Show inclusion map extends to an isometry

[psie]

Homework Statement

Show that when $Y$ is a completion of $X$, then the inclusion map $X\to Y$ extends to an isometry of $\tilde X$ onto $Y$.

Relevant Equations

We say that a complete metric space $Y$ is the completion of $X$ if $X$ is a dense subspace of $Y$. Above $\tilde X$ is the set of equivalence classes of the set of Cauchy sequences in $X$, under the relation $(s_n)\sim(t_n)$ defined by $d(s_n,t_n)\to0$ as $n\to\infty$. The metric on $\tilde X$ is $\rho(\tilde{s},\tilde{t})=\lim_{n\to\infty} d(s_n,t_n)$, where $\tilde s,\tilde t$ are equivalence classes in $\tilde X$.

I'm working an exercise on the completion of metric spaces. This exercise is from Gamelin and Greene's book and part of an exercise with several parts to it. I have already shown that $\sim$ is an equivalence relation, $\rho$ is a metric on $\tilde X$, $(\tilde X,\rho)$ is complete and that $X$ gets mapped onto a dense subset of $\tilde X$ under the map that $x\mapsto \tilde x$, where $\tilde x$ is the equivalence class of the constant sequence $(x,x,\ldots)$.

However, I am really stuck at this last part of the exercise. What confuses me mightily is that the metric on $Y$ seems unspecified. As far as I understand, we want to find a map $f:\tilde X\to Y$ such that $f$ is an isometry and that $f\circ e=i$, where $e:X\to\tilde X$ is the map $x\mapsto \tilde x$ and $i:X\to Y$ the inclusion map. How can we check $f$ is an isometry without knowing the metric on $Y$?

 

[andrewkirk]

We do know some things about the metric on $Y$.
By dint of $Y$ being a completion of $X$, we know that $d_Y(i(x1),i(x2))=d_X(x1,x2)$.

So we only need to consider $d_Y(y,i(x1))$ and $d_Y(y1,y2)$ where $y, y1,y2\in Y - i(X)$.

To do that, we need to define $f$. For $w\in e(X)$ we can do that using the equation you wrote, since the functions $e$ and $i$ are known.

We then need to define $f(w)$ for $w\in \tilde X - e(X)$. Since each such $w$ is a collection of non-convergent Cauchy sequences in $X$, we can map it to the limit in $Y$ of the image under $i$ of such sequences.
We use such mappings to complete our function $f$.

Having completed our function $f$, the fact that it preserves distances will follow naturally from the way we have constructed it.

For $f$ to be an isometry requires it to be a bijection, so we also need to prove there are no elements of $Y$ outside the image of the the completed function $f$, and that no two points in $X$ map to the same point in $Y$, but that should be pretty easy.

 

[psie]

Thanks. I do not follow part of your reasoning.

By dint of $Y$ being a completion of $X$, we know that $d_Y(i(x1),i(x2))=d_X(x1,x2)$.

1. Why is this known? I have shown that the map $x\mapsto\tilde x$ is an isometry, but you seem to be stating that the inclusion map is also an isometry, which I don't see how to show.

So we only need to consider $d_Y(y,i(x1))$ and $d_Y(y1,y2)$ where $y, y1,y2\in Y - i(X)$.

2. I don't understand how this follows from your previous point. Why do we only need to consider $d_Y(y,i(x1))$ and $d_Y(y1,y2)$ where $y,y1,y2\in Y-i(X)$?

To do that, we need to define $f$. For $w\in e(X)$ we can do that using the equation you wrote, since the functions $e$ and $i$ are known.

We then need to define $f(w)$ for $w\in \tilde X - e(X)$. Since each such $w$ is a collection of non-convergent Cauchy sequences in $X$, we can map it to the limit in $Y$ of the image under $i$ of such sequences.
We use such mappings to complete our function $f$.

Having completed our function $f$, the fact that it preserves distances will follow naturally from the way we have constructed it.

3. I don't see from the way we've constructed $f$ that it is an isometry. Here I feel like the metric on $Y$ matters, since $f$ is an isometry if $d_Y(f(\tilde x_1),f(\tilde x_2))=\rho(\tilde x_1,\tilde x_2)$.

For $f$ to be an isometry requires it to be a bijection, so we also need to prove there are no elements of $Y$ outside the image of the the completed function $f$, and that no two points in $X$ map to the same point in $Y$, but that should be pretty easy.

4. I guess we can simply choose the codomain of $f$ to be the image of $f$. How would one show injectivity?

 

[psie]

My main concern was in showing that $f$ is an isometry. I think I've figured it out.

To see that $f$ is an isometry, we simply have to verify that $$d_Y(f(\tilde s),f(\tilde t))=\rho(\tilde s,\tilde t).$$The right-hand side is by definition simply $\lim_{k\to\infty}d(s_k,t_k)$ and since the left-hand side is $d_Y(\lim_{k\to\infty}s_k,\lim_{k\to\infty}t_k)=\lim_{k\to\infty}d_Y(s_k,t_k)=\lim_{k\to\infty}d(s_k,t_k)$, so indeed, we have that $f$ is an isometry by continuity of the metric.

 

[WWGD]

Maybe you can test, verify using the Reals as the Metric Completion of the Rationals. Y-X will consist of the Irrationals; introduced as putative limits of sequences of Rationals that don't converge within the Rationals themselves.

 

[nuuskur]

$Y$ is any completion of $(X,d)$ and that means by definition there is a metric $\rho$ on $Y$ such that there exists an isometry $j:X\to Y$ that is almost surjective (i.e $j(X)$ is dense everywhere in $Y$).

The point of the exercise is to show that any two completions of $(X,d)$ are isometrically isomorphic. And to that end it suffices to show that this abstract version of completion is isomorphic to the quotient space $\overline{X}$, which equipped with the metric $d^*([(x_n)],[(y_n)]) := \lim d(x_n,y_n)$.

1. Define $f:\overline{X}\to Y, [(x_n)] \mapsto \lim j(x_n) \in Y$ and show that $f$ is well defined.
2. Now take two classes $[(x_n)], [(y_n)]\in\overline{X}$ and we can compute:
   $$\begin{align*} \rho (f[(x_n)], f[(y_n)]) &= \rho (\lim j(x_n), \lim j(y_n)) = \rho (\lim (j(x_n),j(y_n))) = \lim \rho (j(x_n),j(y_n)) \\ &= \lim d(x_n,y_n) = d^* ([(x_n)], [(y_n)]). \end{align*}$$
3. Finally, verify that $f\circ e = j$, where $e:X\to\overline{X}$ is the inclusion. Take $x\in X$, then
   $$fe(x) = f[(x)] = \lim j(x) = j(x).$$

All the heavy lifting and epsilonics are contained within the process of defining $\overline{X}$.