#### 4.7.3 Compactness And Open Coverings

In Proposition 4.7.12 it was shown that sequential compactness in ℝ is the same as closed and
bounded. Here we give the traditional definition of compactness and show that this is also
equivalent to closed and bounded.

Definition 4.7.16 A set K is called compact if whenever C is a collection of
open sets such that K ⊆∪C, there exists a finite subset of open sets

⊆C
such that K ⊆∪_{i=1}^{m}U_{i}. In words, it says that every open cover admits a finite subcover.
Proposition 4.7.17 Every closed interval

is compact.
Proof: Suppose not. Then there exists an open cover C which admits no finite subcover.
Consider the two intervals

,. At least one of these fails to have a finite open
cover from

C. Otherwise, there would be a finite open cover of

. Pick the interval which
fails to have a finite open cover from

C. Call this interval

I_{2}. It is half the length of

I_{1} ≡. Now do for

I_{2} exactly what was done for

I_{1}. Split it in half and take the
half which fails to have a finite subcover. Continue obtaining a nested sequence of
closed intervals such that the length of

I_{n} is 2

^{n−1} times the length of

I_{1}. By the
nested interval lemma, there exists a point

p which is in all these intervals. Thus

p ∈ U ∈C. Therefore, there exists

δ > 0 such that

⊆ U. Now for all

n large
enough, the length of

I_{n} which contains

p, is less than

δ. Hence,

I_{n} is contained in

U
contrary to the definition of

I_{n} which required that it admit no finite subcover from

C.

■
Now here is the main result, often called the Heine Borel theorem.

Theorem 4.7.18 Let K be a nonempty set in ℝ. Then K is compact if and
only if K is closed and bounded.

Proof: First suppose K is closed and bounded. Then K ⊆

for suitably large

p.
Thus from Proposition

4.7.17,

is compact. Then if

C is an open cover of

K, it follows
that

D≡C∪ is an open cover of

. It follows that there are finitely many of the
open sets in

D which cover

K. However,

ℝ ∖ K contains no points of

K and so if this
finite set includes this one, you can simply delete it and still have an open cover of

K.

Conversely, suppose that K is compact. Why is it closed and bounded? Suppose first it is
not closed. Then there exists a limit point p which is not in K. By Theorem 4.7.8 there exists
a sequence of distinct points of K

such that lim

_{n→∞}p_{n} =

pK. Then

p is the

only
limit point of this sequence. Hence

is a closed set. Then since p

K,
is an open cover of K but it admits no finite subcover. This is because the sequence of open
sets is increasing and ℝ ∖ C_{n} fails to contain p_{n} ∈ K. Thus K is closed. Why is K
bounded? If not, there exists a sequence

_{n=1}^{∞}⊆ K such that

> n. This
sequence cannot have any limit points. Hence

∪_{j=n}^{∞}k_{j} ≡ C_{n} is a closed set. The
open sets

_{n=1}^{∞} provide an open cover which admits no finite subcover
since it is an increasing sequence of open sets, the

n^{th} of which fails to include

k_{n}.

■
This theorem and the earlier result shows that in ℝ, sequential compactness, compactness
and closed and bounded are all the same thing. The same conclusion can be drawn for
ℂ.