Chapter 3 The Real Numbers - PowerPoint Presentation

Slide1

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 of open sets, then it is contained in the union of some finite number of the sets in F .

(

)

)

(

)

(

)

(

)

(

|

0

|

1

|

2

|

3

A

1

A

2

A

3

A

4













S

If 0  

x  2, then by the Archimedean property 3.3.10(c), there exists p   such that 1/p   x.

Thus

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 subfamily

G

is not an open cover of (0,

2) and (0,

2) is not compact.

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

Example 3.5.2



x

1



x

2



x

3



x

n



 

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 .

(If this intersection were empty, then m –  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  S.

If m is an accumulation point of S, then since S is closed, we have m  S and m = max S.

If m is not an accumulation point of S, then for some   0 we have N *(m; )  S = .

But m is the least upper bound of S, so N (m; )  S  .

Together these imply m  S, so again we have m = max S. Similarly, inf S  S, so inf S = min S. 

Theorem 3.5.5

The Heine-Borel Theorem

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.

Then there would exist a point p  (cl S )\ S.

Thus

{Un : n  } is an open cover of S.

Theorem 3.5.5

The Heine-Borel Theorem

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

U

1

= (–

,

p

–

1

)  (

p

+ 1,

).

U

2

= (–

,

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

U

n

’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.

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 iscompact, let F be an open cover of S.

For each

x  define

S

x = {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

S



y

[

]

S

x

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

[

]

[

]

S

y

[

]

[

]

S

z

=

S

We shall show that m  S and m  S both lead to contradictions.

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 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.

[

]

S



m

(

)

F

0

F

1

, …,

F

k

 x1

 x2

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

Since

x

1

<

m

and

m

= sup

B

, there exist F1, …, Fk in F that cover

But then

F

0

, F1, …, Fk cover , so that x2  B. This contradicts m = sup B.

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  > 0 such that N (m; )  S = .

[

]

]

[

S

S



m

(

)

N

(

m

;

 )

But then Sm –  = Sm + /2.

Since m –   B, we have m +  /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.

Theorem 3.5.6

The Bolzano-Weierstrass Theorem

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.

( )

( )

( )

( )

( )







Theorem 3.5.6

The Bolzano-Weierstrass Theorem

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.

Butso S = {x1, …, xn}. This contradicts S having infinitely many points. 

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

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

Theorem 3.5.7

Let

F = {K :   A } be a family of compact subsets of . Suppose that the intersection of any finite subfamily of F is nonempty. Then  {K :   A }  .

Proof:

For each

  A , let F = \ K . Since each K is compact, it is closed and its complement F is open.

Then every point of

K belongs to some F .

Since K is compact, there exist finitely many indices 1, …, n such that

But

Thus some point in

K

belongs to each

K



, and



 {K :   A }  . 

by

Exercise 2.1.26(d), so a contradiction.

Corollary 3.5.8

The Nested Intervals Theorem

Let

F

= {An : n   } be a family of closed bounded intervals in such that An +1  An for all n  . Then

Proof:

Given any

n

1

< n2 < … < nk in , we have

Thus Theorem 3.5.7 implies that



[ ]

A

1

[

]

A

2

[ ]

A

3

[ ]

A

4





