1. Chapter 3
The Real Numbers

2. Section 3.5
Compact Sets

3. 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
(a) The interval S = (0, 2) is not compact.
To see this, let An = (1/n, 3) for each n  .  
If 0  x  2, then by the Archimedean
property (c), there exists p  
such that 1/p   x. ) ( A4 ) ( A3 ) ( A2
Thus x  Ap and
F = {An : n  } is an open cover for S. ) ( A1 S | 1 2 3 ( )
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.

4. Example 3.5.2
(b) Let S = {x1, …, xn} be a finite set and let F = {A:   A } be any open cover of S. F  x1 x2 x3 xn
  
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.
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 .

5. The Heine-Borel Theorem
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  .
(If this intersection were empty,
then m –  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.
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.

6. The Heine-Borel Theorem
A subset S of is compact iff S is closed and bounded.
Then there
would exist a point p  (cl S )\ S.
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, ).
U3 = (– , p – 1/3 )  ( p + 1/3, ). ) ( ) (
U2 = (– , p – 1/2 )  ( p + 1/2, ). ) (
U1 = (– , p – 1 )  ( p + 1, ). ) [ | p S
Now each Un is an open set and we have = \{p}  S.
Thus {Un : n  }
is an open cover of 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.

7.     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 }.
Examples: [ ]
Sz = S [ ] Sy [ ] Sx
Sw =   w [ ] S  x  y  z d
Since S is closed and bounded, Lemma 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.
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.

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 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 < 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.

9.  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 = . ( )
N (m;  ) [ ] S  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. 

10. 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.

11.  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 (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)
( )
( )
  
( ) 
   S x
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. 

12. 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.
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.
Then every point of K belongs to some F .
Since K is compact, there exist finitely many indices
1, …, n such that
But
by Exercise (d), so a contradiction.
Thus some point in K belongs to each K , and  {K :   A }  . 

13. [ ] [ ] [ ] [ ] 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
[ ] A1
[ ] A2
[ ] A3
[ ] A4 
Proof: Given any n1 < n2 < … < nk in , we have
Thus Theorem implies that 