Theorem 5.2.6: Heine-Borel Theorem
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A set
S of real numbers is compact if and only if every open
cover
C of
S can be reduced to a finite subcovering.
Context
Proof:
First, assume that every open cover
C of
S can be
reduced to a finite subcovering. We will show that
S must
then be closed and bounded, which means by the previous result that
S is compact.
S must be bounded: Take the collection
C = { :
S},
where
=
( - 1,
+ 1).
Then this collection is an open cover of S, and by assumption
can be reduced to a finite subcovering of S. But if
aj1 is the smallest of the centers of
the sets , and
aj2 is the largest one, then S is
contained in the set
( aj1 - 1,
aj2 + 1) and is therefore bounded.
S must be closed: Suppose S was not closed. Then there
exists an accumulation point s of S that is not
contained in S. Since s is an accumulation point of
S we know:
- for any n > 1 there exists
an S
with
| s - an | < 1 / n
because every neighborhood of
s must contain elements from
S. The sequence
{ an } clearly converges to
s. Define the
collection of sets
- C = { comp([s - 1/n, s + 1/n]), n > 0 }
Then each set in
C is open, being the complement of closed
sets. They also cover
S, because the only point that this
collection is missing is the point
s, which is not part of
S. By assumption, a finite subcover already covers
S.
If
N is the largest index of that finite subcovering, then
aN+1 is not part of that subcovering. However,
aN+1 is an element of
S, so that this
subcovering can not cover
S. That is a contradiction, showing
that if
S was not closed, not every covering of
S can
be reduced to a finite subcovering. Hence,
S must be closed.
Now we have to prove the other direction. Assume therefore that
S compact. Let C be any open cover of S. We
need to reduce C to a finite subcover. Since S is
compact, we know it is closed and bounded. Then
a = inf(S) and
b = sup(S)
are both part of
S ( Why ?). Define the
set A as
- A = { x: x [a, b]
and a finite subcollection of C covers
[a, x] S }
Then the set
A is not empty (because
a A).
Define
Since
A is a subset of
[a, b], we know that
a < c b. Suppose
c < b. Since
S is closed, comp(
S) is open.
Therefore, if
c comp(
S) then
there exists an open neighborhood
U of
c that is
contained in
[a, b] (because
c < b) and disjoint
from
S. But then
c can not be the supremum of the
set
A. Therefore, if
c < b then
c S. Then
c must be contained in some set
from the open
cover
C of
S. Choose two points
y and
z in
with
y < c < z. As before, there exists a finite subcollection
of
C whose members cover
[a, y] S.
Then these sets, together with
cover
[a, z] S. But
then
z A,
which means again that
c can not be the upper bound for
A. This means that assuming
c < b leads to a
contradiction, so that
c = b. But that will be exactly what we
need. If
sup(A) = c = b, then let
be that member
of the open cover
C that contains
b. There exists some
open neighborhood
(b - ,
b + )
contained in
. But
b -
is not an upper bound for
A, so there exists
x with
x > b - and
x A. Then
[a, x] S is
covered by a finite number of members of
C. Together with the
set
these sets
form a finite open cover for
S.
We have indeed reduced the open cover of S to a finite
subcovering of S, finishing the proof. I think.