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

Chapter 3
The Real Numbers

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Section 3.5, Slide 1


Section 3.5
Compact Sets

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Section 3.5, Slide 2


Definition 3.5.1
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 .
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) The interval S = (0, 2) is not compact.
•
•
•

•
•
•

(
(
(
S

(

)

A4

)

A3

)

A2

)

A1

|(

|

)|

|

0

1

2

3

{ ABut
...G, A=nk }
n1 , if

If 0 < x < 2, then by the Archimedean
property 3.3.10(c), there exists p ∈ 

such that 1/p  <

x.

F = {An : n ∈

is any finite subfamily of F , and if

An1 ∪ L ∪ Ank = Am =
It follows that the finite subfamily G

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.

Thus x ∈ Ap and

} is an open cover for S.

m = max {n , …, nk}, then
1

( m1 , 3) .

is not an open cover of (0,  2) and (0, 2) is not compact.

Section 3.5, Slide 3


Example 3.5.2
(b) Let S = {x1, …, xn} be a finite set and let F = {Aα: α ∈ A } be any open cover of S.

Aα 2

Aα1
•
x1

Aα n

Aα 3

•
x2

•
x3

A

For each i = 1, …, n, there is a α
set

i

from F

{ A 1 ,..., Aα n }

It follows that the subfamily
α

that

F

•
xn

• • •

contains x

i

, since F

is

an open cover.

also covers S. We conclude that any finite set

is compact.

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

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

.

Section 3.5, Slide 4


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 ε > 0 we have N *(m; ε) ∩ S = ∅.
(If this intersection were empty,

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

then m – ε would be an upper bound of 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.
For each n ∈

Proof: First, let us suppose that S is compact.

S ⊆

Then each I is open and
n

∞

Un =1 I n ,

, let In = (− n, n).

so {I  : n ∈ N} is an open cover of S.
n

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

S ⊆ ( I n1 ∪ L ∪ I nk ) = I m ,

where m = max {n ,…, nk}.
1

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It follows that | x | < m for all x ∈ S, and S is bounded.

Section 3.5, Slide 5


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 ∈

would exist a point p ∈ (cl S )\ S.

)

Then there

, we let U = (– ∞, p – 1/n ) ∪ ( p + 1/n, ∞).
n

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

(

)

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

(

U = (– ∞, p – 1 ) ∪ ( p + 1, ∞).
1

(

)
)|

[
S

p

Now each Un is an open set and we have

∞

Un =1U n
=

\{p} ⊇ S.

Thus {Un : n ∈

}

is an open cover of S.

Since S is compact, there exist n1 < n2 < … < nk in

S ⊆

k

Ui =1U n .

such that 

Furthermore, the Un’s are nested. That is, Um ⊆ Un if m ≤ n.

i

S ⊆ UItnfollows
. that
k
p ∈ (cl S

 

But then S ∩ N ( p; 1/nk) = ∅,

contradicting our choice of

 )\S and showing that S must be closed.

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

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.5, Slide 6


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
For each x ∈

compact, let F be an open cover of S.
Sx = {z ∈ S : z ≤ x}

= S ∩ (–

define

∞, x]

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

and let

Examples:
Sz = S

Sw = ∅

•

[

]

[

Sy

[

]

[

Sx

[

]

[

]

S

d

w

•
x

[

]
]

•

]
S

•

y

z

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

If we can show that B is not bounded above, then it will contain a
B is nonempty.

number z greater than sup S. But then Sz = S, and since z ∈ B, we can conclude that
S is compact.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.5, Slide 7


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.

We shall show that m ∈ S and

m ∉ S both lead to contradictions.
If m ∈ S, then since F

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

F1, …, Fk

F0
)

(
[

S

•

•

•

x1

m

x2

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

x1 < m and m = sup B, there exist F1,

 …, F

S

But then F0, F1, …, Fkxcover

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2

k

in F

that

]

m < x2.

cover

S x1 .

, so that x2 ∈ B. This contradicts m = sup B.

Section 3.5, Slide 8


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 = ∅.

N (m; ε )
)

(
[

S

]

•

[

S

]

m

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. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.5, Slide 9


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
at least one point in

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

contains infinitely many points, then there exists

that is an accumulation point of S.

Section 3.5, Slide 10


Theorem 3.5.6

The Bolzano-Weierstrass Theorem

If a bounded subset S of
one point in

contains infinitely many points, then there exists at least

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
Since S has no accumulation points, given

Heine-Borel Theorem 3.5.5, S is compact.
any x ∈ S, there exists a neighborhood N

 (x) of x such that S ∩

N (x) = {x}.

N(x)

(
S

)

(

•

)

•x

( )

(

•

Now the family {N

)

•
 (x) : x

(

)

•

(

)

•

•••

•••

∈ S} is an open cover of S, and since S is compact there exist

x1, …, x

n in S such that {N (x1), …, N (xn)} covers S.

But

S ∩ [ N ( x1 ) ∪ ×××∪ N ( xn )] = {x1, K , xn },

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

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Section 3.5, Slide 11


Theorem 3.5.7

Let F

= {Kα : α ∈

 A } be a family of compact subsets of

.

Suppose that the

intersection of any finite subfamily of F
Proof: For each α ∈ A , let Fα =
its complement Fα is open.

is nonempty. Then ∩ {Kα : α ∈ A

} ≠ ∅.

\ Kα . Since each Kα is compact, it is closed and

Choose a member K of F and suppose that no point of K

belongs to every Kα . Then every point of K belongs to some Fα . That is, the sets Fα

form an open cover ofSince
K. K is compact, there exist finitely

α1,

 …,  α

K ⊆ ( Fα1 ∪ ×××∪ Fα n ).

n such that

But

Fα1 ∪ ×××∪ Fα n = (
=
by

\ Kα1 ) ∪ ×××∪ (

\ Kα n )

\ ( Kα1 ∩ ×××∩ Kα n ),

K ∩ ( Kα1 ∩ ×××∩ Kα n ) = ∅,

Exercise 2.1.26(d), so

Thus some point in K belongs to each Kα , and

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many indices

a contradiction.



 {Kα : α ∈ A } ≠ ∅. ♦

Section 3.5, Slide 12


Corollary 3.5.8

The Nested Intervals Theorem

Let F = {An : n ∈ 

} be a family of closed bounded intervals in

An +1 ⊆ An for all n ∈

∞

I

A

n =1 n

. Then

≠ ∅.

[

A1

]
[

A2

such that

]
[

A3

]
[

A4

]
•
•

•

Proof: Given any n1 <

n2 < … < nk in

Thus Theorem 3.5.7 implies that

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I

, we have
∞

A
n =1♦ n

I

k

A
i =1 ni

= Ank ≠ ∅.

≠ ∅.

Section 3.5, Slide 13