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Definitions
A subset U of C
is said to be open
if
for each u U there is a disk D(u,r)
such that D(u,r)  U.
A
neighborhood of a point x X C
is any open set U
C
such that x U .
A subset
F of
C
is said to be closed if its complement is open in
C.
A point
s is said to be a limit point
(or
accumulation point)
of a
subset S of
C
if each neighborhood of s contains at least one
point of S other than s.
Example.
The origin is an accumulation point of
C
and
C-{0}.
Theorem. F
is closed if and only if F contains all of its
limits points.
The following theorems are easy to prove:
Theorem.
Let  and
z=x+iy.
Then
 if and only if
 
Theorem.  is
Cauchy if and only if
 and
 are Cauchy.
Using this theorem, since
R
is complete
(every
Cauchy sequence in
R
converges to a point of
R),
it follows that
C
is complete.
Moreover,
C
is connected since the only clopen (closed and open)
subsets of
C
are the empty set
 and
C.
Definitions.
A subset D of
C
is said to be connected if D is not
the union of two nonempty sets A and B such that
 .
A subset D of
C
is called a domain
if
D is open and connected.
A subset D of
C
is said to be convex if the line segment joining
any two points in D lies in D.
We state the following two properties:
Theorem.
Suppose that D is a subset of
C
such that every two points of D can be joined by a
polygon in D, then D is connected.
Theorem.
If D is a domain, then every two points of D
can be joined by a polygon in D.
Definitions.
a) A collection of open sets is an open cover of a
set X if X is contained in the union of
these open sets.
A
subset K of
C
is said to be compact
if
from each open cover
 of K, one can extract a finite
subcover of K.
Definition.
A sequence of complex numbers
 is said to have a limit z
if  .
The following theorem is easy to prove:
Theorem.
Let  and
 Then
 and
Definition.
A sequence of complex numbers
is said to be a Cauchy sequence
if
  N such that
The following theorem is also easy to prove:
Theorem.
is a Cauchy sequence
 and
 are Cauchy sequences.
Since
R
is complete (every Cauchy sequence in
R
converges), it follows from the preceding theorem that
C
is complete.
Definitions.
a) Let  be a sequence of functions defined for
each  C.
We say that converges
to
f(z) if
and for each z
N(which depends on
 and z) such that
b) We say that
converges uniformly to f(z)
on D if
and for each z
N(which depends only on
) such that
Clearly uniform convergence implies convergence
but the converse is not true.
The following theorem extends from real variables:
Theorem (Cauchy Criterion for uniform convergence of
sequences)  converges
uniformly to f on
 C is Cauchy.
Proof.
( ) Given
 ,
Nsuch that
 and
Now 
 Given
,
Nsuch that
Fix z. Then
 is a Cauchy sequence in
C
which is complete. Thus
converges to
 say.
Fixing p and letting q we get
  .
For series of functions we have the
following criterion:
Theorem (Cauchy
Criterion for uniform convergence of series).
Consider
the sequence defined on a set C.
Then
 converges uniformly on D given
  such that
Proof.
( ) Consider the sequence of partial sums
with general term
 Since
converges uniformly on D,
 converges uniformly on D. By
Cauchy Criterion for uniform convergence of sequences,
given such that
 That is,
 .
( ) Using the hypothesis we can write
Again by Cauchy Criterion for uniform
convergence of sequences
 converges uniformly on D and
consequently converges uniformly on D.
Example.
Let
For a fixed n,
  1. Now
 1.
On the other
hand, for a fixed p,
0. Now
0.
This won’t
happen with uniform convergence as the following theorem
shows:
Theorem.
Let a be a limit point of a
subset D of
C.
Suppose that
converges uniformly to f on
D and that
 Then
 converges and
 ; that is,
 
Proof.
The proof
is left as an exercise.
Theorem.
Suppose that
converges
uniformly to f on D and
 is continuous on D. Then f
is continuous on D.
Proof.
Fix
 It is enough to show that f is
continuous at  Let
 . Since
converges uniformly to f on
D,  such that
 In particular, for n, such that
 when
 and
Consequently,
   
when
.
The
following is another useful criterion for uniform
convergence of sequences:
Theorem
(Weierstrass Criterion for uniform convergence of
sequences).
Suppose
 Put
 Then
 uniformly on D
 as
Proof.
Follows
easily from the definition of uniform convergence.
Following
is the Weierstrass corresponding criterion for series
which is frequently quite
useful in
verifying uniform convergence:
Theorem (Weierstrass Criterion for
uniform convergence of series).
Suppose
is a sequence of functions defined on
D such that
Then
 converges converges uniformly on D.
Proof.
If
converges, then, for arbitrary
, and for
 and n large enough:
 So
converges uniformly on D by
Cauchy Criterion for uniform convergence of series.
Example.
Let
 The series
 converges uniformly on [-a,a]
because
 which
is the nth term of a convergent geometric
series.
We turn
now to the connection between uniform convergence and
integration.
We begin
wth the following theorem whose proof is left as an
exercise:
Theorem.
Suppose
that is a sequence of integrable
functions on [a,b] which
converges
uniformly to f on [a,b]. Then f is
integrable and
Following
is an important corollary:
Corollary.
Suppose
that is a sequence of integrable
functions on [a,b] and
 uniformly on [a,b]. Then
(term-by-term integration)
Proof
of Corollary.
uniformly on [a,b] implies that
the sequence of partial sums { )} converges uniformly on [a,b].
Thus
As an
application we consider the following problem which shows
the power of uniform
convergence as well as the care that we must take in
similar situations.
Problem.
Consider the
sequence of functions with general term :
  and
1) Show that
 for every
2) With
 for
 prove that
3) Deduce
that the sequence of functions with general term
 converges uniformly to
 on
 ,
4) Deduce
that the sequence of functions with general term
 converges uniformly to
 on
,
5) Deduce
the value of 
Solution:
1)
 So ln
 Using L’Hopital’s Rule we get
2) Consider
In general, and with an induction proof,
Now
 and so
 is increasing and
Since
  and so
 is increasing. So
It follows
that
 is increasing. Thus
Therefore
 or
Thus
3)
 using 1).
By
Weierstrass Criterion for uniform convergence of sequences,
converges
uniformly to
on
4) Use 3)
and   since
  on
and
5)
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