% Reusable material removed from Appendix A on 2026-07-27.
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% This was the subsection "Profinite topology and ultrametric on a profinite
% set".  Section 3.3 now uses only the standard metric attached to a single
% inverse sequence of finite sets, so this more general discussion is not
% needed in the paper.
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% notation inconsistencies (r versus v) and claims that require a separate
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\subsection{Profinite topology and ultrametric on a profinite set}
In this subsection, let $J$ be a small category, $F\colon J \to \FinSet$ be a functor, and $\{\alpha_j\colon P \to Fj\}_{j\in \ob(J)}$ be a limit cone over the functor
\[
J \xrightarrow{F} \FinSet \xrightarrow{\text{discrete}} \Top.
\]

As a topological space,
$P$ is profinite, i.e. compact Hausdorff and totally disconnected. (This is a well known fact, for example, see \memo{Cite} for a proof.)

\begin{definition}[distinguishing value]
    For two elements $p,q \in P$, we define its \demph{distinguishing value} $r(p,q)\in \N\sqcup\{\infty\}$ by
    \[
    v(p,q) \coloneqq \inf\{|Fj| \in \N\mid j\in \ob(J), \;\alpha_j(p) \neq \alpha_j(q)\}.
    \]
\end{definition}

\begin{lemma}\label{lem:LemmaForUltrametric}
    With the above settings, we have the following:
    \begin{enumerate}
        \item For any $p, q\in P$, we have $v(p,q) = \infty \iff p=q$.
        \item For any $p,q \in P$, we have $v(p,q) = v(q,p)$.
        \item For any $p,q,r$, we have $\min (v(p,q), v(q,r)) \leq v(p,r)$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    To prove $(1)$, we observe that $p=q$ if and only if $\alpha_j(p) = \alpha_j(q)$ for any $j\in \ob(J)$ if and only if $v(p,q) = \infty$.

    $(2)$ is trivial by definition.

    We prove $(3)$. If $p=r$, then $(1)$ implies $v(p,r)=\infty$, and the inequality. If $p\neq r$, let $j\in \ob(J)$ be an object such that $\alpha_j(p)\neq \alpha_j(r)$ and $v(p,r) = |F_j|$. Then, either $\alpha_j(p) \neq \alpha_j(q)$ or $\alpha_j(q) \neq \alpha_j(r)$ holds. Therefore, either $v(p,q) \leq |Fj| = v(p,r)$ or $v(q,r) \leq |Fj| = v(p,r)$ holds. This implies the inequality.
\end{proof}

\begin{proposition}\label{prop:UltrametricOfProfiniteDiagram}
    The function $d(p,q)\coloneqq 2^{-v(p,q)}$ is an ultrametric over the set $P$, that is, $(P, d)$ satisfies the following conditions.
    \begin{enumerate}
        \item For any $p, q\in P$, we have $d(p,q) = 0 \iff p=q$.
        \item For any $p, q\in P$, we have $d(p,q) = d(q,p)$.
        \item For any $p, q, r\in P$, we have $\max(d(p,q),d(q,r)) \geq d(q,r)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    This follows from \Cref{lem:LemmaForUltrametric}.
\end{proof}

In this metric space, the inequality $d(p,q) \leq \frac{1}{2^N}$ holds if and only if $|Fj|<N$ implies $\alpha_j(p) = \alpha_j(q)$ for any $j\in \ob(J)$.

\begin{remark}[This metric does NOT induce the profinite topology in general.]
    Even if the indexing category $J$ is cofiltered, this metric structure does not necessarily induce the profinite topology on $P$.
For example, the Cantor space $2^{\N}$ is the cofiltered limit of the finite products $2^S$ for all finite subsets $S\subset \N$. Every pair of distinct elements $(p,q)$ can be distinguished by $2^{\{k\}}$ with a singleton $\{k\}\subset \N$, so $r(p,q) = 2$ and $d(p,q)=2^{-2}$. This proves that the ultrametric $d$ induces the discrete topology on $2^{\N}$.
\end{remark}

\begin{proposition}
    The identity function $(P,d) \xrightarrow{\id_P} (P, \mathcal{O}_{\text{initial}})$ is continuous.
    Under the following assumption, the continuous function is homeomorphic.
    \begin{itemize}
        \item For any $n\in \N$, there exists an object $t_n \in \ob(J)$ such that, for any $j\in J$ such that $|Fj|<n$, there exists at least one morphism $t_n \to j $ in $J$.
    \end{itemize}
\end{proposition}
\begin{proof}
    Let us take an element $p\in P$ and an open neighborhood $p\in U$ with respect to the initial topology. Then, there exist a finite number of objects $j_1 , \dots, j_n$ such that $p\in \cup_{i=1}^n \alpha_{j_i}^{-1}(\alpha_{j_i}(p))\subset U$. Let $N$ be the maximum value of $|Fj_i|$ among all $1\leq i\leq n$. If $d(p,q)\leq\frac{1}{2^{N+1}}$, then we obtain $\alpha_{j_i}(p) = \alpha_{j_i}(q)$ which implies $q\in \alpha_{j_i}^{-1}\alpha_{j_i}(p)$ and $q\in U$. This completes the proof of the continuity of $(P, d) \to (P, \mathcal{O}_{\text{initial}})$.

    Suppose the assumption in the statement. We prove that the identity function $(P,d) \xrightarrow{\id_P} (P, \mathcal{O}_{\text{initial}})$ is homeomorphic. Take a point $p\in P$ and an arbitrary positive real number $\epsilon>0$. Then let $N$ be the minimum natural number such that $\frac{1}{2^N}< \epsilon$. By the assumption, we take an object $t_N\in \ob(J)$ with the required conditions. We prove that the open subset $p\in \alpha_{t_N}^{-1}\alpha_{t_N}(p) \in \mathcal{O}_{\text{initial}}$ is subsumed by the $\epsilon$-ball of the point $p$. Take an arbitrary element $q\in \alpha_{t_N}^{-1}\alpha_{t_N}(p)$. Then, for any $j\in \ob(J)$ with $|F_j|<N$, we have a morphism $\beta \colon t_N \to j$ in $J$. Therefore, we have $\alpha_j(p) = \beta(\alpha_{t_N}(p))= \beta(\alpha_{t_N}(q)) = \alpha_j(q)$. This proves $d(p,q) \leq \frac{1}{2^N}<\epsilon$.
\end{proof}

\begin{corollary}\label{cor:ProfiniteCompletionCompleteMetric}
    Let $\T$ be an equational theory with only finitely many functional symbols, and $A$ be a $\T$-algebra. Then, its profinite completion $\pro{A}$ is a complete metric space with respect to its distinguishing metric.
\end{corollary}
\begin{proof}
    This follows since $\fQuo(A)$ is filtered, and there are only finitely many isomorphism classes for $n$-element $\T$-algebras.
\end{proof}