Metric Spaces, Completeness & Completion

Functional analysis begins by deciding what it means for approximations to get close. A metric supplies that geometry; completeness says that an internally consistent approximation process does not point to a missing object. The two ideas must be kept separate: topology knows which sequences converge, while completeness depends on the chosen metric.

Metric, balls, and the induced topology #

A metric on a nonempty set $X$ is a map $d:X\times X\to[0,\infty)$ satisfying

$$d(x,y)=0\iff x=y,\qquad d(x,y)=d(y,x),\qquad d(x,z)\le d(x,y)+d(y,z).$$

The open ball $B_r(x)=\{y:d(x,y)<r\}$ defines the metric topology. A sequence $x_n$ converges to $x$ exactly when $d(x_n,x)\to0$. A map $f:(X,d_X)\to(Y,d_Y)$ is continuous at $x$ when small input distance forces small output distance; it is Lipschitz when a single constant works globally:

$$d_Y(f(x),f(y))\le Ld_X(x,y).$$

Metrics can induce the same open sets without assigning the same geometry. On $\mathbb R$, the usual metric and $d'(x,y)=|\arctan x-\arctan y|$ induce the same topology. Nevertheless, $(\mathbb R,d')$ is incomplete: $x_n=n$ is $d'$-Cauchy, but its missing limit would have arctangent $\pi/2$.

Cauchy sequences detect missing limits #

A sequence is Cauchy if

$$\forall\varepsilon>0\ \exists N\ \forall m,n\ge N: d(x_m,x_n)<\varepsilon.$$

Every convergent sequence is Cauchy. The converse defines completeness. This formulation is useful because it refers only to the approximants, not to an already-known limit.

If $X$ is complete, a subspace $A\subseteq X$ with the restricted metric is complete exactly when $A$ is closed. The forward direction uses a convergent sequence in $A$; the reverse direction lets an $A$-Cauchy sequence converge in $X$ and uses closedness to keep the limit in $A$.

Completion adds exactly the missing limits #

Let $\mathcal C$ be the set of Cauchy sequences in $X$. Declare

$$(x_n)\sim(y_n)\quad\Longleftrightarrow\quad d(x_n,y_n)\to0.$$

The completion $\widehat X$ consists of equivalence classes $[(x_n)]$ with

$$\widehat d([(x_n)],[(y_n)])=\lim_{n\to\infty}d(x_n,y_n).$$

The reverse triangle inequality makes the scalar sequence on the right Cauchy, and the equivalence relation makes its limit independent of the chosen representatives. The map $x\mapsto[(x,x,\ldots)]$ is an isometric embedding with dense image. Every Cauchy sequence of classes has a limit in $\widehat X$, so the new space is complete. Any two completions are uniquely isometric once their copies of $X$ are identified.

Canonical examples are $\mathbb Q\hookrightarrow\mathbb R$, finite sequences $c_{00}\hookrightarrow\ell^p$ for $1\le p<\infty$, and smooth/test functions completed under energy norms to produce function spaces used in PDEs.

Product metrics and equivalent norms #

On a finite product, metrics such as

$$d_1(x,y)=\sum_{j=1}^n d_j(x_j,y_j),\qquad d_\infty(x,y)=\max_j d_j(x_j,y_j)$$

induce the same product topology, and the product is complete iff every factor is complete. In finite-dimensional vector spaces, all norms are equivalent, so convergence, compactness, and completeness do not depend on which norm is chosen. This stops being true for infinite-dimensional normed spaces.

Worked diagnostic: is a candidate complete? #

Consider polynomials on $[0,1]$ with the uniform norm. The sequence of Taylor polynomials for $e^x$ is uniformly Cauchy and converges uniformly to $e^x$, which is not a polynomial. Hence the polynomial space is incomplete. Its uniform completion is $C[0,1]$ because polynomials are dense there and $C[0,1]$ is complete in the sup norm.

When diagnosing a space:

1. identify the exact metric or norm;
2. characterize its Cauchy sequences;
3. find the natural ambient complete space;
4. ask whether the candidate is closed in that ambient space;
5. if not, identify the closure—the likely completion.