Analysis with an introduction to proof 5th by steven lay ch03c

If F is a family of open sets whose union contains S, then F is called an open cover of S... F In proving a set is compact we must show that every open cover has a finite subcover.. Pr

Chapter 3 The Real Numbers

Section 3.5 Compact Sets

Definition 3.5.1

(a) The interval S = (0, 2) is not compact.

If F is a family of open sets whose union contains S, then F is called an open cover

of S If G ⊆F and G is also an open cover of S, then G is called a subcover of S

Thus S is compact iff every open cover of S contains a finite subcover

Example 3.5.2

To see this, let An = (1/n, 3) for each n ∈

A set S is said to be compact if whenever it is contained in the union of a family F

) (

) (

) (

) (

|

0

|

1

|

2

|

3

A1 A2 A3 A4

•

•

•

•

•

•

S

If 0 < x < 2, then by the Archimedean

property 3.3.10(c), there exists p ∈ 

such that 1/p Thus x   < x ∈ Ap and

F = {An : n ∈ } is an open cover for S

But if G = is any finite subfamily of F , and if m = max {n1, …, nk}, then

It follows that the finite subfamilyG is not an open cover of (0, 2) and (0, 2) is not compact

1

k

( )

1

k

(b) Let S = {x1, …, xn} be a finite set and let F = {Aα: α∈A } be any open cover of S.

Example 3.5.2

•

For each i = 1, …, n, there is a set from F that contains xi , since F is an open cover

It follows that the subfamily also covers S We conclude that any finite set

is compact

F

In proving a set is compact we must show that every open cover has a finite subcover

It is not sufficient to pick a particular open cover and extract a finite subcover

Because of this, it is often difficult to show directly that a given set satisfies the definition

of being compact Fortunately, the classical Heine-Borel theorem gives us a much easier characterization to use for subsets of

1

3

2

n

i

A α

{ A α A αn

(If this intersection were empty,

then m – ε would be an upper bound of S.)

Lemma 3.5.4

If S is a nonempty closed bounded subset of , then S has a maximum and a minimum

Proof: Since S is bounded above and nonempty, m = sup S exists by the completeness axiom

We want to show that m  If m is an accumulation point of S, then since S is closed, ∈ S

we have m ∈ S and m = max S.

If m is not an accumulation point of S, then for some ε> 0 we have N *(m;  ε ) ∩ S = ∅

But m is the least upper bound of S, so N (m;  ε ) ∩ S ≠ ∅

Together these imply m ∈ S, so again we have m = max S Similarly, inf S ∈ S, so inf S = min S ♦

A subset S of is compact iff S is closed and bounded

Proof: First, let us suppose that S is compact For each n ∈ , let In = (− n, n)

Then each In is open and so {In : n ∈ N} is an open cover of S

Since S is compact, there exist finitely many integers n1,  …, nk such that

where m = max {n1,…,  nk} It follows that | x | < m for all x ∈ S, and S is bounded

1 n ,

S ⊆ U ∞ =

1

k

Then there

would exist a point p∈ (cl S  )\ S.

Thus {Un : n ∈ }

is an open cover of S.

A subset S of is compact iff S is closed and bounded

Proof: Next, we assume that S is compact and suppose that S is not closed.

For each n ∈ , we let Un = (– ∞, p – 1/n ) ∪ ( p + 1/n, ∞ ).

)

p

S

U1 = (– ∞, p – 1 ) ∪ ( p + 1, ∞ ).

U2 = (– ∞, p – 1/2 ) ∪ ( p + 1/2, ∞ ).

U3 = (– ∞, p – 1/3 ) ∪ ( p + 1/3, ∞ ).

Now each Un is an open set and we have = \{p} ⊇ S.

Since S is compact, there exist n1 < n2 < … < nk in such that  Furthermore, the Un’s are nested That is, Um ⊆ Un if m ≤ n

It follows that   But then S ∩ N  (  p; 1 /nk) = ∅ , contradicting our choice of

p  ∈  (cl S )\S and showing that S must be closed

So far, we have shown that if S is compact, then S is closed and bounded.

1 n

n ∞ = U

U

i

k n i

.

k

n

Theorem 3.5.5 The Heine-Borel Theorem

A subset S of is compact iff S is closed and bounded

Proof: For the converse, we suppose that S is closed and bounded To show that S is

compact, let F be an open cover of S. For each x ∈ define

Sx = {z ∈ S : z ≤ x} = S ∩ (– ∞, x]

and let B = {x : Sx is covered by a finite subcover of F }.

Since S is closed and bounded, Lemma 3.5.4 implies that S has a minimum, say d.

Then Sd = {d }, and this is certainly covered by a finite subcover of F Thus d ∈ B and

B is nonempty.

S

y

Sx

d

If we can show that B is not bounded above, then it will contain a number z greater than sup S But then Sz = S, and since z ∈ B, we can conclude that

S is compact.

Examples:

•

w

•

z

Sw = ∅

•

x

Sy

Sz = S

We shall show that m ∈ S and

m ∉ S both lead to contradictions.

A subset S of is compact iff S is closed and bounded

Proof: We have Sx = {z ∈ S : z ≤ x} and B = {x : Sx is covered by a finite subcover of F }

Suppose that B is bounded above and let m = sup B.

If m ∈ S, then since F is an open cover of S, there exists F0 in F such that m ∈ F0.

•

m

F0 F1, …, Fk

•

x1

•

x2

Since F0 is open, there exists an interval [x1,  x2] in F0 such that x1 < m < x2.

Since x1< m and m = sup B, there exist F1, …, Fk in F that cover

But then F0, F1, …, Fk cover , so that x2 ∈ B This contradicts m = sup B.

1.

x

S

2

x

S

Theorem 3.5.5 The Heine-Borel Theorem

A subset S of is compact iff S is closed and bounded

Proof: We have Sx = {z ∈ S : z ≤ x} and B = {x : Sx is covered by a finite subcover of F }

Suppose that B is not bounded above and let m = sup B.

On the other hand, if m ∉ S, then since S is closed there exists an ε> 0 such that

N (m;  ε ) ∩ S = ∅

•

m

N (m;ε)

But then Sm – ε = Sm + ε /2.

Since m – ε ∈ B, we have m +ε /2 ∈ B, which again contradicts m = sup B.

Since the possibility that B is bounded above leads to a contradiction,

we must conclude that B is not bounded above, and hence S is compact ♦

In Example 3.4.15 we showed that a finite set will have no accumulation points

We also saw that some unbounded sets (such as ) have no accumulation points.

As an application of the Heine-Borel theorem, we now derive the classical Bolzano-Weierstrass theorem, which states that these are the only conditions that can allow a set to have no accumulation points.

If a bounded subset S of contains infinitely many points, then there exists

at least one point in that is an accumulation point of S.

( ) ( ) ( ) ( ) ( ) • • •

If a bounded subset S of contains infinitely many points, then there exists at least one point in that is an accumulation point of S.

Proof: Let S be a bounded subset of containing infinitely many points and suppose that S has no accumulation points

Then S is closed by Theorem 3.4.17(a), so by the Heine-Borel Theorem 3.5.5, S is compact Since S has no accumulation points, given any x ∈ S, there exists a neighborhood N  (x) of x such that S ∩ N  (x) = {x}.

N(x)

( )

x

S

Now the family {N  (x)  : x ∈ S} is an open cover of S, and since S is compact there exist x1, …,  xn in S such that {N  (x1), …, N  (xn)} covers S.

But

so S = {x1, …, xn} This contradicts S having infinitely many points ♦

S ∩ N x ∪ ×××∪ N x = x K x

Choose a member K of F and suppose that no point of K belongs to every Kα

That is, the sets Fα form an open cover of K.

Theorem 3.5.7

Let F = {K α  :  α   ∈ A } be a family of compact subsets of Suppose that the

Proof: For each α∈A , let Fα = \ Kα Since each Kα is compact, it is closed and

its complement Fα is open

Then every point of K belongs to some Fα

Since K is compact, there exist finitely many indices

Thus some point in K belongs to each Kα , and  {Kα  : α   ∈ A } ≠ ∅ ♦

1

n

K ⊆ F α ∪ ×××∪ F α

1

n

1

n

Corollary 3.5.8 The Nested Intervals Theorem

Proof: Given any n1 < n2 < … < nk in , we have

Thus Theorem 3.5.7 implies that ♦

[ ]

A1

[ ]

A2

[ ]

A3

[ ]

A4

•

•

•

I

k

I

I