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
such that K ⊆∪i=1mUi. In words, it says that every open cover admits a finite subcover.
Proposition 4.7.17 Every closed interval
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 I2.
It is half the length of
. Now do for
exactly what was done for I1.
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 In
times the length of I1
. 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
. Now for all n
enough, the length of In
which contains p,
is less than δ.
is contained in U
contrary to the definition of In
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
Thus from Proposition 4.7.17
is compact. Then if
is an open cover of K,
is an open cover of
. It follows that there are finitely many of the
open sets in
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
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
is the only
limit point of this sequence. Hence
is a closed set. Then since p
is an open cover of K but it admits no finite subcover. This is because the sequence of open
sets is increasing and ℝ ∖ Cn fails to contain pn ∈ K. Thus K is closed. Why is K
bounded? If not, there exists a sequence
sequence cannot have any limit points. Hence ∪j=n∞kj ≡ Cn
is a closed set. The
provide an open cover which admits no finite subcover
since it is an increasing sequence of open sets, the nth
of which fails to include kn
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