\section{The Hausdorff Completion}
\label{section:completion}


\begin{definition}[Hausdorff Completion]
\label{definition:hausdorff-completion}
    Let $(X, \fU)$ be a uniform space, then there exists a $(\wh X, \iota)$ such that:
    \begin{enumerate}
        \item $(\wh X, \wh \fU)$ is a complete Hausdorff uniform space.
        \item $\iota \in UC(X; \wh X)$.
        \item[(U)] For any complete Hausdorff uniform space $Y$ and Cauchy continuous function $f: X \to Y$, there exists unique $F \in C(\wh X; Y)$ such that the following diagram commutes

        \[
        \xymatrix{
         & \wh X \ar@{->}[rd]^{F} &  \\
        X \ar@{->}[rr]_{f} \ar@{->}[ru]^{\iota} &  & Y
        }
        \]


        Moreover, if $f \in UC(X; Y)$, then $F \in UC(\wh X; Y)$.
    \end{enumerate}
    Moreover,
    \begin{enumerate}
        \item[(4)] For any symmetric entourage $V \in \fU$, let
        \[
        \wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
        \]
        then the family $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$ forms a fundamental system of entourages for $\wh X$. In particular, $\bracs{\wh V \cap \iota(X)| V \in \fU, V \text{ symmetric}}$ is a fundamental system of entourages for $\iota(X)$.

        \item[(5)] $\iota(X)$ is dense in $\wh X$.
    \end{enumerate}



    The pair $(\wh X, \iota)$ is the \textbf{Hausdorff completion} of $X$.
\end{definition}
\begin{proof}[Proof {{\cite[Theorem 2.3.3]{Bourbaki}}}. ]
    (1, Uniform), (4): Let $\wh X$ be the set of all minimal Cauchy filters on $X$. For each $V \in \fU$, let
    \[
    \wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
    \]

    and $\wh \fB = \bracsn{\wh V| V \in \fU, V \text{ symmetric}}$, then
    \begin{enumerate}
        \item[(FB1)] Let $\wh U, \wh V \in \wh \fB$. By \autoref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \subset U \cap V$. In which case, for any $(\fF, \mathfrak{G}) \in \wh W$, there exists $E \in \fF \cap \mathfrak{G}$ with $E \times E \subset W \subset U \cap V$. Thus $\wh W \subset \wh U \cap \wh V$.
        \item[(UB1)] Let $\wh U \in \wh \fB$ and $\fF \in \wh X$, then since $\fF$ is Cauchy, there exists $E \in \fF$ such that $E \times E \subset U$, so $(\fF, \fF) \in \wh U$.
        \item[(UB2)] Let $\wh U \in \wh \fB$. By \autoref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $W \in \fU$ such that $W \circ W \subset U$. For any $\fF, \mathfrak{G}, \mathfrak{H} \in \wh X$ such that $(\fF, \mathfrak{G}), (\mathfrak{G}, \mathfrak{H}) \in \wh W$, there exists $W$-small sets $E \in \fF \cap \mathfrak{G}$ and $F \in \fF \cap \mathfrak{H}$. Since $\mathfrak{G}$ is a filter, $E \cap F \ne \emptyset$ by (F2) and (F3). By \autoref{lemma:small-intersect}, $E \cup H$ is $W \circ W$-small and thus $U$-small. Using (F1), $E \cup H \in \fF \cap \mathfrak{H}$, so $(\fF, \mathfrak{H}) \in \wh U$. Therefore $\wh W \circ \wh W \subset U$.
    \end{enumerate}
    By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\wh \fU \supset \wh \fB$. Moreover, $\wh \fB$ consists of symmetric entourages by construction.

    (1, Hausdorff): It is sufficient to show that $\Delta$ is closed and use (6) of \autoref{definition:hausdorff}. Let $(\fF, \mathfrak{G}) \in \ol{\Delta}$, then $(\fF, \mathfrak{G}) \in U$ for all $U \in \fU$ closed. Let $\fB = \bracs{F \cup G| F \in \fF, G \in \mathfrak{G}}$, then
    \begin{enumerate}
        \item[(FB1)] For any $F \cup G, F' \cup G' \in \fB$, $(F \cup G) \cap  (F' \cup G') \supset (F \cap F') \cup (G \cap G') \in \fB$.
        \item[(FB2)] By (F3), $\emptyset \not\in \fF \cup \mathfrak{G}$, so $\emptyset \not\in \fB$.
    \end{enumerate}
    Thus $\fB$ is a filter base by \autoref{proposition:filterbasecriterion}, and the filter $\mathfrak{H}$ generated by $\fB$ is contained in $\fF$ and $\mathfrak{G}$. By \autoref{proposition:goodentourages}, for every $U \in \fU$, there exists a $U$-small set $E \in \fF \cap \mathfrak{G} \subset \fB \subset \mathfrak{H}$. So $\mathfrak{H} \subset \fF, \mathfrak{G}$ is a Cauchy filter. By minimality of $\fF$ and $\mathfrak{G}$, $\fF = \mathfrak{G} = \mathfrak{H}$.

    (2): For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter by (1) of \autoref{proposition:cauchyfilterlimit}. Define $\iota: X \to \wh X$ by $x \mapsto \cn(x)$. Let $\wh U \in \wh \fU$ and $(\cn(x), \cn(y)) \in \wh U$, then there exists a $U$-small set $E \in \cn(x) \cap \cn(y)$. By (V1), $(x, y) \in E \times E \in U$.

    Conversely, if $(x, y) \in U$, then $E = U(x) \cap U(y) \in \cn(x) \cap \cn(y)$ by (F2), and $E$ is $U$-small by symmetry of $U$.

    Thus $(\iota \times \iota)^{-1}(\wh U) = U$, $(\iota \times \iota)^{-1}: \wh \fU \to \fU$ is a bijection, and $\iota \in UC(X; \wh X)$.

    (4): Let $\fF \in \wh X$ and $\wh U \in \wh \fU$. Since $\fF$ is a Cauchy filter, there exists a $U$-small set $E \in \fF$. Using \autoref{proposition:cauchyinterior}, assume without loss of generality that $E$ is open. Let $x \in E$, then $E \in \cn(x)$, so $(\fF, \iota(x)) \in \wh \fU$, and $X$ is dense in $\wh X$ by (3) of \autoref{definition:closure}.

    (1, Complete): Using \autoref{lemma:complete-dense}, it is sufficient to show that every Cauchy filter in $\iota(X)$ converges in $\wh X$.

    Let $\wh \fF \in 2^{\iota(X)}$ be a Cauchy filter, then since $\iota: X \to \iota(X)$ is surjective, $\iota^{-1}(\wh \fF)$ is a filter base by \autoref{proposition:preimage-filterbase}.

    For any $U \in \fU$, there exists a $\wh U$-small set $\wh E \in \wh \fF$. Since $U = (\iota \times \iota)^{-1}(\wh U)$, $\iota^{-1}(\wh E) \subset U$, so $\iota^{-1}(\wh \fF)$ is a Cauchy filter base.

    Let $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$ be a minimal Cauchy filter, then $\iota(\mathfrak{G})$ is a Cauchy filter base by \autoref{proposition:imagecauchy}.

    Let $U \in \fU$. By \autoref{proposition:cauchyinterior}, there exists a $U$-small open set $E \in \mathfrak{G}$. For any $x \in E$, $E \in \cn(x)$ by \autoref{lemma:openneighbourhood}, so $(\cn(x), \mathfrak{G}) = (\iota(x), \mathfrak{G}) \in \wh U$, and $\iota(E) \subset \wh U(\mathfrak{G})$. Since $E \in \mathfrak{G}$, $\wh U(\mathfrak{G}) \supset \iota(E) \in \iota(\mathfrak{G})$, so $\iota(\mathfrak{G})$ converges to $\mathfrak{G}$.

    Now, given that $\mathfrak{G} \subset \iota^{-1}(\wh \fF)$, $\iota(\mathfrak{G}) \subset \iota(\iota^{-1}(\fF)$, so $\iota(\iota^{-1}(\fF))$ converges to $\mathfrak{G}$ as well. Since $\fF$ is a filter on $\iota(X)$, $\iota(\iota^{-1}(\fF)) = \fF$, thus $\fF$ is convergent.

    (U): Let $(Y, \mathfrak{V})$ be a complete Hausdorff uniform space and $f \in UC(X; Y)$. For each $\fF \in \wh X$, let $F(\fF) = \lim_{x, \fF}f(x)$.  For any $x \in X$, $F(\iota(x)) = \lim_{y \to x}f(y) = f(x)$, so $f = F \circ \iota$. Since $\fF$ is a Cauchy filer, $f \in UC(X; Y)$, and $Y$ is a complete Hausdorff uniform space, the limit exists and is unique. Thus $F: \iota(X) \to Y$ is well-defined.

    Let $U \in \mathfrak{V}$, then there exists a symmetric entourage $V \in \fU$ such that $(f(x), f(y)) \in U$ for all $(x, y) \in V$, then for any $(\iota(x), \iota(y)) \in \wh V \cap (\iota(X) \times \iota(X))$,  $(F(\iota(x)), F(\iota(y))) \in U$, so $F$ is uniformly continuous. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\td F \in C(\wh X; Y)$ such that $\td F|_{\iota(X)} = F|_{\iota(X)}$, and any such $\td F$ is uniformly continuous.
\end{proof}

\begin{lemma}
\label{lemma:completion-of-hausdorff}
    Let $(X, \fU)$ be a Hausdorff uniform space, then the canonical map $\iota: X \to \wh X$ is an embedding.
\end{lemma}
\begin{proof}
    Since the mapping $(\iota \times \iota)^{-1}$ is a bijection between two bases of uniformities of $X$ and $\wh X$, it is sufficient to show that $\iota$ is injective. To this end, observe that for any $x, y \in X$, $\cn(x) = \cn(y)$ if and only if $x = y$ by (4) of \autoref{definition:hausdorff}.
\end{proof}

\begin{definition}[Associated Hausdorff Uniform Space]
\label{definition:associated-hausdorff}
    Let $(X, \fU)$ be a uniform space, then there exists $(X', i)$ such that:
    \begin{enumerate}
        \item $X'$ is a Hausdorff uniform space.
        \item $i \in UC(X; X')$.
        \item For any pair $(Y, f)$ satisfying (1) and (2), there exists a unique $f' \in UC(X'; Y)$ such that the following diagram commutes

        \[
        \xymatrix{
        X' \ar@{->}[rd]^{F} &  \\
        X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y
        }
        \]

    \end{enumerate}
    known as the \textbf{Hausdorff uniform space associated with} $(X, \fU)$.
\end{definition}
\begin{proof}[Proof {{\cite[Proposition 2.8.16]{Bourbaki}}}. ]
    Let $(\wh X, \iota)$ be the Hausdorff completion of $X$, $X' = \iota(X)$, and $i = \iota$, then $(X', i)$ satisfies (1) and (2).

    (U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using \autoref{lemma:completion-of-hausdorff}, identify $Y$ as a subspace of $\wh Y$. By (U) of the Hausdorff completion, there exists a unique $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:

    \[
    \xymatrix{
    \wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
    X' \ar@{->}[u]^{\iota} \ar@{->}[r]^{F} & Y' \ar@{->}[u]_{\iota} \\
    X \ar@{->}[r]_{f} \ar@{->}[u]^{i} & Y \ar@{=}[u]
    }
    \]


    Since $\ol F(X') \subset \iota(Y) = Y'$, $F = \ol F|_{X'} \in UC(X'; Y')$ is continuous. By \autoref{lemma:completion-of-hausdorff}, $\iota \in UC(Y; Y')$ is a homeomorphism. Upon identifying $Y$ with $Y'$, $F \in UC(X'; Y)$ is the desired map.
\end{proof}

\begin{proposition}
\label{proposition:hausdorff-uniform-factor}
    Let $X, Y$ be uniform spaces, $X', Y'$ be their associated Hausdorff uniform spaces, and $\wh X, \wh Y$ be their Hausdorff completions, then there exists a unique $F \in UC(X'; Y')$ and $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:

    \[
    \xymatrix{
    \wh X \ar@{->}[r]^{\overline{F}} & \wh Y \\
    X' \ar@{->}[u] \ar@{->}[r]^{F} & Y' \ar@{->}[u] \\
    X \ar@{->}[r]_{f} \ar@{->}[u] & Y \ar@{->}[u]
    }
    \]

\end{proposition}
\begin{proof}
    By (U) of \autoref{definition:hausdorff-completion} and \autoref{definition:associated-hausdorff}.
\end{proof}


\begin{proposition}
\label{proposition:initial-completion}
    Let $X$ be a set, $\seqi{Y}$ be uniform spaces, $\seqi{f}$ with $f_i: X \to Y_i$ for each $i \in I$. Let $\fU$ be the initial uniformity on $X$ induced by $\seqi{f}$, and $(\wh X, \iota_X)$ be the Hausdorff completion of $X$.
    \begin{enumerate}
        \item For each $i \in I$, let $(\wh Y_i, \iota_i)$ be the Hausdorff completion of $Y_i$, then there exists a unique $F_i \in UC(\wh X; \wh Y_i)$ such that the following diagram commutes
        \[
        \xymatrix{
        \wh X \ar@{->}[r]^{F_i} & \wh Y \\
        X \ar@{->}[r]_{f_i} \ar@{->}[u] & Y \ar@{->}[u]
        }
        \]

        \item The uniformity of $\wh X$ is the initial uniformity induced by $\seqi{F}$.
        \item There exists a unique $F \in UC(X; \prod_{i \in I}\wh Y_i)$ and $\ol{F} \in UC(\wh X; \prod_{i \in I}\wh Y_i)$ such that the following diagram commutes

        \[
        \xymatrix{
         & \prod_{i \in I} \wh Y_i \ar@{->}[rd]^{\pi_i} &  \\
         & \wh X \ar@{->}[r]^{F_i} \ar@{->}[u]_{\overline{F}} & \wh Y_i \\
        X \ar@{->}[ruu]^{F} \ar@{->}[rr]_{f_i} \ar@{->}[ru]_{\iota_X} &  & Y_i \ar@{->}[u]_{\iota_i}
        }
        \]


        Moreover, $\ol{F}(\wh X) = \overline{F(X)}$, and $\ol{F}$ is an embedding.
    \end{enumerate}
    In particular, by \autoref{proposition:dense-product}, there is a natural isomorphism
    \[
    \prod_{i \in I}\wh X_i \iso \wh{\prod_{i \in I}X_i}
    \]

    induced by extending the identity on $\pi_{i \in I}X_i$.
\end{proposition}
\begin{proof}
    (1): By (U) of \autoref{definition:hausdorff-completion}.

    (2), (3): By (U) of the \hyperref[product]{definition:product-topology}, there exists $f \in UC(X; \prod_{i \in I}Y_i)$ such that the following diagram commutes:

    \[
    \xymatrix{
     & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} &  \\
    X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} &  & Y_i
    }
    \]


    First suppose that $X$ and $\seqi{Y}$ are Hausdorff. For each $i \in I$, $\iota_i \circ \pi_i \in UC(\prod_{i \in I} Y_i; \wh Y_i)$, so by (U) of \autoref{proposition:product-complete}, there exists a unique $\iota_P \in UC(\prod_{i \in I} Y_i; \prod_{i \in I} \wh Y_i)$ such that the following diagram commutes

    \[
    \xymatrix{
    \prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
    \prod_{i \in I}Y_i \ar@{->}[r]^{\pi_i} \ar@{^{(}->}[u]^{\iota_P} & Y_i \ar@{^{(}->}[u]_{\iota_i}
    }
    \]


    for all $i \in I$. By \autoref{lemma:completion-of-hausdorff}, each $\iota_i \in UC(Y_i; \wh Y_i)$ is an embedding, so \autoref{proposition:product-topology-embedding} implies that $\iota_P$ is an embedding as well.

    Since $X$ has the initial topology, $f: X \to \prod_{i \in I}Y_i$ is an embedding by \autoref{proposition:initial-product-topology}. Thus the composition $\iota_P \circ f$ is an embedding. As $\prod_{i \in I}\wh Y_i$ is complete by \autoref{proposition:product-complete} and \autoref{proposition:product-hausdorff}, $\overline{\iota_P \circ f(X)} \subset \prod_{i \in I}\wh Y_i$ is a complete Hausdorff uniform space. Let $Z$ be a complete Hausdorff uniform space and $g \in UC(X; Z)$, then \autoref{theorem:uniform-continuous-extension} implies that there exists a unique $G \in UC(\overline{\iota_P \circ f(X)}; Z)$ such that the following diagram commutes

    \[
    \xymatrix{
    \prod_{i \in I}\wh Y_i &  \\
    \overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{->}[rd]^{G} &  \\
    X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[r]_{g} & Z
    }
    \]


    Thus $(\overline{\iota_P \circ f(X)}, \iota \circ f)$ satisfies (1), (2), and (U) of the \hyperref[Hausdorff completion]{definition:hausdorff-completion}. Therefore $\wh X$ may be identified as a subspace of $\prod_{i \in I}\wh Y_i$ as follows:

    \[
    \xymatrix{
    \prod_{i \in I}\wh Y_i \ar@{->}[r]^{\pi_i} & \wh Y_i \\
    \overline{\iota_P \circ f(X)} \ar@{^{(}->}[u] \ar@{=}[r] & \wh X \\
    X \ar@{^{(}->}[u]^{\iota_P\circ f} \ar@{->}[ru]_{\iota_X} &
    }
    \]


    In which case, $\wh X$ must be equipped with the initial topology induced by the projection maps.

    Now assume that $X$ and $Y$ are arbitrary. Let $X'$ and $\seqi{Y'}$ be the Hausdorff spaces associated with $X$ and $\seqi{Y}$, respectively. By \autoref{proposition:hausdorff-uniform-factor}, there exists $\seqi{f'}$ such that the following diagram commutes

    \[
    \xymatrix{
    X' \ar@{->}[r]^{f_i'} & Y_i' \\
    X \ar@{->}[u] \ar@{->}[r]_{f_i} & Y_i \ar@{->}[u]
    }
    \]


    for all $i \in I$. By (5) of \autoref{definition:hausdorff-completion}, there is a correspondence between the uniformities of $X$ and $X'$, and $Y$ and $Y'$. Thus $X'$ is equipped with the initial uniformity genereated by $\seqi{f'}$.
\end{proof}