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.
DEFINITION 1:
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
.
DEFINITION 2:
A subsequence
of points in a metric space R is called a Cauchy sequence (or a fundamental sequence ) if it satisfies the Cauchy criterion.
THEOREM 1:
Every convergent sequence
is fundamental.
Proof 1:
If
converges to a limit x, then, given any
, there is an integer
such that
for all
.
But, then

for all
. QED.
DEFINITION 3:
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.
Example 1:
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.
Example 2:
The completeness of the real line R is familiar from elementary analysis:
Example 3:
The completeness of the Euclidean n-space
follows from that of
. In fact, let
where 
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
.
Let
where 
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.
Example 4:
Let
be a Cauchy sequence in the function space
introduced earlier. Then, given any
, there is an
such that
….I
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.
Example 5:
Next, let
be a sequence in the space
so that

, where 
Suppose further that
is a Cauchy sequence. Then, given any
there is a
such that
…let us call this II.
if
.
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.
QED.
Example 6.
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
, if 
, if 
, if 
then
is a fundamental sequence in
since

However,
cannot converge to a function in
. In fact, consider the discontinuous function
, when 
, 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
.
QED.
7.2 The nested sphere theorem.
A sequence of closed spheres ![S[x_{1}, r_{1}], S[x_{2}, r_{2}], \ldots, S[x_{n}, r_{n}], \ldots S[x_{1}, r_{1}], S[x_{2}, r_{2}], \ldots, S[x_{n}, r_{n}], \ldots](https://s0.wp.com/latex.php?latex=S%5Bx_%7B1%7D%2C+r_%7B1%7D%5D%2C+S%5Bx_%7B2%7D%2C+r_%7B2%7D%5D%2C+%5Cldots%2C+S%5Bx_%7Bn%7D%2C+r_%7Bn%7D%5D%2C+%5Cldots&bg=%23f6f5ed&fg=222222&s=0&c=20201002)
in a metric space R is said to be nested (or decreasing) if
![S[x_{1}, r_{1}] \supset S[x_{2}, r_{2}] \supset \ldots \supset S[x_{n}, r_{n}] \supset \ldots S[x_{1}, r_{1}] \supset S[x_{2}, r_{2}] \supset \ldots \supset S[x_{n}, r_{n}] \supset \ldots](https://s0.wp.com/latex.php?latex=S%5Bx_%7B1%7D%2C+r_%7B1%7D%5D+%5Csupset+S%5Bx_%7B2%7D%2C+r_%7B2%7D%5D+%5Csupset+%5Cldots+%5Csupset+S%5Bx_%7Bn%7D%2C+r_%7Bn%7D%5D+%5Csupset+%5Cldots&bg=%23f6f5ed&fg=222222&s=0&c=20201002)
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
.
Then, 
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,
.
QED.
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
, where 
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.
QED.
COROLLARY TO Baire’s theorem:
A complete metric space R without isolated points is uncountable.
Proof:
Every single element set
is nowhere dense in R.
QED.
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
.
Example 1.
Clearly,
if R is already complete. (Quiz: homework).
Example 2:
The space of all real numbers is the completion of the space of all rational numbers.
THEOREM 4:
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 
if 
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, 
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
, 
Then,
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

That is, 
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
, 
Then, clearly 
(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
.
QED.
Example.
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.
Regards,
Nalin Pithwa.
Mathematics Hothouse shares:
Like this:
Like Loading...