) = \bigcup_{k} S_k(f)$
is a countable union of finite sets,
hence at most countable.
But each stage's contribution
is itself finite,
so the spectral data of~$f$
is a coherent family of finite objects---not
an independently existing infinite collection.
The function~$f$ is determined
by its finite-stage approximations,
and each approximation involves
only finitely many basis elements.