\section{The Hausdorff Completion}
\label{section:tvs-complete}

\begin{definition}[Hausdorff Completion of TVS]
\label{definition:tvs-completion}
    Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
    \begin{enumerate}
        \item $\wh E$ is a complete separated TVS.
        \item $\iota \in L(E; \wh E)$.
        \item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
    \end{enumerate}

    Moreover,
    \begin{enumerate}
        \item[(4)] $\iota(E)$ is dense in $\wh E$.
    \end{enumerate}

    The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
\end{definition}
\begin{proof}
    All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the \hyperref[Hausdorff completion]{definition:hausdorff-completion}.

    Using \autoref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \autoref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
    \[
    \xymatrix{
    \widehat E \times \widehat E \ar@{->}[r] & \widehat E &  & K \times \widehat E \ar@{->}[r] & \widehat E \\
    E \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u] &  & K \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u]
    }
    \]


    By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
\end{proof}

\begin{remark}[{{\cite[Section I.1]{SchaeferWolff}}}]
\label{remark:hausdorff-completion-field}

	The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
\end{remark}