\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Complete Metric Spaces
Dense Sets
Needed by:
None.
Links:
Sheet PDF
Graph PDF

Metric Completions

Why

We can always work with complete metric spaces.

The justification is that we can always, given an incomplete metric space, construct a larger metric space which contains a subset isomorphic to the original one.

Result

Let $(A, d)$ be an incomplete metric space. There exists a complete metric space $(B, d')$ with $C \subset B$ such that $(A, d)$ and $(C, d')$ are isometric and the image under the isometry of $C$ is dense in $B$.
Copyright © 2023 The Bourbaki Authors — All rights reserved — Version 13a6779cc About Show the old page view