What is the Cauchy completion of a metric space?

Question

I was wondering what the "Cauchy" completion of a metric space is. I can't find any helpful information on Google. Feel free to post any links to sources you find relevant.

Answer 1

Intuitively, you add enough points so that all Cauchy sequences are convergent.

Concretely, if $(X,d)$ is a metric space, you define $\overline X$ to be the quotient $\tilde X/R$, where $\tilde X$ is the set of all Cauchy sequences of $X$, and $R$ is the equivalence relation $$ (x_n)\,R\,(y_n)\iff\ d(x_n,y_n)\to0. $$ On $\overline X$ one defines the metric $$ d'(\,(x_n),(y_n)\,)=\lim_nd(x_n,y_n).$$ This distance is well-defined because one can use the triangle inequality and the fact that $\{x_n\}$ and $\{y_n\}$ are Cauchy to show that the sequence of numbers $\{d(x_n,y_n)\}$ is Cauchy.

Now one can embed $X\hookrightarrow \overline X$ by mapping $x\longmapsto (x)$, i.e. each $x$ maps to the class of all Cauchy sequences that converge to it. This embedding is clearly isometric, and dense: if $(x_n)\subset X$ is a Cauchy sequence and $\varepsilon>0$, there exists $n_0$ such that $d(x_n,x_m)<\varepsilon$ for all $n,m\geq n_0$; if $y=x_{n_0}$ then $$d((y),(x_n))=\lim_nd(y,x_n)=d (x_{n_0},x_n)<\varepsilon.$$

The only nontrivial thing is to check that $\overline X$ is complete. It is not actually complicated, it is cumbersome to write because we need to deal with sequences of sequences. If $(A_m)$ is a Cauchy sequence in $\overline X$, then each $A_m$ is the class of a Cauchy sequence $(A_{mn})_n\subset X$. So, for every $r\in\mathbb N$, there exists $n_r$ such that $$d'(A_m,A_\ell)<\frac1{2^r},\ \ \text{ for all } m,\ell\geq n_r.$$In turn, using the definition of $d'$, this means that there exists $m_r\in\mathbb N$, with $m_r\geq \max\{m_{r-1},n_r\}$, such that $$\tag{1} d(A_{n_r,k},A_{n_{r+1},k})<\frac1{2^r},\ \ \text{ for all }k\geq m_r,$$ and such that $$\tag{2} d(A_{n_r,h},A_{n_r,j})<\frac1{2^r},\ \ \ \text{ for all }h,j\geq m_r. $$ (this, because the sequence $A_{n_r}$ is Cauchy).

Note that $(1)$ implies, by the triangle inequality, that $$\tag3 d(A_{n_r,k},A_{n_{r+\ell},k})\leq\sum_{j=1}^\ell d(A_{n_{r+j-1},k},A_{n_{r+j},k}) <\sum_{j=1}^\ell \frac1{2^{r+j}}<\frac1{2^{r-1}}. $$

Now consider the sequence $(A_{n_r,m_r})_r\subset X$. This sequence is Cauchy, since for any $s\geq r$ \begin{align} d(A_{n_r,m_r},A_{n_s,m_s})&\leq d(A_{n_r,m_r},A_{n_r,m_s})+d(A_{n_r,m_s},A_{n_s,m_s})\\ \ \\ %&\leq d(A_{n_r,m_r},A_{n_r,m_s})+\frac1r\\ \ \ &\leq \frac1{2^r}+\frac1{2^{r-1}}<\frac1{2^{r-2}}. \end{align} (this first estimate by $(2)$, and the second one by $(3)$).

So the sequence $(A_{n_r,m_r})_r$ is Cauchy, and it is not hard to check that its class in $\overline X$ is the limit of $(A_m)_m$.

Answer 2

A completion of a metric space $X$ is another metric space $X^*$ with the following property:

1. $X$ embeds isometrically as a subspace of $X^*$,

2. $X^*$ is complete,

3. $X$ is dense in $X^*$.

Every metric space can be completed in a way that circumvents the cumbersome computations of the Cauchy/Hausdorff completion definition.

Explicitly, given any metric space $(X,d)$, one can embed it as a subspace of the metric space $\widehat{X}=\hom(X,\mathbb R)$, of continuous functions to $\mathbb R$ with the uniform metric, which one easily checks is complete. Then the closure of $X$ in $\widehat{X}$ is the desired completion, since closed subsets of complete spaces are complete.

The embedding is given by assigning $x\in X$ to the continuous function $\hat{x} : X\longrightarrow \mathbb R$ such that $\hat x(y) = d(x,y)$. This is clearly injective. To see this is continuous, note that by the triangle inequality $\sup_{z\in X}|\hat{x}(z)-\hat{y}(z)|\leqslant d(x,y)$, and this is attained by choosing $z=x$, so $d_\infty(\hat x,\hat y) = d(x,y)$ and the embedding above is, in fact, an isometry.

To see that $\widehat{X}$ is complete, suppose that $(f_n)$ is a sequence of functions that is uniformly Cauchy. Because for each $z\in X$ the sequence $(f_n(z))$ is Cauchy in $\mathbb R$, we can define $f(z) = \lim f_n(z)$ and show $(f_n)$ converges uniformly to $f$, which shows that $f$ is continuous and hence belongs to $\widehat{X}$.

Given $\varepsilon>0$ we can find $N$ such that $\sup_{z\in X}|f_m(z)-f_n(z)|<\varepsilon$ for $m,n>N$, and hence we estimate $|f(z)-f_n(z)| \leqslant |f(z)-f_m(z)|+\varepsilon$. For each fixed $z$ we now deduce, by letting $m\to \infty$, that $|f(z)-f_n(z)| \leqslant \varepsilon$ for $n>N$, which shows $f_n\to f$ uniformly.