TOPOLOGY IN C

Sessions

 

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 XC  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 sequencedefined on a setC. Then

converges uniformly on Dgiven 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 wonít 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

convergesconverges 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 LíHopitalís 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)