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ANALYSIS II

Proof of Compactness Characterizatons

__Compactness Characterization
Theorem__

Suppose that K is a subset of a metric space
X, then the following are equivalent:

- K is compact,
- K satisfies the Bolzanno-Weierstrass
property (i.e., each infinite subset of K
has a limit point in K),
- K is sequentially
compact (i.e., each sequence from K has a
subsequence that converges in K).

__Defn__
A set K in a metric
space X is said to be *totally bounded*,
if for each > 0 there
are a finite number of open balls with radius which
cover K. Here the centers of the balls and the total number will depend
in general on .
__Defn__ A
set D is said to be *dense *in
a set A if each neighborhood of each point x of A contains a member from
D. A set A in a metric space is called *separable
*if it has a countable dense
subset.

(__Compactness the
Bolzanno-Weierstrass property__)

Suppose K is compact, but that A is an infinite
subset of K with no limit point in K. But K is closed since it is compact,
so the derived set of A is empty and A is therefore closed. In particular,
each point of A must be isolated. Hence for each x in A there is a ball
B_{x} which contains only a single member from A (namely x). The
collection of these balls (together with the complement of A),

**{**B_{x}
**|** x belongs to A**}**
** {**A^{c}**}**

is an open cover for K. In any finite
subcover (which must exist by the compactness of K), one of the balls must
contain an infinite number of elements of A since A is infinite. **Contradiction**,
since each one of these balls has exactly one member from A.

(__Bolzanno-Weierstrass property
Sequential Compactness__)

Suppose {x_{n}}_{n} is
a sequence in K and K has the Bolzanno-Weierstrass property. If the range
of the sequence is finite, we are done since one of the values must be
repeated infinitely often and then that subsequence converges since it
is a constant sequence. In the case the range of the sequence is infinite,
we apply the B-W property to obtain a limit point x_{0} in K. A
subsequence is then easily constructed which converges to x_{0}
.

__(Sequential Compactness Completeness__)
[**Homework**]

__(Sequential Compactness Totally
Bounded__)
[**Homework**]

(Hint: Suppose not, then by induction there exists >
0 and a sequence of points x_{1},
..., x_{n},... such that x_{n+1}
does not belong to the B(x_{0})
... B(x_{n})
. This will give rise to a contradiction since all members of the sequence
are at least units apart and
there can be no convergent subsequence. )

__(Sequential Compactness
separable and that the topology has a countable base__)

K is totally bounded so there are a finite
number of balls B_{1}(x_{1,1}),
..., B_{1}(x_{1,n(1)}), of radius
1 which cover K. Continuing with each natural number k, we find a finite
number of balls B_{1/k}(x_{k,1}),
..., B_{1/k}(x_{k,n(k)})
which cover K. The collection of all centers then forms a countable dense
subset of K. [**Homework:
Prove this**] Moreover, for each point
x_{0} in K and each neighborhood *O*
of x_{0} , there is a ball from this countable list containing
x_{0} which is contained in *O*,
i.e. the topology of K has a `*countable
base*'. Notice, in fact, that
each open set can be written as a union of this countable collection of
open balls.

(__Sequential Compactness
each open cover of K has a countable subcover__)

Let *C*
be an open cover of K. Each point x in K belongs to some *O*_{x}
in *C* . There is a countable
base (of balls) {B_{n}}_{n} for the topology so there
exists a natural number n(x) so that x belongs to B_{n(x)} which
is contained in *O*_{x}. For each of these balls B_{n}
which arise in this way, select one such *O*_{x} and (abusing
our notation and thereby our sensibilities) label it *O*_{n}
. This constructs the desired countable subcover of *C*
.

(__Sequential Compactness Compactness__)

Let *C*
be an open cover for a sequentially compact set K. Let {*O*_{1}
, ..., *O*_{n} , ...} be a countable subcover of *C*
which covers K. If {*O*_{1} , ..., *O*_{k}
} is not a finite subcover of K, then there is a member of K, x_{k+1}
say, which lies outside the union of this finite collection. By sequentially
compactness, there is a subsequence {y_{j}} of {x_{k}}
which converges to some y_{0} in K. **Contradiction**, since
y_{0} will be in one of these open sets and by the definition of convergence,
all but a finite number of the y_{j}'s will also belong to this same
set. But this is impossible by our construction.

__Note__
If one looks closely, we have in fact shown another important characterization
of compactness in metric spaces:

__Theorem__
A set K in a metric space is compact if
and only if it is complete and totally bounded.

[**Extra
Credit Homework**]

*Robert Sharpley Feb 18 1998*