# Metric Space- Complete Sets, Dense Sets and Compact Sets

**METRIC SPACE- COMPLETE SETS, DENSE SETS AND COMPACT SETS**

Objective

To study the complete sets, dense sets and compact sets in the metric space.

Modules

Module I- Complete sets

Module II- Dense sets

Module III- Compact sets

**Module I- Complete sets**

Definition 1

A sequence is said to be a Cauchy sequence if for, there exists a positive integer so that implies.

Noted that every convergent sequence is a Cauchy sequence. In the real set , we know that a Cauchy sequence is the same as a convergent sequence. However it is always not the case. For example, take , the rational number. Then a Cauchy sequence is not a convergent sequence anymore, since rational sequence may converge to an irrational number, which is not in X. This leads to the following definition.

Definition 2

The set S is said to be complete if all the Cauchy sequence in S is a convergent sequence.

Clearly, a compact set is complete. A closed subset of a complete set is complete. Let us see an interesting theorem of complete metric spaces.

**Theorem 1 (Contractive mapping theorem)**

Suppose, where X is complete, has the property that for all, where. Then f is continuous on X, and there exists unique

x so that.

Proof.

The continuity of f is trivial. Now pick, and define. Then

and for ,

.

As the last inequality tends to 0 when n → ∞, we know is a Cauchy sequence and hence converges to. By the continuity of f, we

have.

The uniqueness of the fixed point x can be seen by supposing there is another fixed point y, and. This forces, as .

This theorem can be seen in such an interesting way: When you are using a computer and reduce a maximized window to a smaller size, a unique point on the screen is fixed!

We are now going to reach a more advanced theorem: the Baire’s Theorem. Before reaching that, let us go over the dense sets first.

**Module II- Dense sets**

Definition 3

Supposein the metric space X. Then S is dense in T if forand, there existssuch that.

For example, the rational set is dense in . As every point of T is a limit point of the dense subset S, we see that S = T. Conversely, ifand S = T, then S is dense in T. It is also not difficult to see that if *S is dense in T and T is dense in U, then S is dense in U*.

Now let us turn to the concept of nowhere dense sets.

Definition 4

The set S is said to be nowhere dense if any open set contains an open ball that is disjoint with S.

For example, a finite set and the integer set in R are nowhere dense. Any open

set and the rational set in R are not nowhere dense. It is easy to see that if a set is nowhere dense, so are its subsets.

Theorem 2

S is nowhere dense if and only if S is nowhere dense.

Proof.

Suppose *S is nowhere dense. An open U contains an open ball B which is disjoint to S. Then no point of B is the limit point of S, since otherwise B *would contain some points of *S. Hence S is disjoint to B. This shows S *is nowhere dense. The converse is trivial.

**Theorem 3(the Baire’s Theorem)**

Suppose is open dense in a

complete metric space X for all integer i. Then is dense in X.

Proof.

Let and. We need to show there is such that contains y. Asis dense, there exists such thatcontains y. Also we can choose aso thatis a subset of the open. Then for each, there exists and so contains and is a subset of

with. By the choice of we have. Now for,

.

The last inequality tends to 0 when. This showsis a Cauchy sequence, hence the completeness of X gives the existence of the limit x. As is a sequence in the closed set, so. Thus for all n and of course in the set. Finally, as, we

have.

We see then contains y and the proof is completed.

Corollary

The complete metric space X cannot be the countable union of nowhere dense sets.

Proof.

Let, where is nowhere dense. It is no doubt that. Note that is open dense. Now taking complement of both sides gives. However by theorem 5.7, the right side of the equality is dense, hence cannot be empty.

**Module III- Compact sets**

Definition 5

The set is compact if any infinite sequence in has convergent subsequence in.

Compactness in

**Theorem 4 (Heine-Borel theorem)**

The closed interval is compact in.

Proof.

Let be an open cover of the set. Suppose there is no finite subcover, and we seek obtain a contradiction. Divide the set into and , one of the two (and possibly both) does not admit a finite subcover: call it . Repeat this process of dividing to get nested intervals of length which do not admit finite subcovers.

Now is an increasing sequence bounded above by , while is an decreasing bounded below by .They therefore both converge, say, and with since. In fact, because

and.

This limit is in the set and is therefore covered by some open set. This implies that we can surround with an -ball

.

But implies that there is an such that; similarly for, so that. But was not supposed to be covered by a finite number of. This implies that there must be a finite subcover to start with.

Corollary

In, a set is compact if and only if it is closed and bounded.

Proof.

We know that a compact set is closed and bounded. Conversely be a closed and bounded set. By its boundedness . Hence is a closed subset of a compact set and this implies that it is itself compact.

Corollary

Letbe a continuous function. Then its image is bounded and the function attains its bounds, i.e.,

such that .

Proof.

We know that is compact. Therefore its continuous image is also compact and bounded. i.e., . i.e., , Moreover compact sets are closed and so contain their limit points. In particular it contains. i.e., for some .

More generally, a continuous function on a compact set will be bounded and attain its bounds.

**Compactness in a general metric space**

We have seen that in the space of real numbers, a subset is compact if and only if it is closed and bounded. But this is not true for every metric space. Instead, we can prove the following.

Theorem 5

A set is compact if and only if it is complete and totally bounded.

Proof.

The proof will be divided into several parts:

First we shall prove that every infinite subset of a compact set K has a limit point in K.

Let A be a subset of a compact set K, and suppose it has no points in K. This means that any is not a limit point, and so we can find ballswhich do not contain elements of A unless they are at the centre. In particular this is an open cover of K. But K is compact and must have a finite subcover of these,

.

Therefore, .

So A is a finite set. Equivalently, infinite subset of K must have at least one limit point in K.

The property that every infinite subset of a set has a limit point is termed as the Bolzano-Weistraproperty (BW- compact for short). So we have shown here that compact sets are BW- compact.

Now, we shall prove that K is BW- compact implies K is complete.

Let be a Cauchy sequence in a given BW- compact set K. If the set is infinite, then it has a limit point. Now we can deduce that there is a subsequence of which converges to x, and sinceis Cauchy, that the whole sequence converges to x.

Otherwise, the set is finite and the sequence must repeat itself. The fact that the sequence is Cauchy implies that the sequence must eventually be constant for , so that it converges to in K. Since K is compact, it is totally bounded. Therefore, K is compact implies K is complete and totally bounded.

Conversely assume that K is complete and totally bounded. Suppose K is covered by open sets, and that no finite number of these open sets is enough to cover K. Since K is totally bounded, we can find a finite number of balls of radius 1 which cover K,

.

If each of these balls had a finite subcover from the open cover, then so would K. So at least one of these balls cannot be covered by a finite number of s. Let us call this ball.

Now consider, also totally bounded. We can again cover it with a finite number of balls of radius, one of which does not have a finite subcover, say. Repeat this process to get a nested sequence of balls, none of which have a finite subcover. The sequence is Cauchy since (for), and K is complete, hence in K.

But x is covered by some open set . Hence

.

Moreover, since and, we can find an N such that and, so that, which contradicts the way that the balls were chosen.

Note that complete subsets are always closed and totally bounded sets are always bounded. Hence the statement K compact implies K closed and bounded is really just a special case of this theorem. In general, however, closed and bounded sets are not complete and totally bounded, although we have seen that this is true in.

Theorem 6

A set K is compact if and only if every finite subset of K has a limit point in K.

Proof.

One implication has already been proved. More over we have also shown that BW-compactness implies completeness, so we only need to prove that BW-compact sets are totally bounded to show the converse.

Let K be a (non empty) BW-compact set. Let and let be any point in K.

Either covers K or if not, there is some pointwith. Repeating the argument for, either cover K or else they omit some point, say. Continuing like this we get a sequence of points each covered by satisfying.

Suppose this process does not stop and we end up with an infinite set. This has a limit point, since K is BW-compact, which implies that some of these are close to within. But by the choice of,. This leads to a contradiction unless the process actually stops in a finite number of steps and the cover the whole of K.

**Assignment questions**

- Show that the interval (a,b] is incomplete, but [a,b] is complete.
- Show that there does not exist a function f: R→R which is continuous on exactly the rational number but not the irrational. (Hint: Use Baire’s Theorem).
- (i) Show that the Cantor set K is nowhere dense.

(ii) Define Kq= {qK: q∈ Q}. Is the set closed? Is it Open?

- Show that a finite set of points is BW-compact.
- Show that any closed and bounded subset of is compact.

Quiz questions

- A Cauchy sequence is the same as a convergent sequence in
- the set of real numbers
- the set of rational numbers
- the set of irrational numbers
- Suppose is open dense in a complete metric space X for all integer i. Then
- is dense in X.
- is dense in X.
- is nowhere dense in X.
- A set is compact if and only if it is
- complete and bounded
- complete and not totally bounded
- complete and totally bounded

Answers

- a
- b
- c

Glossary

Metric space: For a set X (more than 1 element), suppose there is a function d, which maps XX to , and satisfy the following three properties:

For x,y and zX,

(i) d(x,y) 0 and d(x,y) =0 if and only if x=y;

(ii) d(x,y)=d(y,x);

(iii) d(x,y)d(x,z)+d(y,z).(triangle inequality)

Then the pair (X,d) is called a metric space, and d is called distance function, or metric.

Open ball: For xX, define Bd(x,r)={yX:d(x,y)<r}. Bd(x,r) is called an open ball centered at x with radius r.

Convergence: In the metric space *(X,d), we said a sequence xn converges to x*, if for every , there exists an integer K such that nk implies d(xn,x) <.

Limit point: A point x is said to be the limit point of a set S if there exists a sequence xn which converges to x, where xnS and xnx .

Closed set: The set S is said to be closed if S contains all its limit points.

Continuity: Suppose *f is a function from X to Y with metric dx and dy respectively. f *is said to be continuous at x in X if for any , there exists such that dx(x,y)< implies dy(f(x),f(y))< .

Bounded: A set is bounded when it can be covered by a ball.

A set is totally bounded when it can be covered by a finite number of - balls, however small .

Cover: Let be a collection of some open sets. If, then we say that is an open cover of.

Subcover: is called a subcover of with respect to if and.

Summary

A sequence is said to be a Cauchy sequence if for, there exists a positive integer so that implies.

The set S is said to be complete if all the Cauchy sequence in S is a convergent sequence.

**Contractive mapping theorem: **Suppose, where X is complete, has the property that for all, where. Then *f is continuous on X, and there exists unique x *so that.

Supposein the metric space X. Then S is dense in T if forand, there existssuch that.

The set S is said to be nowhere dense if any open set contains an open ball that is disjoint with S.

S is nowhere dense if and only if S is nowhere dense.

**Baire’s Theorem: **Suppose is open dense in a complete metric space X for all integer i. Then is dense in X.

The complete metric space X cannot be the countable union of nowhere dense sets.

The set is compact if any infinite sequence in has convergent subsequence in.

**Heine-Borel theorem: **The closed interval is compact in.

In, a set is compact if and only if it is closed and bounded.

Letbe a continuous function. Then its image is bounded and the function attains its bounds, i.e.,

such that .

A set is compact if and only if it is complete and totally bounded.

A set K is compact if and only if every finite subset of K has a limit point in K.

**Frequently Asked Questions (FAQs)**

- Suppose is nonempty closed in a complete metric space X for all n and Define. If, showsis nonempty. Is it true that if X is not complete?

Answer:

For each x in K, there exists an open setand a real numbersuch that. Then is an open cover of K, so by compactness of K there exists finitely many pointsin K such that . Take, then for each x in K, if we take an index i such that,we will have, from, that. (Note our choice of r is independent of our x in K.) Alternatively one can argue by contradiction. If our assertion is false, then for each positive integer j there exists such that is not subset of any. Since K is compact, the sequencehas a convergent subsequencewith limit x in K. Now for this x there exists and r > 0 such that . For k sufficiently large, since, we have. This contradicts our choice of .

- Show that the Cantor set K is nowhere dense.

Answer:

For each point x in K there exists an open ballcentered at x such that its closure. The family of all , where *x runs through the whole K, is an open cover of K. By compactness of K *there exists a finite subcoverof K. Now each of theis a closed ball and hence compact. Hence if we set , then H, being a finite union of compact sets, is compact, and clearly.

- Prove that any closed subset of a compact set is compact.

Answer:

Suppose, where is open. Let, which is open. Then and together form an open cover of. Since is compact, from this open cover there exists a finite subcover, which means. Hence. We have found a finite subcover of.

REFERENCE

- W. Rudin, Principles of Mathematical analysis, McGraw Hill book company, Auckland (1985).
- P. K. Jain and K. Ahmad, Metric Spaces, Narosa, New Delhi (1993).
- J. R. Munkers, Topology, Prentice hall of India (1991).
- G. F. Simmons,
*Introduction to topology and Modern Analysis*, McGraw Hill (1963). - B. K. Tyagi, Metric Spaces, Cambridge University Press, new Delhi (2010).