Prove Set of Bounded Continuous Functions is Cauchy

October 16, 2022 Post a Comment

20. The Category of Complete Metric Spaces

20.1 Definition:

A sequence in a Metric Space $\QTR{Large}{(M,d)}$ is called a Cauchy Sequence

if for any 0 there exists an n such that if n then

20.2 Lemma:

In the setting of 20.1 , every Cauchy sequence is bounded. In particular there is a number $\QTR{Large}{b>}$ 0 and an such that

Proof:

Select n such that if n $_{1}$ then 1. Now let and select any

$\QTR{Large}{b>\{}$ 1

20.3 Lemma:

In the setting of 20.1 , if then is a Cauchy Sequence.

Proof:

This quickly follows from the observation that

hence if then

20.4 Examples and Observations:

In general, the converse to 20.3 is not true. Consider, for example, the open interval $\QTR{Large}{(}$ 0,1 $\QTR{Large}{)}$ and the sequence clearly is Cauchy but does not have a limit.

It is also the case that Cauchy sequences are not preserved under mapping by continuous functions. For example, consider $\QTR{Large}{f:(}$ 0,1 , given by

and the Cauchy sequence .

on the other hand
In , every Cauchy sequence converges.

Proof:

Since is bounded, so is . let and be the greatest lower bound and least upper bounds respectively. On checks that
Assignment: Due April 13 . Fill in the details.

20.5 Definition:

A Metric Space is called complete if every Cauchy sequence converges. In particular,

is a complete metric space.

20.6 Theorem:

Let $\QTR{Large}{(M,d)}$ be a complete Metric Space and let be closed, then

$\QTR{Large}{(A,d)}$ is a complete Metric Space.

Proof:

Any Cauchy sequence in $\QTR{Large}{A}$ converges in $\QTR{Large}{M}$ , but $\QTR{Large}{A}$ is closed so it contains all its limit points.

20.7 Theorem:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and , we can define a metric on by setting

Then if $\QTR{Large}{(M,d)}$ and are complete so is .

Proof:

After checking that is a metric, the essential point is the proof is to show that if is a Cachy sequence for then so is and for $\QTR{Large}{d}$ and respectively. Hence and converge to $\QTR{Large}{x}$ and $\QTR{Large}{y}$ respectively. Finally, converges to $\QTR{Large}{(x,y)}$ .

Assignment: Due April 13 . State and prove the converse.

We would like to discuss the Category of Complete Metric Spaces. Unsettled is the appropriate choice of morphisms. Since we wish these to be structure preserving 20.4.1 tells us that continuous functions might be a poor choice since, in general, they do do preserve Cauchy sequences. There is an important sub-class of continuous functions which do preserve Cauchy sequences and, in fact, are the continous functions on an important sub-Category of Complete Metric Spaces.

20.8 Definition:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and and a map we say that $\QTR{Large}{f}$ is

uniformly continuous if for all $\QTR{Large}{x\in }$ $\QTR{Large}{M}$ ,given any 0 there is a 0 such that Note that this differs for continuity in that is independent of $\QTR{Large}{x}$ .

Remarks:

We will show, for example, that every continuous function $\QTR{Large}{f:[}$ 0,1 0,1 $\QTR{Large}{]}$ is uniformly continuous.
To verify that complete metric spaces and uniformly continuous maps form a category we need to check the the composition of uniformly continuous maps is uniformly continuous.

20.9 Theorem:

In the setting of 20.8, if is a Cauchy sequence in $\QTR{Large}{M}$ then is a Cauchy sequence in .

Proof:

Given 0 we need to find n such that if n then

But since is uniformly continuous we can choose 0 such that for all $\QTR{Large}{x}$ . Select n such that if n then

We will also want to look at the following uniformly continuous functions.

20.10 Definition:

Given Metric Spaces $\QTR{Large}{(M,d)}$ and ,a map is said to be an isometry if for all ,

Notes:

Given Metric Spaces $\QTR{Large}{(M,d)}$ one first observes that if 0 then is also a Metric Space . Moreover the two Metric Spaces have the same Cauchy Sequences.
In the setting of 1., Given a second metric such that for all we have

then $\QTR{Large}{d}$ and have the same Cauchy Sequences.