VII. Complete Metric Spaces

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 \{ x_{n}\} of points in a metric space R with metric \rho is said to satisfy the Cauchy criterion if given any \epsilon >0, there is an integer N_{\epsilon} such that \rho(x_{n}, x_{n^{'}})<\epsilon for all n,n^{'}> N_{\epsilon}.

DEFINITION 2:

A subsequence \{ x_{n}\} 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 \{ x_{n}\} is fundamental.

Proof 1:

If \{ x_{n} \} converges to a limit x, then, given any \epsilon>0, there is an integer N_{\epsilon} such that

\rho(x_{n}, x) \leq \frac{\epsilon}{2} for all n > N_{\epsilon}.

But, then

\rho(x_{n}, x_{n^{'}}) \leq \rho(x_{n},x)+\rho(x_{n^{'}},x)<\epsilon

for all n, n^{'} >N_{\epsilon}. 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 \rho(x,y)=0, if x=y; let \rho(x,y)=1, when x \neq y. Then, the Cauchy sequence in R are just the “stationary sequences,” that is, the sequences \{ x_{n}\} 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 \Re^{n} follows from that of \Re^{1}. In fact, let

x^{(p)} = (x_{1}^{(p)}, x_{2}^{(p)}, \ldots, x_{n}^{(p)}) where p = 1, 2, \ldots

be fundamental sequence of points in \Re^{n}. Then, given \epsilon >0, there exists an N_{\epsilon} such that

\Sigma_{n=1}^{\infty} (x_{k}^{(p)}-x_{k}^{(q)})^{2} < \epsilon^{2}

for all p, q > N_{\epsilon}. It follows that

|x_{k}^{(p)}-x_{k}^{(q)}|<\epsilon for k=1,2,\ldots, n for all p,q > N_{\epsilon}, that is, each \{x_{k}^{(p)} \} is a fundamental sequence in \Re^{1}.

Let x = (x_{1}, \ldots, x_{n}) where x_{k} = \lim_{p \rightarrow \infty} x_{k}^{(p)}

Then, obviously \lim_{p \rightarrow \infty} x^{(p)} = x.

This proves the completeness of \Re^{n}. The completeness of the spaces R_{0}^{n} and R_{1}^{n} introduced in earlier examples/blogs is proved in almost the same way. (HW: supply the details). QED.

Example 4:

Let \{ x_{n}(t)\} be a Cauchy sequence in the function space C_{[a,b]} introduced earlier. Then, given any \epsilon >0, there is an N_{\epsilon} such that

|x_{n}(t) - x_{n^{'}}(t)|< \epsilon….I

for all n, n^{'} > N_{\epsilon} and all t \in [a,b]. It follows that the sequence \{ x_{n}(t)\} 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 n^{'} \rightarrow \infty in I, we find that

|x_{n}(t) - x(t)|\leq \epsilon for all n > N_{\epsilon} and all t \in [a,b], that is, \{ x_{n}(t)\} converges in the metric space C_{[a,b]} to a function x(t) \in C_{[a,b]}. Hence, C_{[a,b]} is a complete metric space.

Example 5:

Next, let x^{(n)} be a sequence in the space l_{2} so that 

x^ {(n)} = (x_{1}^{(n)}, x_{2}^{(n)}, \ldots, x_{k}^{(n)}, \ldots)

\Sigma_{k=1}^{\infty}(x_{k}^{(n)})^{2} < \infty, where n = 1, 2, , \ldots

Suppose further that \{ x^{(n)}\} is a Cauchy sequence. Then, given any \epsilon > 0 there is a N_{\epsilon} such that

\rho^{2}(x^{(n)},x^{(n^{'})}) = \Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k}^{(n^{'})})^{2}< \epsilon…let us call this II.

if n, n^{'} > N_{\epsilon}.

It follows that (x_{k}^{(n)}-x_{k}^{(n^{'})})^{2} < \epsilon (for k =1, 2, \ldots)

That is, for every k the sequence \{ x_{k}^{(n)}\} is fundamental and hence, convergent. Let

x_{k} = \lim_{n \rightarrow \infty} x_{k}^{(n)}, x = (x_{1}, x_{2}, \ldots, x_{k}, \ldots)

Then, as we now show, x is itself a point of l_{2} and moreover, \{ x^{(n)}\} converges to x in the l_{2} metric space, so that l_{2} is a complete metric space.

In fact, the Cauchy criterion here implies that \Sigma_{k=1}^{M}(x_{k}^{(n)}-x_{k}^{(n^{'})})^{2}<\epsilon for any fixed M. …let us call this III.

Holding n fixed in III, and taking the limit as n^{'} \rightarrow \infty, we get

\Sigma_{k=1}^{M}(x_{k}^{(n)}-x_{k})^{2} \leq \epsilon….call this IV.

Since IV holds for arbitrary M, we can in turn take the limit of IV as M \rightarrow \infty, obtaining

\Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k})^{2} \leq \epsilon.

But, as we have learnt earlier in this series of blogs, the convergence of the two series \Sigma_{k=1}^{\infty}(x_{k}^{(n)})^{2} and \Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k})^{2} implies that of the series \Sigma_{k=1}^{\infty}x_{k}^{2}.

This proves that x \in l_{2}. Moreover, since \epsilon is arbitrarily small, III implies that

\lim_{n \rightarrow \infty}\rho( x^{(n)} ,x ) = \lim_{n \rightarrow \infty} \sqrt{\Sigma_{k=1}^{\infty} (x_{k}^{(n)}-x_{k})^{2}} = 0

That is, \{ x^{(n)}\} converges to x in the l_{2} metric space, as asserted.

QED.

Example 6.

Consider the space C_{[a,b]}^{2}. To recap: consider the set of all functions continuous on the closed interval [a,b] with the distance metric defined by: \rho(x,y) = (\int_{a}^{b}|x(t)-y(t)|^{2}dt)^{\frac{1}{2}}.

It is easy to show that the space C_{[a,b]}^{2} is incomplete. If

\phi_{n}(t) = -1, if -1 \leq t \leq -\frac{1}{n}

\phi_{n}(t) = nt, if -\frac{1}{n} \leq t \leq \frac{1}{n}

\phi_{n}(t) = 1, if \frac{1}{n} \leq t \leq 1

then \{ \phi_{n}(t)\} is a fundamental sequence in C_{[a,b]}^{2} since

\int_{-1}^{1}(\phi_{n}(t)-\phi_{n^{'}}(t))^{2} dt \leq \frac{2}{\min{ \{ n,n^{'}}\}}

However, \{ \phi_{n}(t)\} cannot converge to a function in C_{[-1,1]}^{2}. In fact, consider the discontinuous function

\psi(t) = -1, when t \leq 0

\psi(t) = 1, when t \geq 0.

Then, given any function f \epsilon C_{[-1,1]}^{2}, it follows from Schwarz’s inequality (obviously still valid for piecewise continuous functions) that

(\int_{-1}^{1}(f(t)-\psi(t))^{2}dt)^{\frac{1}{2}} \leq (\int_{-1}^{1}(f(t)-\phi_{n}(t))^{2}dt)^{\frac{1}{2}} + (\int_{-1}^{1}(\phi_{n}(t) - \psi(t))^{2}dt)^{\frac{1}{2}}

But the integral on the left is non-zero, by the continuity of f, and moreover, it is clear that

\lim_{n \rightarrow \infty}\int_{-1}^{1}(\phi_{n}(t)-\psi(t))^{2}dt = 0

Therefore, \int_{-1}^{1}(f(t)-\phi_{n}(t))^{2}dt cannot converge to zero as n \ rightarrow \infty.

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

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

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 \{ S_{n}\} = \{ S_{[x_{n}, r_{n}]}\} of closed spheres in R such that r_{n} \rightarrow 0 as n \rightarrow \infty has a non empty intersection

\bigcap_{n=1}^{\infty}S_{n}.

Proof of the nested theorem:

Part I: Assume that R is complete and that if \{ S_{n}\} = \{ S[x_{n}, r_{n}] \} is any nested sequence of closed spheres in R such that r_{n} \rightarrow 0 as n \rightarrow \infty, then the sequence \{ x_{n}\} of centres of the spheres is fundamental because

\rho(x_{n}, x_{n^{'}}) < r_{n} for n^{'}>n and r_{n} \rightarrow 0 as n \rightarrow \infty. Therefore, \{ x_{n} \} has a limit. Let

x = \lim_{n \rightarrow \infty} x_{n}.

Then, x \in \bigcap_{n=1}^{\infty}S_{n}

Not only that, we can in fact say that S_{n} contains every point of the sequence \{ x_{n}\} except possibly the points x_{1}, x_{2}, \ldots, x_{n-1} and hence, x is a limit point of every sphere S_{n}. But, S_{n} is closed, and hence, x \in S_{n} 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 \{ x_{n}\} be any fundamental sequence in R. Then, x has a limit point in R. To see this, use the fact that \{x_{n}\} is fundamental to choose a term x_{n_{1}} of the sequence \{ x_{n} \} such that

\rho(x_{n}, x_{n_{1}}) < \frac{1}{2} for all n \geq n_{1}, and let S_{1} be the closed sphere of radius 1 with centre x_{n_{1}}. Then, choose a term x_{n_{1}} of \{ x_{n}\} such that n_{2} > n_{1} and \rho(x_{n}, x_{n_{1}}) < \frac{1}{2^{2}} for all n > n_{2}, and let S_{2} be the closed sphere of radius \frac{1}{2} with centre x_{n_{2}}.

Continue this construction indefinitely, that is, once having chosen terms x_{n_{1}}, x_{n_{2}}, \ldots, x_{n_{k}} (where n_{1}<n_{2}< \ldots n_{k}), choose a term x_{n_{k+1}} such that n_{k+1}>n_{k} and

\rho(x_{n}, x_{n_{k+1}}) < \frac{1}{2^{k+1}} for all n \geq n_{k+1}.

Let S_{k+1} be the closed sphere of radius \frac{1}{2^{n}} with centre x_{n_{k+1}}, and so on. This gives a nested sequence S_{n} 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 \{ x_{n}\}. But, if a fundamental sequence contains a subsequence converging to x, then the sequence itself must converge to x (HW quiz). That is,

\lim_{n \rightarrow \infty}x_{n} =x.

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 S \subset R contains another sphere S^{'} such that S^{'} \bigcap A = \phi. (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 R = \bigcup_{n=1}^{\infty}A_{n} ….call this VI.

where every set A_{n} is nowhere dense in R. Let S_{0} \subset R be a closed sphere of radius 1. Since A_{1} is nowhere dense in S_{0}, being nowhere dense in R, there is a closed sphere S_{1} of radius less than \frac{1}{2} such that S_{1} \subset S_{0} and S_{1} \bigcap A_{1} =\phi. Since A_{2} is nowhere dense in S_{1}, being nowhere dense in S_{0}, there is a closed sphere S_{2} of radius less than \frac{1}{3} such that S_{2} \subset S_{1} and S_{2} \bigcap A_{2} = \phi, and so on. In this way, we get a nested sequence of closed spheres \{ S_{n}\} with radii converging to zero such that

S_{n} \bigcap A_{n} = \phi, where n=1,2,3,\ldots

By the nested sphere theorem, the intersection \bigcap_{n=1}^{\infty}S_{n} contains a point x. By construction, x cannot belong to any of the sets A_{n}, that is,

x \notin \bigcup_{n=1}^{\infty}A_{n}

It follows that R \neq \bigcup_{n=1}^{\infty} A_{n} 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 \{ x\} 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 [R], a complete metric space R^{*} is called a completion of R if R \subset R^{*} and [R]=R^{*}, that is, if R is a subset of R^{*} everywhere dense in R^{*}.

Example 1.

Clearly, R^{*}=R 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 x \in R into itself.

Proof of Theorem 4:

(The proof is somewhat lengthy but quite straight forward).

First , we prove the uniqueness showing that if R^{*} and R^{**} are two completions of R, then there is a one-to-one mapping x^{**} = \phi(x^{*}) onto R^{**} such that \phi(x)=x for all x \in R and

\rho_{1}(x^{*}, y^{*}) = \rho_{2}(x^{**}, y^{**})….call this VII.

(y^{**}=\phi(y^{*})), where \rho_{1} is the distance metric in R^{*} and \rho_{2} the distance metric in R^{**}. The required mapping \phi is constructed as follows: Let x^{*} be an arbitrary point of R^{*}. Then, by the definition of a completion, there is a sequence \{ x_{n}\} of points of R converging to x^{*}. The points of the sequence \{ x_{n}\} also belong to R^{**}, where they form a fundamental sequence (quiz: why?). Therefore, \{ x_{n}\} converges to a point x^{**} \in R^{**} since R^{**} is complete. It is clear that x^{**} is independent of the choice of the sequence \{ x_{n}\} converging to the point x^{*} (homework quiz: why?). If we set \phi(x^{*})=x^{**}, then \phi is the required mapping. In fact, \phi(x)=x for all x \in R, since if x_{n} \rightarrow x \in R, then obviously x = x^{*} \in R^{*}, x^{**}=x. Moreover, suppose x_{n} \rightarrow x^{*}, y_{n} \rightarrow y^{*} in R^{*}, while x_{n} \rightarrow x^{**}, y_{n} \rightarrow y^{**}. Then, if \phi is the distance in R,

\rho_{1}(x^{*},y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) = \lim_{n \rightarrow \infty}(x_{n}, y_{n})…call this VIII.

While at the same time, \rho_{2}(x^{**}, y^{**})=\lim_{n \rightarrow \infty}\rho_{2}(x_{n}, y_{n})=\lim_{n \rightarrow \infty}\rho(x_{n}, y_{n})….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 \{x_{n} \} and \{\overline{x}_{n} \} in R are equilvalent and write \{ x_{n}\} \sim  \{ \overline{x}_{n}\}

if \lim_{n \rightarrow \infty} \rho(x_{n}, \overline{x}_{n})=0

As anticipated by the notation and terminology, \sim is reflective, symmetric and transitive, that is, \sim 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 R^{*}. Then, we define the distance between two arbitrary points x^{*}, y^{*} by the formula

\rho_{1}(x^{*}, y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) ….call this IX.

where \{ x_{n}\} is any “representative” of x^{*} (namely, any Cauchy sequence in the class x^{*}) and \{ y_{n}\} is any representative of y^{*}.

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 \{x_{n} \} \in x^{*}, \{ y_{n} \} \in y^{*}, and satisfies the three properties of a distance metric function. Given any \epsilon >0, it follows from the triangle inequality in R (this can be proved with a little effort: homework quiz) that

|\rho(x_{n}, y_{n}) - \rho(x_{n^{'}},y_{n^{'}})| = |\rho(x_{n},y_{n}) -\rho(x_{n^{'}},y_{n}) + \rho(x_{n^{'}},y_{n}) - \rho(x_{n^{'}}, y_{n^{'}})|

That is, \leq |\rho(x_{n},y_{n}) -\rho(x_{n^{'}}, y_{n}) | + |\rho(x_{n^{'}}, y_{n}) - \rho(x_{n^{'}}, y_{n^{'}}) |

that is, \leq \rho(x_{n}, x_{n^{'}}) + \rho(y_{n}, y_{n^{'}}) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon….call this X

for all sufficiently large n and n^{'}.

Therefore, the sequence of real numbers \{ x_{n}\} = \{ \rho(x_{n},y_{n})\} is fundamental and hence, has a limit. This limit is independent of the choice of \{x_{n} \} \in x^{*}, \{ y_{n}\} \in y^{*}. In fact, suppose that

\{ x_{n}\}, \{ \overline{x}_{n}\} \in x^{*}, \{ y_{n}\}, \{ \overline{y}_{n} \in y^{*}

Then,

|\rho(x_{n}, y_{n}) - \rho(\overline(x)_{n}, \overline{y}_{n})| \leq \rho(x_{n}, \overline{x}_{n}) + \rho(y_{n}, \overline{y}_{n}) by a calculation analogous to X. But,

\lim_{n \rightarrow \infty} \rho(x_{n}, \overline{x}_{n}) = \lim_{n \rightarrow \infty}(y_{n}, \overline{y}_{n})=0

since \{ x_{n} \} \sim \{ \overline{x}_{n}\} and \{ y_{n} \} \sim \{ \overline{y}_{n}\}, and hence,

\lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) = \lim_{n \rightarrow \infty} \rho(\overline{x}_{n}, \overline{y}_{n}).

As for the three properties of a metric, it is obvious that

\rho_{1}(x^{*}, y^{*}) = \rho_{1}(y^{*}, x^{*}), and the fact that

\rho_{1}(x^{*}, y^{*}) =0 if and only if x^{*}=y^{*} is an immediate consequence of the definition of equivalent Cauchy sequences.

To verify the triangle inequality in R^{*}, we start from the triangle inequality:

\rho(x_{n}, z_{n}) \leq \rho(x_{n}, y_{n}) + \rho(y_{n}, z_{n})

in the original space R, and then take the limit as n \rightarrow \infty, obtaining

\lim_{n \rightarrow \infty} rho(x_{n}, z_{n}) \leq \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}+ \lim_{n \rightarrow \infty} \rho(y_{n}, z_{n})

That is, \rho_{1}(x^{*}, z^{*}) \leq \rho_{1}(x^{*}, y^{*}) + \rho_{1}(y^{*}, z^{*})

We now come to the crucial step of showing that R^{*} is a completion of R. Suppose that with every point x \in R, we associate the class x^{*} \in R^{*} of all Cauchy sequences converging to x. Let

x = \lim_{n \rightarrow \infty} x_{n}, y = \lim_{n \rightarrow \infty} y_{n}

Then, clearly \rho(x,y) = \lim_{n \rightarrow \infty}(x_{n}, y_{n})

(the above too can be proven with a slight effort: HW quiz); while, on the other hand,

\rho_{1}(x^{*}, y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) by definition. Therefore,

\rho(x,y) = \rho_{1}(x^{*}, y^{*}) and hence, the mapping of R into R^{*} carrying x into x^{*} is isometric. Accordingly, we need no longer distinguish between the original space R and its image in R^{*}, in particular between the two metrics \rho and \rho_{1}. In other words, R can be regarded as a subset of R^{*}. The theorem will be proved once we succeed in showing that

(i) R is everywhere dense in R^{*}, that is, [R] = R;

(2) R^{*} is complete.

Towards that end, given any point x^{*} \in R^{*} and any \epsilon >0, choose a representative of x^{*}, namely a Cauchy sequence \{ x_{n}\} in the class x^{*}. Let N be such that \rho(x_{n}, x_{n^{'}}) < \epsilon for all n, n^{'} >N. Then,

\rho(x_{n}, x^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, x_{n^{'}}) \leq \epsilon if n >N, that is, every neighbourhood of the point x^{*} contains a point of R. It follows that [R] = R.

Finally, to show that R^{*} is complete, we first note that by the very definition of R^{*}, any Cauchy sequence \{ x_{n}\} consisting of points in R converges to some point in R^{*}, namely to the point x^{*} \in R^{*} defined by \{ x_{n}\}. Moreover, since R is dense in R^{*}, given any Cauchy sequence x_{n}^{*} consisting of points in R^{*}, we can find an equivalent sequence \{ x_{n}\} consisting of points in R. In fact, we need only choose x_{n} to be any point of R such that \rho(x_{n}, x_{n}^{*}) < \frac{1}{n}. The resulting sequence \{ x_{n}\} is fundamental, and, as just shown, converges to a point x^{*} \in R^{*}. But, then the sequence x_{n}^{*} also converges to x^{*}.

QED.

Example.

If R is the space of all rational numbers, then R^{*} is the space of all real numbers, both equipped with the distance \rho(x,y) = |x-y|. 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.

Purva building, 5A
Flat 06
Thakur Complex, Near Dimple Arcade
Mumbai , Maharastra 400101
India

Problem Set based on VI. Convergence, open and closed sets.

Problem 1.

Give an example of a metric space R and two open spheres S(x,r_{1}), S(y, r_{2}) in R such that S(x, r_{1}) \subset S(y,r_{2}) although r_{1}> r_{2}.

Problem 2:

Prove that every contact point of a set M is either a limit point of M or is an isolated point of M.

Comment. In particular, [M] can only contain points of the following three types:

a) Limit points of M belonging to M.

b) Limit points of M which do not belong to M.

c) Isolated points of M.

Thus, [M] is the union of M and the set of all its limit points.

Problem 3:

Prove that if x_{n}\rightarrow x and y_{n} \rightarrow y as n \rightarrow \infty then \rho(x_{n},y_{n}) \rightarrow \rho(x,y).

Hint : use the following problem: Given a metric space (X,\rho) prove that |\rho(x,z)-\rho(y,u)| \leq |\rho(x,y)|+|\rho(z,u)|

Problem 4:

Let f be a mapping of one metric space X into another metric space Y. Prove that f is continuous at a point x_{0} if and only if the sequence \{ y_{n}\} = \{ f(x_{n})\} converges to y=f(x_{0}) whenever the sequence x_{n} converges to x_{0}.

Problem 5:

Prove that :

(a) the closure of any set M is a closed set.

(b) [M] is the smallest closed set containing M.

Problem 6:

Is the union of infinitely many closed sets necessarily closed? How about the intersection of infinitely many open sets? Give examples.

Problem 7:

Prove directly that the point \frac{1}{4} belongs to the Cantor set F, although it is not the end point of any of the open interval deleted in constructing F. Hint: The point \frac{1}{4} divides the interval [0,1] in the ratio 1:3 and so on.

Problem 8:

Let F be the Cantor set. Prove that

(a) the points of the first kind form an everywhere dense subset of F.

(b) the numbers of the form t_{1}+t_{2} where t_{1}, t_{2} \in F fill the whole interval [0,2].

Problem 9:

Given a metric space R, let A be a subset of R, and x \in R. Then, the number \rho(A,x) = \inf_{a \in A}\rho(a,x) is called the distance between A and x. Prove that

(a) x \in A implies \rho(A,x)=0 but not conversely

(b) \rho(A,x) is a continuous function of x (for fixed A).

(c) \rho(A,x)=0 if and only if x is a contact point of A.

(d) [A]=A \bigcup M, where M is the set of all points x such that \rho(A,x)=0.

Problem 10:

Let A and B be two subsets of a metric space R. Then, the number \rho(A,B)= \inf_{a \in A, b \in B}\rho(a,b) is called the distance between A and B. Show that \rho(A,B)=0 if A \bigcap B \neq \phi, but not conversely.

Problem 11:

Let M_{K} be the set of all functions f in C_{[a,b]} satisfying a Lipschitz condition, that is, the set of all f such that |f(t_{1}-f(t_{2})| \leq K|t_{1}-t_{2}| for all t_{1}, t_{2} \in [a,b], where K is a fixed positive number. Prove that:

a) M_{K} is closed and in fact is the closure of the set of all differentiable functions on [a,b] such that |f^{'}(t)| \leq K

(b) the set M = \bigcup_{K}M_{K} of all functions satisfying a Lipschitz condition for some K is not closed;

(c) The closure of M is the whole space C_{[a,b]}

Problem 12:

An open set G in n-dimensional Euclidean space R^{n} is said to be connected if any points x,y \in G can be joined by a polygonal line(by a polygonal line we mean a curve obtained by joining a finite number of straight line segments end to end.) lying entirely in G. For example, the open disk x^{2}+y^{2}<1 is connected, but not the union of the two disks x^{2}+y^{2}<1, (x-2)^{2}+y^{2}<1 (even though they share a contact point). An open subset of an open set G is called a component of G if it is connected and is not contained in a larger connected subset of G. Use Zorn’s lemma to prove that every open set G in R^{n} is the union of no more than countably many pairwise disjoint components.

Comment: In the case n=1, that is, the case on the real line, every connected open set is an open interval, possibility one of the infinite intervals (-\infty, \infty), (a, \infty) or (-\infty, b). Thus, theorem 6 (namely: Every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals) on the structures of open sets on the line is tantamount to two assertions:

(i) Every open set on the line is the union of a finite or countable number of components.

(ii) Every open connected set on the line is an open interval.

The first assertion holds for open sets in R^{n} (and, in fact, is susceptible to further generalizations), while the second assertion pertains specifically to the real line.

Cheers,

Happy analysis !!

Nalin Pithwa