$f$ 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.
-
$\begingroup$ How do you define transcendental? $\endgroup$– Alex YoucisMar 7, 2013 at 5:23
-
$\begingroup$ which is not a polynomial function, let say our $f$ is such a function. $\endgroup$– MyshkinMar 7, 2013 at 5:28
-
$\begingroup$ Related $\endgroup$– mrfMar 7, 2013 at 5:42
1 Answer
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.
-
-
-
-
$\begingroup$ please tell me what do u want to say at the last line? I mean just before "Thus, $f^{-1}(w)$ is infinite." $\endgroup$– MyshkinMar 7, 2013 at 9:55
-
$\begingroup$ Clearly the set of points in $B_1'$ that map to $w$ is infinite--if there were only finitely many, then we could find a $B_n'$ small enough that doesn't contain any preimage of $w$, and this would contradict that $w$ is in the image of every $B_n'$. $\endgroup$ Mar 7, 2013 at 9:56