Reference: Introductory Real Analysis by Kolmogorov and Fomin. Translated by Richard A. Silverman. Dover Publications.
Available on Amazon India and Amazon USA. This text book can be studied in parallel with Analysis of Walter Rudin.
7.1. Definition and examples:
The reader is presumably already familiar with the notion of completeness of the real line. (One good simple reference for this could be: Calculus and analytic geometry by G B Thomas. You can also use, alternatively, Advanced Calculus by Buck and Buck.)The real line is, of course, a simple example of a metric space. We now make the natural generalisation of the notion of completeness to the case of an arbitrary metric space.
A sequence of points in a metric space R with metric is said to satisfy the Cauchy criterion if given any , there is an integer such that for all .
A subsequence of points in a metric space R is called a Cauchy sequence (or a fundamental sequence ) if it satisfies the Cauchy criterion.
Every convergent sequence is fundamental.
If converges to a limit x, then, given any , there is an integer such that
for all .
for all . QED.
A metric space R is said to be complete if every Cauchy sequence in R converges to an element of R. Otherwise, R is said to be incomplete.
Let R be the “space of isolated points” (discrete metric space) defined as follows: Define , if ; let , when . Then, the Cauchy sequence in R are just the “stationary sequences,” that is, the sequences all of whose terms are the same starting from some index n. Every such sequence is obviously convergent to an element of R. Hence, R is complete.
The completeness of the real line R is familiar from elementary analysis:
The completeness of the Euclidean n-space follows from that of . In fact, let
be fundamental sequence of points in . Then, given , there exists an such that
for all . It follows that
for for all , that is, each is a fundamental sequence in .
Then, obviously .
This proves the completeness of . The completeness of the spaces and introduced in earlier examples/blogs is proved in almost the same way. (HW: supply the details). QED.
Let be a Cauchy sequence in the function space introduced earlier. Then, given any , there is an such that
for all and all . It follows that the sequence is uniformly convergent. But the limit of a uniformly convergent sequence of continuous functions is itself a continuous function (see Problem 1 following this Section). Taking the limit as in I, we find that
for all and all , that is, converges in the metric space to a function . Hence, is a complete metric space.
Next, let be a sequence in the space so that
Suppose further that is a Cauchy sequence. Then, given any there is a such that
…let us call this II.
It follows that (for )
That is, for every k the sequence is fundamental and hence, convergent. Let
Then, as we now show, x is itself a point of and moreover, converges to x in the metric space, so that is a complete metric space.
In fact, the Cauchy criterion here implies that for any fixed M. …let us call this III.
Holding n fixed in III, and taking the limit as , we get
….call this IV.
Since IV holds for arbitrary M, we can in turn take the limit of IV as , obtaining
But, as we have learnt earlier in this series of blogs, the convergence of the two series and implies that of the series .
This proves that . Moreover, since is arbitrarily small, III implies that
That is, converges to x in the metric space, as asserted.
Consider the space . To recap: consider the set of all functions continuous on the closed interval with the distance metric defined by: .
It is easy to show that the space is incomplete. If
then is a fundamental sequence in since
However, cannot converge to a function in . In fact, consider the discontinuous function
, when .
Then, given any function , it follows from Schwarz’s inequality (obviously still valid for piecewise continuous functions) that
But the integral on the left is non-zero, by the continuity of f, and moreover, it is clear that
Therefore, cannot converge to zero as .
7.2 The nested sphere theorem.
A sequence of closed spheres
in a metric space R is said to be nested (or decreasing) if
Using this concept, we can prove a simple criterion for the completeness of R:
THEOREM 2: The Nested Sphere Theorem:
A metric space R is complete if and only if every nested sequence of closed spheres in R such that as has a non empty intersection
Proof of the nested theorem:
Part I: Assume that R is complete and that if is any nested sequence of closed spheres in R such that as , then the sequence of centres of the spheres is fundamental because
for and as . Therefore, has a limit. Let
Not only that, we can in fact say that contains every point of the sequence except possibly the points and hence, x is a limit point of every sphere . But, is closed, and hence, for all n.
Conversely, suppose every nested sequence of closed spheres in R with radii converging to zero has a non empty intersection, and let be any fundamental sequence in R. Then, x has a limit point in R. To see this, use the fact that is fundamental to choose a term of the sequence such that
for all , and let be the closed sphere of radius 1 with centre . Then, choose a term of such that and for all , and let be the closed sphere of radius with centre .
Continue this construction indefinitely, that is, once having chosen terms (where ), choose a term such that and
for all .
Let be the closed sphere of radius with centre , and so on. This gives a nested sequence of closed spheres with radii converging to zero. By hypothesis, these spheres have a non empty intersection, that is, there is a point x in all the spheres. This point is obviously the limit of the sequence . But, if a fundamental sequence contains a subsequence converging to x, then the sequence itself must converge to x (HW quiz). That is,
7.3 Baire’s theorem:
We know that a subset A of a metric space R is said to be nowhere dense in R if it is dense in no (open) sphere at all, or equivalently, if every sphere contains another sphere such that . (Quiz: check the equivalence).
This concept plays an important role in the following:
THEOREM 3: Baire’s Theorem:
A complete metric space R cannot be represented as the union of a countable number of nowhere dense sets.
Proof of Theorem 3: Baire’s Theorem:
Suppose the contrary. Let ….call this VI.
where every set is nowhere dense in R. Let be a closed sphere of radius 1. Since is nowhere dense in , being nowhere dense in R, there is a closed sphere of radius less than such that and . Since is nowhere dense in , being nowhere dense in , there is a closed sphere of radius less than such that and , and so on. In this way, we get a nested sequence of closed spheres with radii converging to zero such that
By the nested sphere theorem, the intersection contains a point x. By construction, x cannot belong to any of the sets , that is,
It follows that contrary to VI.
Hence, the representation VI is impossible.
COROLLARY TO Baire’s theorem:
A complete metric space R without isolated points is uncountable.
Every single element set is nowhere dense in R.
7.4 Completion of a metric space:
As we now show, an incomplete metric space can always be enlarged (in an essentially unique way) to give a complete metric space.
DEFINITION 4: Completion of a metric space:
Given a metric space R with closure , a complete metric space is called a completion of R if and , that is, if R is a subset of everywhere dense in .
Clearly, if R is already complete. (Quiz: homework).
The space of all real numbers is the completion of the space of all rational numbers.
Every metric space R has a completion. This completion is unique to within an isometric mapping carrying every point into itself.
Proof of Theorem 4:
(The proof is somewhat lengthy but quite straight forward).
First , we prove the uniqueness showing that if and are two completions of R, then there is a one-to-one mapping onto such that for all and
….call this VII.
(), where is the distance metric in and the distance metric in . The required mapping is constructed as follows: Let be an arbitrary point of . Then, by the definition of a completion, there is a sequence of points of R converging to . The points of the sequence also belong to , where they form a fundamental sequence (quiz: why?). Therefore, converges to a point since is complete. It is clear that is independent of the choice of the sequence converging to the point (homework quiz: why?). If we set , then is the required mapping. In fact, for all , since if , then obviously , . Moreover, suppose , in , while , . Then, if is the distance in R,
…call this VIII.
While at the same time, ….call this VIII-A. But VIII and VIII-A imply VII.
We must now prove the existence of a completion of R. Given an arbitrary metric space R, we say that two Cauchy sequences and in R are equilvalent and write
As anticipated by the notation and terminology, is reflective, symmetric and transitive, that is, is an equivalence relation. Therefore, the set of all Cauchy sequences of points in the space R can be partitioned into classes of equivalent sequences. Let these classes be the points of a new space . Then, we define the distance between two arbitrary points by the formula
….call this IX.
where is any “representative” of (namely, any Cauchy sequence in the class ) and is any representative of .
The next step is to verify that IX is indeed a distance metric. That is, also to check that IX exists, independent of the choice of sequence , , and satisfies the three properties of a distance metric function. Given any , it follows from the triangle inequality in R (this can be proved with a little effort: homework quiz) that
that is, ….call this X
for all sufficiently large and .
Therefore, the sequence of real numbers is fundamental and hence, has a limit. This limit is independent of the choice of , . In fact, suppose that
by a calculation analogous to X. But,
since and , and hence,
As for the three properties of a metric, it is obvious that
, and the fact that
if and only if is an immediate consequence of the definition of equivalent Cauchy sequences.
To verify the triangle inequality in , we start from the triangle inequality:
in the original space R, and then take the limit as , obtaining
We now come to the crucial step of showing that is a completion of R. Suppose that with every point , we associate the class of all Cauchy sequences converging to x. Let
(the above too can be proven with a slight effort: HW quiz); while, on the other hand,
by definition. Therefore,
and hence, the mapping of R into carrying x into is isometric. Accordingly, we need no longer distinguish between the original space R and its image in , in particular between the two metrics and . In other words, R can be regarded as a subset of . The theorem will be proved once we succeed in showing that
(i) R is everywhere dense in , that is, ;
(2) is complete.
Towards that end, given any point and any , choose a representative of , namely a Cauchy sequence in the class . Let N be such that for all . Then,
if , that is, every neighbourhood of the point contains a point of R. It follows that .
Finally, to show that is complete, we first note that by the very definition of , any Cauchy sequence consisting of points in R converges to some point in , namely to the point defined by . Moreover, since R is dense in , given any Cauchy sequence consisting of points in , we can find an equivalent sequence consisting of points in R. In fact, we need only choose to be any point of R such that . The resulting sequence is fundamental, and, as just shown, converges to a point . But, then the sequence also converges to .
If R is the space of all rational numbers, then is the space of all real numbers, both equipped with the distance . In this way, we can “construct the real number system.” However, there still remains the problem of suitably defining sums and products of real numbers and verifying that the usual axioms of arithmetic are satisfied.