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$f$ is a transecendental entire funtion I need to show $\{w : f^{-1}(w) \text{ is infinite}\}$ is dense in $\mathbb{C}$ I have no idea how to prove it, please help.

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How do you define transcendental? –  Alex Youcis Mar 7 '13 at 5:23
    
which is not a polynomial function, let say our $f$ is such a function. –  Une Femme Douce Mar 7 '13 at 5:28
    
Related –  mrf Mar 7 '13 at 5:42

1 Answer 1

up vote 3 down vote accepted

Here are the facts that you will need to know for the following (short proof):

Theorem: The only entire functions $f$ with a pole at infinity (i.e. such that $\displaystyle f\left(\frac{1}{z}\right)$ has a pole at $0$) are polynomials

Remark: You may know this result as "the only meromorphic functions on $\mathbb{P}^1$ are rational".

$\text{ }$

Theorem(Open Mapping Theorem:) Every holomorphic function is an open mapping

$\text{ }$

Theorem(Baire Category Theorem): In a complete metric space, the countable intersection of open dense subsets is dense.

You can find proofs for the first two in any standard complex analysis text, and the last in any standard point-set tex.


Ok, since $f$ is not a polynomial we know that $f$ has an essential singularity at $\infty$. In other words, $\displaystyle f\left(\frac{1}{z}\right)$ has an essential singularity at $0$ (this is the first theorem). Let $B_n'$ be the punctured disk of radius $\displaystyle \frac{1}{n}$ centered at $0$. Since $0$ is an essential singularity the Casorati-Weierstrass theorem tells us that each $f(B_n')$ is dense in $\mathbb{C}$. But, by the open mapping theorem we also know that $\displaystyle f(B_n')$ is open in $\mathbb{C}$. Thus, $\left\{f(B_n')\right\}$ is a countable collection of dense open subsets of $\mathbb{C}$, and thus by the Baire Category Theory $\displaystyle \bigcap_n f(B_n')$ is dense in $\mathbb{C}$. But, note that

$$\displaystyle \bigcap_n f(B_n')\subseteq\left\{w\in\mathbb{C}:f^{-1}(w)\text{ is infinite}\right\}$$

Indeed, if $\displaystyle w\in\bigcap_n f(B_n')$ then there exists a point $z_n$ in each $B_n'$ such that $f(z_n)=w$. Since any given point is in finitely many $B_n'$ we see that $\{z_n\}$ is infinite. Thus, $f^{-1}(w)$ is infinite.

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could you explain the first line? –  Une Femme Douce Mar 7 '13 at 5:32
    
and the last line too. –  Une Femme Douce Mar 7 '13 at 5:34
    
@CityOfGod I've edited. –  Alex Youcis Mar 7 '13 at 5:38
    
Thank you very much –  Une Femme Douce Mar 7 '13 at 5:43
    
please tell me what do u want to say at the last line? I mean just before "Thus, $f^{-1}(w)$ is infinite." –  Une Femme Douce Mar 7 '13 at 9:55

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