Reference: Principles of Mathematical Analysis by Walter Rudin

2.31 Definition: By an open cover of a set E in a metric space X we mean a collection of open subsets of X such that .

2.32 Definition A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.

More explicitly, the requirement is that if is an open cover of K, then there are finitely many indices such that

The notion of compactness is of great importance in analysis, especially in connection with continuity.

It is clear that every finite set is compact. The existence of a large class of infinite compact sets in will follow with Theorem 2.41

We observed earlier (section 2.29) that if then E may be open relative to Y without being open relative to Y. The property of being open thus depends on the space in which it is embedded. Note that the same is true of the property of being closed.

Compactness, however, behaves better as we shall see now. To formulate the next theorem, let us say, temporarily that K is compact relative to X if the requirements of Definition 2.32 are met.

2.33 Theorem: Suppose . Then K is compact relative to X if and only if K is compact relative to Y.

By the virtue of the theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without any paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces or of closed spaces (every metric space X is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric spaces.

Proof:

Suppose K is compact relative to X, and let be a collection of sets, open relative to Y, such that . By theorem 2.30 there are sets open relative to X, such that for all x, and since K is compact relative to X, we have

(22)…..

for some choice of finitely many indices . Since , the above equation 22 implies that

(23)

This proves that K is compact relative to Y.

Conversely, suppose K is compact relative to Y and let be a collection of open subsets of X which covers K, and put . Then (23) will hold for some choice of of ; and since (23) implies (22).

This completes the proof. QED.

2.34 Theorem Compact subsets of metric spaces are closed.

Proof:

Let K be a compact subset of a metric space X. We shall prove that the complement of K is an open subset of qX.

Suppose , and . If , let $V_{q}$ and be neighbourhoods of p and q, respectively, of radius less than . Since K is compact, there are finitely many points in K such that

If , then V is a neighbourhood of p which does not intersect W. Hence, so that p is an interior point of . Thus, we have proved the theorem. QED.

2.35 Theorem Closed subsets of compact sets are compact.

Proof: Suppose , F is closed (relative to X), and K is compact. Let be an open cover of F. If is adoined to , we obtain an open cover of K. Since K is compact, there is a finite subcollection of which covers K, and hence, F. If is a member of , we may remove it from and still retain an open cover of F. We have thus proved that a finite subcollection of covers F.

PS: Remark: Note the technique of proof here.

Corollary: If F is closed and K is compact, then is compact.

Proof: Theorem 2.24b and 2.34 show that is closed since . Theorem 2.35 shows that is compact. Note: Theorem 2.24b is as follows: For any collection of closed sets, is closed. Theorem 2.34 just proved above says that compact subsets of metric spaces are closed.

2.36 Theorem If is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of is nonempty, then is nonempty.

Proof:

PS: This proof of Rudin is not v clear to me. If one of my readers can help, I would be obliged.

2.37 Theorem: If E is an infinite subset of a compact set K, then E has a limit point in K.

2.37 Proof: If no point of K were a limit point of E, (proof by contradiction), then each would have a neighbourhood which contains at most one point of E (namely, q, if ). (This follows just from the definition of limit point). It is clear that no finite subcollection of can cover E and the same is true of K, since . This contradicts the compactness of K. (recall definition of a compact set).

2.38 Theorem If is a sequence of intervals in , such that where , then is not empty.

2.38 Proof: If let E be the set of all . Then E is nonempty and bounded above by . Let x be the sup of E. If m and n are positive integers, then

so that for each m. Since it is obvious that , we see that for m=1,2,3….

QED.

2.39 Theorem: Let k be a positive integer. If is a sequence of k-cells such that (n=1,2,3,…) then is not empty.

Proof: Let consist of all points such that

() (where )

and put

For each j, the sequence satisfies the hypotheses of previous theorem 2.38. Hence there are real numbers with such that

with and

Setting we see that for . Thus the theorem is proved. QED.

2.40 Theorem Every k-cell is compact.

Proof:

Let I be a k-cell, consisting of all points such that with .

Put

Then if .

Suppose, to get a contradiction, that there exists an open cover of I which contains no finite subcover of I. Put . The intervals and then determine k-cells whose union is I. At least one of these sets , call it , cannot be covered by any finite subcollection of (otherwise I could be so covered). We next subdivide and continue the process. We obtain a sequence with the following properties:

(a)

(b) is not covered by any finite subcollection of

(c) if and , then

By (a) and Theorem 2.39 there is a point which lies in every . For some x, . Since is open, there exists such that implies that . If n is so large that (there is such an n, for otherwise for all positive integers n, which is absurd since R is archimedean), then implies that which contradicts (b).

This completes the proof.

QED.

The equivalence of (a) and (b) in the next theorem is known as the Heine Borel Theorem.

2.41 Theorem: If a set E in has one of the following three properties then it has the other two:

(a) E is closed and bounded;

(b) E is compact.

(c) Every infinite subset of E has a limit point in E.

Proof:

Let us assume (a) is true.

Then for some k-cell I, and (b) follows from Theorems 2.40 and 2.35. Theorem 2.37 shows that (b) implies (c). It remains to be shown that (c) implies (a).

If E is not bounded, then E contains points with

where

The set S consisting of these points is infinite and clearly has no limit point in hence has none in E. Thus (c) implies that E is bounded.

If E is not closed, then there is a point which is a limit point of E but not a point of E. For there are points of E such that . Let S be the set of these points . Then S is infinite (otherwise would have a constant positive value for infinitely many n). S has as a limit point, and S has no other limit point in . For if , , then

for all but finitely many n, this shows that y is not a limit point of S (use the following Theorem 2.20: if p is a limit point of E, then every neighbourhood of p contains infinitely many points of E).

Thus, S has no limit points in E, hence E must be closed if (c) holds.

QED.

Remarks: (b) and (c) are equivalent in any metric space (prove this as an exercise ) but that (a) does not, in general imply, (b) and (c) both. Examples are furnished in Exercise 16 of this chapter and by the space which is discussed in Chapter 11.

2.42 Theorem (Weierstrass) : Every bounded infinite subset of has a limit point in .

Proof: Being bounded, the set E in question is a subset of a k-cell . By Theorem 2.40, I is compact, and so E has a limit point in I, by Theorem 2.37 (If E is an infinite subset of a compact set K, then E has a limit point in K).

Perfect Sets:

2.41 Theorem: Let P be a nonempty perfect set in . Then P is uncountable.

Proof:

((Note: Definition of a perfect set: E is perfect if E is closed and if every point of E is a limit point of E. ))

Since P has limit points, P must be infinite. Suppose P is countable and denote the points of P by We shall construct a sequence of neighbourhoods, as follows:

Let be any neighbourhood of . If consists of all such that , the closure of is the set of all such that . (note this) (this is true or makes sense : we have used the fact that P is also closed, being perfect so the limit point of P can belong to E itself which is the case when ).

Suppose has been constructed so that is not empty. Since every point of P is a limit point of P (note that here we have used the other part of the definition a perfect set), there is a neighbourhood such that (i) (ii) (iii) is not empty. By (iii) satisfies our induction hypothesis, and the construction can proceed.

Put . Since is closed and bounded. is compact. And, by definition of , it is an infinite set. Since is closed and bounded, is compact. Since no point of P lies in . But each is nonempty, by (iii) and by (i), this contradicts the Corollary to theorem 2.36 (If is a sequence of nonempty compact sets such that (where ) then is not empty. )

Corollary : Every interval where is uncountable. In particular, the set of all real numbers is uncountable.

2.44 The Cantor Set: The set which we are now going to construct shows that there exist perfect sets in which contain no segment.

Let be the interval .. Remove the segment and and let be the union of the intervals

and

Remove the middle thirds of these intervals, and let be the union of the intervals

and and and

Continuing in this way, we obtain a sequence of compact sets such that

(a)

(b) is the union of intervals each of length

The set

is called the Cantor set. P is clearly compact, and Theorem 2.36 shows that P is not emtpy. Also, we have shown that P is closed being compact. Now, we have to prove that every point of P is a limit point of P: we can also show that P contains no isolated points: let us therefore look at the construction of P:

No segment of the form

(24)

where k and m are positive integers, has a point in common with P. Since every segment contains a segment of the form (24) if , P contains no segment.

To show that P is perfect, it is enough to show that P contains no isolated point. Let and let S be any segment containing x. Let be that interval of which contains x. Choose n large enough so that . Let be an endpoint of such that .

It follows from the construction of P that . Hence, x is a limit point of P and P is perfect.

QED.

One of the most interesting properties of the Cantor set is that it provides us with an example of an uncountable set of measure zero. (the concept of measure will be discussed in Chapter 11).

CONNECTED SETS:

2.45 Definition: Two subsets A and B of a metric space X are said to be separated if both and are both empty; that is, if no point of A lies in the closure of B and no point of B lies in the closure of A.

A set is said to be connected if E is not a union of two nonempty separated sets.

2.46 Remark: Separated sets are of course disjoint, but disjoint sets need not be separated. For example, the interval and the segment are not separated since 1 is a limit point of . However, the segments and are separated.

The connected subsets of the line have a particularly simpe structure.

2.47 Theorem: A subset E of the real line is connected if and only if it has the following property: if , and , then .

Proof:

If there exist , , and soe such that , then , where

and

Since and , they are separated. Hence E is not connected.

To prove the converse (note we prove the contrapositive of the converse) suppose E is not connected. So is separated. Then there are nonempty separated sets A and B such that . Choose and and assume WLOG that . Define

By theorem 2.28 (Let E be a nonempty set of real numbers which is bounded above Let . Then if E is closed), hence . In particular .

If , it follows that and .

If , then , hence there exists such that and . Then and .

QED.

Cheers,

Nalin Pithwa