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.
-
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 $C$.
-
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.
-
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.
Best Answer
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.