Cauchy Completeness: Complete Metric Space, Banach Space, and Hilbert Space

Can one turn the last sentence around? In other words, is every Cauchy sequence a convergent sequence? Put differently, if $\{ f_n\}$ is a Cauchy sequence, is it true that $\{ f_n\}$ has a limit? What is, in fact, meant by this question is whether $\{ f_n\}$ has a limit in the same space to which the elements $f_n$ belong.

The answer is this: in a finite dimensional complex vector space, a Cauchy sequence always has a limit in that vector space; in other words, a finite dimensional vector space is complete. However, such a conclusion is no longer true for many familiar infinite dimensional vector spaces.

Example: Consider the inner product space

$$C[a,b] = \{ f(x)\colon f~\textrm{is~continuous~on~} a\le x\le b\}$$

with inner product $\langle f,f\rangle = \int^b_a \overline{f} f~dx~ =\Vert f\Vert^2$ .

Claim: $C[a,b]$ is ``incomplete''.

Discussion: Consider the sequence of continuous functions

$$u_k(x)=\frac{1}{2} +\frac{1}{\pi}\textrm{arctan}\,kx\qquad -1\le x\le 1.$$

Figure 1.5: Discontinuous function as a limit of continuous functions.

From Fig. 1.5 we see that:

1. $\lim\limits_{k,p\to\infty} \Vert u_k-u_p\Vert^2=\lim\limits_{k,p\to \infty} \int^1_{-1} (u_p-u_k)^2\,dx =0$ ; in other words, the sequence $\{ u_k\}$ is a Cauchy sequence.
2. For fixed $x$
   $$\lim\limits_{k\to\infty} u_k(x) =v(x)= \begin{cases} 1 &0<x\le 1\\ \frac{1}{2} &x=0\\ 0 &-1\le x<0 \end{cases}$$
   which is a discontinuous function, i.e., $v\not\in C[-1, 1]$ . Furthermoe, we say that the sequence of functions $\{u_k~:k=1,2,\cdots\}$ converges pointwise to the function $v$ .
3. $\lim\limits_{k\to\infty} \Vert v-u_k\Vert^2=\lim\limits_{k\to\infty} \int^1_{-1}(v(x)-u_k)^2\,dx =0$ .
4. One can show that $\not\exists$ any continuous function $w$ such that
   $$\Vert w-u_k\Vert\to 0~~\textrm{as}~~k\to\infty\,.$$

We say that $C[a,b]$ , the space of continuous square integrable functions $(\int^b_a \vert f\vert^2\,dx <\infty)$ , is Cauchy incomplete, or $C[a,b]$ is Cauchy incomplete relative to the given norm $\Vert f \Vert =\sqrt{\langle f,f\rangle}$ . This is so because we have found a Cauchy sequence of functions $\{ u_n\}$ in the inner product space $C[a,b]$ with the property that

$$\lim_{n\to \infty } u_n=v\not\in C[a,b]\,.$$

In other words, the limit of the Cauchy sequence does not lie in the inner product space. This is just like the set of rationals which is not extensive enough to accomodate the norm $\vert x-y\vert$ : there are holes (but no gaps) in the space. These holes are the irrational numbers, which are not detected by $\vert\cdots\vert$ when it is used to determine whether or not an infinite sequence is convergent.

Example: The $1$ -dimensional vector space of rationals over the field of rationals is Cauchy incomplete.

Problem: Why is it that the real line equipped with the distance

$$d(x,y)=\vert \textrm{arctan}\,x - \textrm{arctan}\,y\vert$$

is an incomplete metric space?

In order to remove this incompleteness deficiency, one enlarges the space so that it includes the limit of any of its Cauchy sequences. A space which has been enlarged in this sense is said to be Cauchy complete. This enlarged space is called a complete metric space.

The Cauchy completion of the rationals are the reals. The Cauchy completion of an inner product space is a Hilbert space The Cauchy completion of a normed linear space is a Banach space.