Prove That Cauchy Sequence Function is Uniformly Continuous

 [ 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 in X is said to be Cauchy if

• for any ε>0, there exists N such that whenever m, n > N, we have

Let's list some basic results.

Proposition 1. A convergent sequence is Cauchy.

Proof.

Suppose in the metric space (X,d). For any ε>0, there existsN such that whenevern >N, we have Thus, wheneverm,n >N, we have:

 ♦

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

There's nothing to prove here sinceY inherits the distance function fromX. 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 are sequences of metric spaces respectively, then is a Cauchy sequence of X × Y if and only if each of is Cauchy in the respective metric space.

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

Proof

First suppose and are Cauchy. For any ε>0,

Thus, whenm,n > max(M,N), we have – for any metric d onX ×Y in the above list –

For the converse, we use the fact that

and

for any one of three choices ofd. ♦

One might ask if it's necessary to consider all three metrics onX ×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 homeomorphismf :R ⁺ →R ⁺ of the space of positive reals given byf(x) = 1/x. The sequence is Cauchy, but the resulting sequence is not. Put in another way, we can define two metrics onR ⁺ via and Then the sequence is Cauchy under the first metric but not the second.

The same example also tells us:

Definition. A function of metric spaces is said to beCauchy-continuous if whenever is a Cauchy sequence in X, 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 ofR.

Definition. A map of metric spaces is said to beuniformly continuous if

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

Take the functionf :R ⁺ →R ⁺ given byf(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 existx andx' 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 integern > 1/δ and let We now have:

but

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

Theorem 4. If is a uniformly continuous map of metric spaces, then it is Cauchy-continuous.

Proof.

Let ε>0. Then:

Thus, whenm,n >N, we get So is indeed a Cauchy sequence inY. ♦

The following are some easy properties of uniformly continuous functions.

Proposition 5.

Proof.

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

and

for any one of the three d. ♦

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

Exercises

1. Prove that if is a Cauchy sequence in a metric space X, then every subsequence is also Cauchy.
2. Prove that if is a convergent sequence in a topological spaceX, then every subsequence is also convergent.
3. Suppose is a Cauchy sequence in a metric spaceX. Prove that if a subsequence of is convergent, so is the entire sequence.
4. Prove that a Cauchy-continuous map of metric spaces is continuous.
5. Find a Cauchy-continuousf which is not uniformly continuous.

Answers (Highlight to Read)

3. Suppose (x_n ) is Cauchy, and the subsequence indexed byn[1] <n[2] < n[3] … converges tox. For each ε>0, pickN such that (i) whenm,n >N, d(x_m ,x_n ) < ε/2 and (ii) whenk>N,d(x_n[k] , x) < ε/2. Thus, when n>N, we haved(x_n ,x) ≤d(x_n ,x_n[k] ) +d(x_n[k] , x) < ε for somek>N. This proves that (x_n ) →x.
   
4. 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 andy _2n-1 =x, forn = 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).
5. Takef :R →R, given byf(x) =x ².
   • Not uniformly continuous since we can set ε=2; ifx=n,y=n + (1/n), then |x–y| = 1/n, but |f(x)-f(y)| > 2.
   • Cauchy-continuous since any Cauchy sequence (x_n ) inR must be convergent so (f(x_n )) is also convergent, and hence Cauchy.

Summary

We have three related notions of continuity (for maps of metric spaces).

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.

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Source: https://mathstrek.blog/2013/02/07/topology-cauchy-sequences-and-uniform-continuity/

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