I have to prove that every metric space is isometric to a dense subset of a complete metric space.

My proof: Let $X$ be the metric space, and $\{p\}$ the set of limits of all the cauchy sequences in $X$. Then $X\bigcup\{p\}$ is a complete metric space, and $X$ is isometric to $X$, which is a dense subset of $X\bigcup\{p\}$.

Is this proof correct? Because my book has a huge 2-page proof, which surely employs different arguments. A potential flaw in my argument might be the unsubstantiated statement that $X\bigcup \{p\}$ is a metric space if $X$ is a metric space.

Thanks in advance!

  • $\begingroup$ What does "the set of limits of all the cauchy sequences in X" mean? E.g. what is an element of this set? $\endgroup$ – Pete L. Clark Aug 12 '13 at 6:15
  • $\begingroup$ Let $\{x_i\}$ be a cauchy sequence in $X$, converging to limit $l_i$, which may or may not be within $X$. Then $l_i\in\{p\}$. I'm assuming the existence of such a limit, regardless of whether it is inside $X$ or not. Is this in invalid assumption? For example, $3,3.14,3.141,\dots$ does not have a limit in $\Bbb{Q}$, but it does have a limit (in $\Bbb{R})$. $\endgroup$ – fierydemon Aug 12 '13 at 6:18
  • $\begingroup$ $@$Ayush: In the example of $\mathbb{Q}$ you give, the place where your limit lives is a complete metric space into which your space is isometrically embedded. Are you assuming that an arbitrary metric space can be isometrically embedded in a complete space? If so, then your reasoning is fine but the result is almost trivial (take the closure!). If not, then aren't you not assuming the very thing you are trying to prove? $\endgroup$ – Pete L. Clark Aug 12 '13 at 6:22
  • $\begingroup$ Is the statement (albeit trivial) that any metric space can be embedded in a complete metric space incorrect? $\endgroup$ – fierydemon Aug 12 '13 at 6:25
  • $\begingroup$ Note that Prahlad's answer contains the standard way to proceed. Roughly speaking you take the limit of the Cauchy sequence to be the Cauchy sequence itself and then correct for the fact that different Cauchy sequences may have the same limit (in any completion). Then you need to check a bunch of details. You haven't proved anything yet: this one important hole aside, none of your statements are justified! $\endgroup$ – Pete L. Clark Aug 12 '13 at 6:26

Well, the limits of some of the Cauchy sequences in $X$ may not exist (because $X$ needs not be a complete space), so you need to specify what you mean by "the set of limits". For example, suppose your space consists precisely of points $x_n, n\in\mathbb N$, with $$d(x_n,x_{n+k})=\frac1{2^n}+\frac1{2^{n+1}}+\dots+\frac1{2^{n+k-1}}.$$ There is no limit $p$ to add to $X$.

Of course, you can say "Ah, well, then pick a new point $p$ and declare it to be the limit of the sequence". OK, fine. What is the distance between $x_{17}$ and $p$, in that case?

Now, consider the same example as in the first paragraph, and note that the sequence $x_3,x_6,x_9,\dots$ is Cauchy. Your description says we need to add a new limit point $p'$ corresponding to it. Is $p'\ne p$?

This suggests that the problem is slightly more complicated than anticipated: If two Cauchy sequences are to have the same limit, we better make sure that we add the same limit point for both of them. How do we know that two different sequences ought to have the same limit?

Let's say that we manage to solve all those obstacles, that is: We indeed add to $X$ new points, one for each Cauchy sequence in $X$, making sure that if two distinct Cauchy sequences are to converge to the same thing, they indeed do. We also somehow manage to extend the distance function so the new space is metric, and $X$ is dense in it. How do we know that there is not a Cauchy sequence of new points for which we haven't yet added a limit point? Because if there are such sequences, then we ought to iterate the procedure. For how long? Does it ever end? Now, if there are no such sequences, that is certainly part of what the proof needs to show.

All that being said, you are on the right track. First we need to deal with what things we are to add. Of course, new points, and the new points can be anything we want, but let's try to choose something specific so we can keep track of them. A natural thing to do is to exploit the fact that if two Cauchy sequences are to have the same limit, then we better add the same point as limit of both of them. One way to take advantage of this is to introduce an equivalence relation on Cauchy sequences, saying that two such sequences are equivalent "if they are to have the same limit". Then as the new points we can just add the equivalence classes of this equivalence relation.

How do we check that two distinct sequences have the same limit? Luckily, this is easy: Say the sequences are $x_1,x_2,\dots$ and $y_1,y_2,\dots$ Define a new sequence by $$z_1=x_1,z_2=y_1,z_3=x_2,z_4=y_2,z_5=x_3,\dots$$ Then the two sequences are equivalent iff the new sequence so described is Cauchy. It does not matter here whether there are repetitions in this sequence of $z_i$. (Naturally, there are things to verify here, mainly that this is indeed an equivalence relation.)

A small problem at this point is that some sequences may be Cauchy and already have a limit in $X$. The easiest solution is to ignore them. It may not be the prettiest solution, because now your space consists of two rather different creatures: Elements of $X$, and equivalence classes of Cauchy sequences of elements of $X$, that do not converge to an element of $X$. But it is a fine solution (meaning: It works). The standard approach is to avoid this separation of creatures, and simply take as the new space the collection of all equivalence classes of Cauchy sequences. (If a sequence converges to $x\in X$, we identify its equivalence class with $x$, so rather than the isometry being inclusion, at the end we have something a tad more elaborate.)

Now comes the second problem: How do we make this thing into a metric space? Luckily, there is an easy solution as well: If $x_n\to p$, then for any $k$, we have that $d(x_k,x_n)\to_{n\to\infty}d(x_k,p)$. So we can use this as the way to define the new distances: Given an equivalence class $p$, let $x\in X$. We define $d(x,p)$ as $\lim_{n\to\infty}d(x,x_n)$, where $ x_1,x_2,\dots$ is some Cauchy sequence in the equivalence class $p$. OK. Maybe not so easy: We need to check that this definition gives us a positive real number (as opposed to $0$ [excluded since the $x_n$ do not converge in $X$, so $x$ better not be their limit], or $+\infty$, or to the case when the limit does not exist). We also need to check that this number is independent of the sequence $x_1,x_2,\dots$ we picked. A similar idea gives us how to define $d(p,q)$ when both $p,q$ are equivalence classes.

Of course, one still needs to verify this indeed gives us a metric space.

Finally, we need to check that this is complete, and $X$ is dense in it. But I'll stop here, as I'm pretty sure the construction in your book is following the same lines.

(There are rather different presentations of the construction, that look superficially different and start from very different ideas, but the space obtained at the end is essentially unique, in the sense that any two constructions will be isometric via an isomorphism that identifies their copies of $X$. Naturally, this also takes an argument.)


Another possibility is to embed $X$ into a complete metric space. For example,

Lemma: Any metric space is isometric to a subspace of a complete normed space.

Proof. Let $(X,d)$ be a metric space and $F=C_b(X)$ be the set of continuous bounded functions $X \to \mathbb{R}$ endowed with the sup norm; then $F$ is a Banach space. Fix $x_0 \in X$. Then $\phi : x \mapsto d(x, \cdot)-d(x_0,\cdot)$ is an isometry from $X$ into $F$. $\square$

Notice that the image of $X$ by the isometry $X \hookrightarrow F$ is dense in the closure $\mathrm{cl}_F(X)$ of $X$ in $F$; moreover, $\mathrm{cl}_F(X)$ is complete.

  • $\begingroup$ How can you make that last inference that the image of $X$ by the isometry is dense in the closure of $X$ in $F$? $\endgroup$ – MadcowD Dec 3 '15 at 10:42
  • $\begingroup$ This is essentially true by definition: if $A \subset B$, then we say that $A$ is dense in $B$ precisely when $\overline{A}=B$. In particular, any subspace is dense in its closure. $\endgroup$ – Seirios Dec 5 '15 at 10:23

I am not sure what you mean by the set $\{p\}$. If you look at the limits of all cauchy sequences, then some of these limits would lie outside $X$, right?

The usual way to do this is to let $Y$ by the space whose elements are cauchy sequences in $X$. Define an equivalence relation on $Y$ by saying that $(x_n) \sim (y_n)$ iff $$ \lim_{n\to\infty} d(x_n,y_n) = 0 $$ Define a metric on the quotient $Y/\sim$ by $$ \delta([x_n],[y_n]) = \lim_{n\to\infty} d(x_n,y_n) $$ Show that this function $\delta$ is well-defined, and that it is a metric on $Y/\sim$. Now $Y/\sim$ is a complete metric space, and $X$ sits isometrically inside $Y/\sim$, by $$ x \mapsto [(x,x,x,\ldots)] $$ I assume that the book does something like this?

  • $\begingroup$ Yes the limit points can be outside $X$, no problem. Nowhere have I assumed that $\{p\}\subset X$. $\endgroup$ – fierydemon Aug 12 '13 at 6:20
  • $\begingroup$ @Ayush: But what does it mean for a sequence in a metric space to converge to a point outside of the space? $\endgroup$ – Pete L. Clark Aug 12 '13 at 6:33
  • $\begingroup$ Well, you can't just choose points outside of $X$. $X$ is your universe, and there is nothing outside it. $\endgroup$ – Prahlad Vaidyanathan Aug 12 '13 at 6:33

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