Topology: Cauchy Sequences and Uniform Continuity

[ Updated on 8 Mar 13 to include Cauchy-continuity and added answers to exercises. ]

We wish to generalise the concept of Cauchy sequences to metric spaces. Recall that on an intuitive level, a Cauchy sequence is one where the elements get “closer and closer”.

Definition. Let (X, d) be a metric space. A sequence $(x_n)$ in X is said to be Cauchy if

for any ε>0, there exists N such that whenever m, n > N, we have $d(x_m, x_n) < \epsilon.$

Let’s list some basic results.

Proposition 1. A convergent sequence is Cauchy.

Proof.

Suppose $(x_n) \to a$ in the metric space (X, d). For any ε>0, there exists N such that whenever n > N, we have $d(x_n, a)<\epsilon/2.$ Thus, whenever m, n > N, we have:

$d(x_m, x_n) \le d(x_m, a) + d(a, x_n) < \frac\epsilon 2 + \frac\epsilon 2 = \epsilon.$ ♦

Proposition 2. Let Y be a metric subspace of (X, d). Then a sequence $(x_n)$ in Y is Cauchy if and only if it’s Cauchy in X.

There’s nothing to prove here since Y inherits the distance function from X. However, this serves to highlight the fact that while a sequence from Y which converges in X may not be convergent in Y, the same problem doesn’t hold for Cauchy sequences, i.e. being Cauchy is a property of the sequence itself, regardless of the ambient space.

Proposition 3. If $(x_n), (y_n)$ are sequences of metric spaces $(X, d_X), (Y, d_Y)$ respectively, then $(x_n, y_n)$ is a Cauchy sequence of X × Y if and only if each of $(x_n), (y_n)$ is Cauchy in the respective metric space.

For the metric of X × Y, we can pick any one of the following:

$((x, y), (x', y')) \mapsto \sqrt{d_X(x, x')^2 + d_Y(y, y')^2}$ ;

$((x, y), (x', y')) \mapsto d_X(x, x') + d_Y(y, y')$ ;

$((x, y), (x', y')) \mapsto\max(d_X(x, x'), d_Y(y, y'))$ .

Proof

First suppose $(x_n)$ and $(y_n)$ are Cauchy. For any ε>0,

there exists M such that when m, n > M, we have $d_X(x_m, x_n) < \frac\epsilon 2;$
there exists N such that when m, n > N, we have $d_Y(y_m, y_n) < \frac\epsilon 2.$

Thus, when m, n > max(M, N), we have – for any metric d on X × Y in the above list – $d((x_m, y_m), (x_n, y_n)) < \epsilon.$

For the converse, we use the fact that

$d((x,y), (x',y')) \ge d_X(x, x')$ and $d((x,y), (x', y')) \ge d_Y(y, y')$

for any one of three choices of d. ♦

One might ask if it’s necessary to consider all three metrics on X × Y since they all give rise to the same topology anyway. And this is where we drop the bombshell.

The concept of Cauchy sequences actually relies heavily on the metric and not just the underlying topology. In other words, it’s possible for two metrics on the same space to be topologically equivalent, but a sequence is Cauchy in one and not the other.

For example, consider the homeomorphism f : R⁺ → R⁺ of the space of positive reals given by f(x) = 1/x. The sequence $x_n = 1/n$ is Cauchy, but the resulting sequence $f(x_n) = n$ is not. Put in another way, we can define two metrics on R⁺ via $(r, r') \mapsto |r-r'|$ and $(r, r')\mapsto |r^{-1} - r'^{-1}|.$ Then the sequence $x_n = 1/n$ is Cauchy under the first metric but not the second.

The same example also tells us:

Definition. A function $f:(X, d_X) \to (Y, d_Y)$ of metric spaces is said to be Cauchy-continuous if whenever $(x_n)$ is a Cauchy sequence in X, $(f(x_n))$ is a Cauchy sequence in Y.

Warning. Not all continuous functions are Cauchy-continuous.

To rectify that, we need a stronger form of continuity.

Uniform Continuity

The answer to our problem is the following definition. We had already seen it earlier in the case of R.

Definition. A map $f:(X, d_X)\to (Y, d_Y)$ of metric spaces is said to be uniformly continuous if

for any ε>0, there exists δ>0 such that whenever $x, x'\in X$ satisfies $d_X(x, x')<\delta,$ we have $d_Y(f(x), f(x')) < \epsilon.$

Clearly, a uniformly continuous function is also continuous (at every point of X). But here’s an example where the converse is not true.

Take the function f : R⁺ → R⁺ given by f(x) = 1/x as before. This is clearly continuous. To show that it’s not uniformly continuous, negating the definition means we need to find an ε>0 such that for any δ>0, there exist x and x’ such that |x – x’| < δ but |f(x) – f(x’)| ≥ ε.

This is not too hard: set ε=1. Now for any δ>0, pick a positive integer n > 1/δ and let $x = \frac 1 n, x' = \frac 1 {n+1}.$ We now have:

$d(x, x') = |x - x'| = \frac 1 {n(n+1)} < \frac 1 n <\delta$ but $d(f(x), f(x')) = 1.$

Next, the main result we’d like to prove is:

Theorem 4. If $f:(X, d_X) \to (Y, d_Y)$ is a uniformly continuous map of metric spaces, then it is Cauchy-continuous.

Proof.

Let ε>0. Then:

by uniform continuity of f, there exists δ>0 such that whenever $x, x'\in X$ satisfies $d_X(x, x')<\delta$ , we have $d_Y(f(x), f(x')) < \epsilon;$
there exist N such that when m, n > N, we get $d_X(x_m, x_n) < \delta.$

Thus, when m, n > N, we get $d_Y(f(x_m), f(x_n)) < \epsilon.$ So $(f(x_n))$ is indeed a Cauchy sequence in Y. ♦

The following are some easy properties of uniformly continuous functions.

Proposition 5.

If Y is a metric subspace of (X, d), then the inclusion map $Y\hookrightarrow X$ is uniformly continuous.

If $f:X\to Y$ and $g:Y\to Z$ are uniformly continuous functions of metric spaces, then so is $g\circ f:X\to Z.$

The projection maps $p_1: X\times Y\to X$ and $p_2 : X\times Y\to Y$ are uniformly continuous, where the metric on X × Y is one of the three in proposition 3.

Proof.

The first two statements are obvious. The last follows from the inequality we saw earlier:

$d((x, y), (x', y')) \ge d_X(x, x')$ and $d((x,y), (x', y'))\ge d_Y(y, y'),$

for any one of the three d. ♦

Finally, we end this article with some exercises on Cauchy and convergent sequences.

Exercises

Prove that if $(x_n)$ is a Cauchy sequence in a metric space X, then every subsequence is also Cauchy.
Prove that if $(x_n)$ is a convergent sequence in a topological space X, then every subsequence is also convergent.
Suppose $(x_n)$ is a Cauchy sequence in a metric space X. Prove that if a subsequence of $(x_n)$ is convergent, so is the entire sequence.
Prove that a Cauchy-continuous map $f:(X, d_X)\to (Y, d_Y)$ of metric spaces is continuous.
Find a Cauchy-continuous f which is not uniformly continuous.

Answers (Highlight to Read)

Suppose (x_n) is Cauchy, and the subsequence indexed by n[1] < n[2] < n[3] … converges to x. For each ε>0, pick N such that (i) when m, n > N, d(x_m, x_n) < ε/2 and (ii) when k>N, d(x_n[k], x) < ε/2. Thus, when n>N, we have d(x_n, x) ≤ d(x_n, x_n[k]) + d(x_n[k], x) < ε for some k>N. This proves that (x_n) → x.
By theorem 6 in the previous article, it suffices to show that for any convergent (x_n) → x in X, we have (f(x_n)) → f(x). Now construct a new sequence (y_n) by interspersing (x_n) and x : y_2n = x_n and y_2n-1 = x, for n = 1, 2, … . Then (y_n) still converges to x, so it’s Cauchy; since f is Cauchy-continuous, (f(y_n)) is Cauchy too. But (f(y_n)) has a subsequence (f(x), f(x), … ), so by Q3, (f(y_n)) → f(x).
Take f : R → R, given by f(x) = x².
- Not uniformly continuous since we can set ε=2; if x=n, y=n + (1/n), then |x–y| = 1/n, but |f(x)-f(y)| > 2.
- Cauchy-continuous since any Cauchy sequence (x_n) in R must be convergent so (f(x_n)) is also convergent, and hence Cauchy.

Summary

We have three related notions of continuity (for maps $f:(X, d_X)\to (Y, d_Y)$ of metric spaces).

$\left\{\begin{matrix}\text{Uniform}\\ \text{continuity}\end{matrix}\right\}\implies \left\{\begin{matrix} \text{Cauchy-}\\ \text{continuity}\end{matrix}\right\}\implies \left\{\text{Continuity}\right\}.$

All implications are non-reversible since there’re counter-examples. The first two are only defined for maps between metric spaces while the last is a topological property.