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Section 2.4 Complete Metric Spaces

Definition 2.4.1. Cauchy sequences.
A sequence \(x_n\) in a metric space \((M,d)\) is Cauchy if, for every \(\eps>0\text{,}\) there is an \(N>0\) such that \(d(x_n,x_m)<\eps\) for every \(n,m>N\text{.}\)
Of course every converging sequence is cauchy but the inverse is not necessarily true.
The set of rationals \(\bQ\) with \(d(q_1,q_2)=|q_1-q_2|\) is a metric space but the sequence
\begin{equation*} q_1=3,q_2=3.1,q_3=3.14,q_3.141,\dots \end{equation*}
built using the sequence of digits of \(\pi\) in its decimal expression is obviously Cauchy by construction but also obviously does not converge in \(\bQ\text{.}\)
Definition 2.4.3. Complete metric spaces.
We say that \((M,d)\) is complete if every Cauchy sequence of \(M\) converges.
For instance, \(\bQ\) with its standard distance is not complete. On the other side, \(\bQ\) lies inside \(\bR\text{,}\) that is a metric space with respect to "the same" distance function (it would be more correct to say that the two distance functions agree on \(\bQ\)) and is complete. The theorem below shows that this is a general fact: given every incomplete metric space \((M,d)\text{,}\) there is always a complete metric space \((\hat M,\hat d)\) such that \(M\) injects into \(\hat M\) so that the restriction of \(\hat d\) to \(M\) is equal to \(d\text{.}\) Step 1: \(\hat M\text{.}\) First we need a candidate for \(\hat M\text{.}\) Since we are asking for Cauchy sequences to converge and each such limit point would be common to all Cauchy sequences that differ by a sequence converging to zero, it makes sense to consider the set
\begin{equation*} \cC = \{(x_1,x_2,\dots) : x_i\text{ is Cauchy}\}, \end{equation*}
to say that
\begin{equation*} (x_1,x_2,\dots) \sim (y_1,y_2,\dots)\text{ if }x_n-y_n\to0 \end{equation*}
and finally set
\begin{equation*} \hat M = \cC/\sim. \end{equation*}
Step 2: \(\hat d\text{.}\) The most natural candidate for \(\hat d\) is
\begin{equation*} \hat d(x,y) = \lim d(x_n,y_n). \end{equation*}
We need to show that this quantity is well defined and does not depend on the Cauchy sequence representative chosen in the equivalence classes of \(x\) and \(y\text{.}\)

The reason why the limit exists is that \(d(x_n,y_n)\) is a Cauchy sequence since, by the triangular inequality,
\begin{equation*} d(x_n,y_n)-d(x_m,y_m) \leq d(x_n,x_m)+d(x_m,y_n)-d(x_m,y_m) \leq d(x_n,x_m) + d(y_n,y_m) \end{equation*}
and both \(x_n\) and \(y_n\) are Cauchy.

If now \(x'_n\) and \(y'_n\) are equivalent respectively to \(x_n\) and \(y_n\text{,}\) so that \(d(x_n,x'_n)\to0\) and \(d(y_n,y'_n)\to0\text{,}\) then, by the triangular inequality,
\begin{equation*} \eqalign{ &d(x_n,y_n) - d(x'_n,y'_n) \leq d(x_n,x'_n) + d(x'_n,y_n) - d(x'_n,y'_n) \leq\\ &\leq d(x_n,x'_n) + d(x'_n,y'_n) + d(y'_n,y_n) - d(x'_n,y'_n) = d(x_n,x'_n) + d(y'_n,y_n), } \end{equation*}
so that \(d(x_n,y_n)\) and \(d(x'_n,y'_n)\) must have the same limit. Hence, \(\hat d\) is well defined.

Step 3: \((\hat M,\hat d)\) is complete. Let \(\xi_n=(x_{n,m})\) be a Cauchy sequence of equivalence classes of Cauchy sequences of \(M\text{,}\) namely, for each \(n_0\text{,}\) the sequence \(\xi_{n_0}=(x_{n_0,m})\) is Cauchy. We need to prove that \(\xi_n=(x_{n,m})\) converges to a Cauchy sequence. Since \(\xi_n\) is Cauchy for every \(n\text{,}\) then for every \(n\) there is a \(N_n>0\) such that \(d(x_{n,j},x_{n,k})<\frac{1}{n}\) for all \(j,k\geq N_n\text{.}\) We set \(u_n=x_{n,N_n}\text{.}\)

Then, for every \(k>0\text{,}\)
\begin{equation*} \eqalign{ d(u_n,u_m) &\leq d(u_n,x_{n,k}) + d(x_{n,k},x_{m,k}) + d(x_{m,k},u_m)\\ &= d(x_{n,N_n},x_{n,k}) + d(x_{n,k},x_{m,k}) + d(x_{m,k},x_{m,N_m}) \leq \frac{1}{n} + d(x_{n,k},x_{m,k}) + \frac{1}{m}, } \end{equation*}
so that, taking \(k\to\infty\text{,}\)
\begin{equation*} d(u_n,u_m) \leq d(x_{n,N_n},x_{n,k}) + d(x_{n,k},x_{m,k}) + d(x_{m,k},x_{m,N_m}) \leq \frac{1}{n} + \hat d(\xi_{n},\xi_{m}) + \frac{1}{m}. \end{equation*}
Since \(\xi_n\) is Cauchy with respect to \(\hat d\text{,}\) this shows that \(u_n\) is Cauchy with respect to \(d\text{.}\)

Moreover, \(\xi_n\to u\text{.}\) Indeed
\begin{equation*} \hat d(\xi_n,u) = \lim d(\xi_{n,k},u_k) \end{equation*}
and
\begin{equation*} d(\xi_{n,k},u_k) \leq d(\xi_{n,k},u_n) + d(u_n,u_k) = d(\xi_{n,k},x_{n,N_n}) + d(u_n,u_k)\text{.} \end{equation*}
For every \(\eps>0\) there is a \(K>0\) such that \(d(u_n,u_k)<\eps/2\) for all \(n,k\geq K\text{.}\) Moreover, by choosing also \(k\geq N_n\text{,}\) we get that \(d(\xi_{n,k},x_{n,N_n})<\frac{1}{n}\text{.}\) Hence, \(d(\xi_{n,k},u_k)\) can be made arbitrarily small by taking \(n\) and \(k\) large enough.

Step 4: \(( M, d)\) injects isometrically into \((\hat M,\hat d)\text{.}\) An isometric injection is given by \(i(x_0) = (x_0,x_0,\dots)\text{,}\) namely each point of \(M\) is sent to the corresponding constant sequence.

Step 5: \(\overline{i(M)}=\hat M\text{.}\) Let \(\xi\in\hat M\) and define the sequence \(\eta_n\) of constant sequences such that, for each \(n>0\text{,}\) \(\eta_{n,m}=\xi_n\) for every \(m>0\text{.}\) Notice that \(\eta_n\in i(M)\) for every \(n>0\text{.}\) Then
\begin{equation*} \hat d(\eta_n,\xi) = \lim_{m\to\infty} d(\eta_{n,m},\xi_m) = \lim_{m\to\infty} d(\xi_{n},\xi_m) \end{equation*}
and, since \(\xi\) is Cauchy, this means that, for every \(n\text{,}\) we can make \(\hat d(\eta_n,\xi)\) arbitrarily small by taking \(n\) large enough, namely \(\eta_n\to\xi\text{.}\) Hence, \(M\) is dense in \(\hat M\text{.}\)