How can you prove that a nonconstant function has a countable number of zeros?

[sbashrawi]

Homework Statement

Let f be a nonconstant function. Prove that f has at most countably many zeros

Homework Equations

The Attempt at a Solution

 

[Dick]

What do you know about analytic functions where the zeros have a accumulation point? Can an uncountable set not have a accumulation point?

 

[sbashrawi]

I don't know what you mean by that since zeros of ananlytic function are isolated points and non of them is a limit point.

 

[HallsofIvy]

That was exactly Dick's point! If there were an uncountable set of zeros, there would have to be an accumulation point, contradicting the fact that the zeros are isolated.

 

[sbashrawi]

You mean that we can prove it by contradicion.

Let S be the set of zeros of f and suppose that it is uncountable.

then we can get a sequence of these zeros convergent to a point in z.

This gives that the zeros are not isolated.

but, if this is what you mean, how can we be sure that we have a convergent series in S.

 

[Dick]

You mean that we can prove it by contradicion.

Let S be the set of zeros of f and suppose that it is uncountable.

then we can get a sequence of these zeros convergent to a point in z.

This gives that the zeros are not isolated.

but, if this is what you mean, how can we be sure that we have a convergent series in S.

Right. So the trick is to show there is a convergent sequence. Divide the plane up into regions of finite area that cover the whole plane. Like squares of edge size one centered on each point m+n*i, where m and n are integers. Can you show at least one of those squares contains an infinite number of zeros?

 

[sbashrawi]

suppose that f doesn't have caountably many zeros.
then it has infinitely many zeros in one of this partition.
Infinitely many zeros in a bounded set implies an accumulation point of zeroes, and an accumulation point of zeros for an analytic function implies that that function is zero everywhere.
contradiction

So f has at most countably many zeros.

 

[Dick]

suppose that f doesn't have caountably many zeros.
then it has infinitely many zeros in one of this partition.
Infinitely many zeros in a bounded set implies an accumulation point of zeroes, and an accumulation point of zeros for an analytic function implies that that function is zero everywhere.
contradiction

So f has at most countably many zeros.

Ok. You didn't make it terribly clear why one element of the partition contains infinitely many zeros. Do you know why?

 

[sbashrawi]

I think this comes from the infinitely many zeros ( uncountable assumption)

 

[Dick]

I think this comes from the infinitely many zeros ( uncountable assumption)

Still unclear. You can have one zero in each square. That's an infinite number of zeros. Sure, it's the uncountable assumption but how does it work?

 

[3029298]

http://math.nyu.edu/student_resources/wwiki/index.php/Complex_Variables:_2006_January:_Problem_5" using Taylor's theorem.