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Below, is the way I learned the completion theorem. I am wondering if this whole rigmarole has any nice geometric intuition. To get a starting point, let us think of our initial space as

$$M=\{(x,y)\in\mathbb{R}^2|x^2+y^2\le 1\}\backslash\{(0,0)\}.$$

Then what would its completion $C$ will look like in terms of the machinery used in the proof via equivalence classes?

I already know that every subset of a complete metric space is complete if and only if it is closed. So, please just don't tell me that the closure of $M$ is what I am looking for. Let us think in terms of equivalence classes.


Definition 1. A mapping $i:M\to N$ is said to be an isometry iff it is surjective and for every $x,\,y\in M$ we have $d_M(x,y)=d_N(i(x),i(y))$.

Definition 2. A completion of a metric space $(M,d_M)$ is a complete metric space $(C,d_C)$ which has a metric subspace $(N,d_N)$ which is dense in $C$ and is isometric with $M$. It is in this sense that we say $C$ is the smallest complete metric space containing $M$.

Theorem. Every metric space has a completion. Furthermore, this completion is unique up to an isometry. This means that any two completions are isometric. This is called the universal property of a completion.

The proof is a little bit lengthy so here is a sketch of different steps of the proof.

  1. Let $\mathscr{C}$ to be the set of all Cauchy sequences in $M$. Define the relation of being co-Cauchy on $\mathscr{C}$. Show that this relation is an equivalence relation on $\mathscr{C}$. Define $C$ as the set of all resulting equivalence classes. Show that the mapping $d_C:C\times C \to \mathbb{R}$ defined by $d_C([a],[b]):=\lim_{n\to\infty}d_M(a_n,b_n)$ is well-defined and a metric on $C$.

  2. Consider the mapping $i:M\to i(M)\subset C$ which takes every point $x\in M$ to the equivalence class $[a]$ corresponding to the constant sequence $a:\mathbb{N}\to M$ defined as $a_n=x$. This makes sense since every constant sequence is Cauchy. Verify that $M$ and $i(M)$ are isometric and $i(M)$ is dense in $C$ that is $\text{clr}\,i(M) = C$. Show that $C$ is complete.

  3. The last step is to show that every two completions are isometric. Let $(C,d_C)$ and $(E,d_E)$ be any two completions. Then, there are isometries $i:M\to i(M)\subset C$ and $j:M\to j(M)\subset E$ such that $\text{clr}\ i(M)=C$ and $\text{clr}\ j(M)=E$. Verify that $i(M)$ and $j(M)$ are isometric by the map $g:=j\circ i^{-1}:i(M)\to j(M)$. Take any equivalence class $[a]\in C$ and let $\mathcal{A}:\mathbb{N}\to i(M)$ be a sequence of equivalence classes converging to it. Define $f([a]):=\lim_{n\to\infty}(g\circ\mathcal{A})_n$. Show that $f:C\to E$ is well-defined and is an isometry.

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Firstly, let us consider what a completion of a subset of $\mathbb{R}^n$ should be:

Consider some subset $A \subseteq \mathbb{R}^n$, and then define the set

$$\tilde{A}:= A \cup \{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\} \subseteq \mathbb{R}^n$$

consisting of all the points in $A$ together with its limit points (where the sequence and limit is considered in $\mathbb{R}^n$). This makes $\tilde{A}$ closed, and thus since $\mathbb{R}^n$ is complete, it makes $\tilde{A}$ complete. In your case this $\tilde{A}$ is the unit ball in $\mathbb{R}^2$.

Since the embedding of $A$ into $\tilde{A}$ should be an isometry, there is only one choice for it, when considering elements of $A$, which $d$ itself.

Since $\tilde{d}$ is necessarily continuous, there is only one choice of metric for elements in $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\}$; and that is

$$\tilde{d}(\lim_{n \rightarrow \infty} x_n, \lim_{n \rightarrow \infty} y_n) := \lim_{n \rightarrow \infty} d(x_n, y_n).$$

Here it was crucial that there was an ambient space ($\mathbb{R}^n$) to makes this construction; in particular to ensure the existence of the limit points of the sequences $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}} \subseteq A\}$.


Now for a more general case: Let some metric space $A$ be given.

The first step is to identify what $\tilde{A}$ should be: When there is no ambient space, the expressions $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\}$ are not well defined. However, recall that any convergent sequence is Cauchy. Hence may (naively) define the completion to be the set $A$ together with such elements which we identify as those "to which some Cauchy sequence converges" - i.e. the elements we adjoin are the Cauchy sequences themselves. In order to make this a metric space, the limit of any Cauchy sequence needs to be unique though, so two Cauchy sequences which "converge" to "the same point" should induce the same element. This is where the equivalence relation comes in. Since any constant sequences converges to itself, the $A$ is in bijection with the set of equivalence classes with a constant sequence.

All this amounts to the set

$$ \tilde{A} := \{(x)_{n \in \mathbb{N}} : \text{constant}, x \in A\} \cup \{(x_n)_{n \in \mathbb{N}} : \text{Cauchy}, \forall n \in \mathbb{N}: x_n \in A\} $$

On the first set, the newly defined metric is the same as the original one (which is formally expressed by the map $i$). On the second set, we use the continuity of the metric to define the distance between two adjoint elements.

How should one now think of the completion geometrically? Visualize where points are that are limit points of sequences in $A$ and then add them to the set.


By the way, a similar (but different) construction is used when defining the real numbers via Dedekind Cuts.

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  • $\begingroup$ (+1) That's a good point about the existence of an ambient space! I was trying to give an example that we don't know its ambient space! But I failed to do so! :) $\endgroup$ Mar 14, 2020 at 12:15
  • $\begingroup$ By the way, I don't think Dedekind cuts are related here. Using the same techniques of Cauchy equivalence classes is a way of making reals from rationals. Dedekind approach didn't include any sequence stuff. $\endgroup$ Mar 14, 2020 at 12:23
  • $\begingroup$ Yeah, I think you are right there. They are in the same spirit, but not really the same construction. $\endgroup$ Mar 14, 2020 at 12:27
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    $\begingroup$ @H.R. Dedekind made an order completion (adding suprema for sets that did not have them yet), the sequence construction is Cantor's and is a metric completion. $\endgroup$ Mar 14, 2020 at 12:27
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Here is an interesting geometric intuition about the completion thoerem! I found this on a note by Brent Nelson. However, as links may disappear over time, I prefer to write it down here for future readers of this post.

We will imagine our initial space $M$ as a 2-dimensional amorphous blob, lying flat on the ground. If we assume $M$ is not initially complete, then this blob will have lots of tiny pinpricks/holes in it which represent missing points (points that will eventually appear in $M$). Indeed, the moral of our construction is that a metric space really only fails to be complete if it is missing "points" that its Cauchy sequences want to converge to. To produce the completion of $M$, we have to find a way to close up these pinpricks, but our only resource is the space $M$ itself. So, make a copy of our blob and place it directly on top of the previous copy. Here we should imagine they have a certain thickness to them, like a sheet of paper, so that our second copy is literally resting on top of the first. Mathematically, we have constructed $M^2$, which we can think of as length two sequences of elements in $M$. Make another copy of $M$ and set it on the top of the stack. This gives $M^3$. Iterating this procedure will obtain an infinitely tall stack of copies of our metric space, which corresponds to $M^\infty$, the set of sequences $(x_n)_{n\in\mathbb{N}}$ with $x_n\in M$, for all $n\in\mathbb{N}$. Each sequence corresponds to choosing a point from each copy in the stack, roughly forming an vertical path up to the "top" of the stack. The Cauchy sequences are those paths that eventually start to straighten out (i.e. eventually do not jump around too much horizontally). To visualize the equivalence relation on the set of Cauchy sequences, imagine we climb under the stack and are looking up through the bottom (initial) copy. Then two Cauchy sequences (vertical paths) are equivalent if they get closer and closer together as they near the top (which is infinitely high up). In particular, if we pick out a point on our bottom copy of $M$, say $x\in M$, and stare straight up through it, we are seeing the constant sequence $(x)_{n\in\mathbb{N}}$. Now, if we attempt to look up through one of the pinpricks we started with, it will be very difficult to see that it actually remains a pinprick all the way to the top. In fact, due to the infinite height of the stack it will appear to close up (imagine looking down a very deep well, or up a very tall skylight). But this is precisely saying that, up to our equivalence relation, we have managed to plug all the pinpricks in $M$ and therefore made it complete. Hence this infinite procedure was necessary, since any finite stack of copies would have still had open pinpricks.

It will be a good exercise if you try to imagine each step of the proof by this intuition!

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The completion of your space will be $C=\{(x,y): x^2+y^2 \le 1\}$. You get back the point $(0,0)$ as the class of all sequences $(x_n, y_n)$ whose norm tends to $0$. The characterisation also says so: your $M$ is dense in $C$ and $C$ is compact hence complete. Unicity does the rest.

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Geometrically, we can picture set $M$ as it is in $\mathbb{R}^2$, so it is a closed disc with a hole. A guess would be that the completion is just the closed disc with the hold filled in; for this you can construct a cauchy sequence in $M$ to do that. Because $\mathbb{R}^2$ is complete, cauchy sequences on the disc are convergent sequences in $\mathbb{R}^2$ so you just have to prove that all cauchy sequences in $M$ either converges at the origin or inside $M$ (because $M$+'hole' is closed in $\mathbb{R}^2$); it is automatically the smallest.

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