0

Let $f$ be an entire function. Let $Z(f)$ be the set of all zeros of $f$ which is further uncountable. We have to show that $Z(f)$ has a limit point in $\mathbb{C}$.

My Approach :

Suppose, for all $z \in \mathbb{C}$, $z$ is not a limit point of $Z(f)$. Then every point of $Z(f)$ is an isolated point. This implies there is an open ball , centered at $w$ and having some positive radius $r_w$, which doesn't contain any point of $Z(f)$ except $w$, for every $w$ in $Z(f)$. We denote this open ball by $\cal B_w$. Let $\cal X$ be the set of all such balls, i.e., $\cal X=\lbrace B_w:w\in$ $ Z(f)$, $\cal B_w$ $\cap$ $ Z(f)\setminus\lbrace w \rbrace=\phi \rbrace$.

My initial thought was to establish the set $\cal X$ is countable via the fact that the set $\lbrace r_1+i\cdot r_2:r_1,r_2 \in \mathbb{Q} \rbrace$ is dense in $\mathbb{C}$ and each that open ball definitely contains a point $r_1+i\cdot r_2$ for some $r_1,r_2 \in \mathbb{Q}$, and hence get the contradiction. But I'm having trouble to prove the bijection. We know that $\cal B_w$ $\cap$ $ Z(f)\setminus\lbrace w \rbrace=\phi$ but we can't know that $\cal B_{w_1} \cap \cal B_{w_2} = \phi$. I mean there can be points which has rational number as a real and imaginary part, and which are members of both open balls. So, there may be problem in injection of the map.

Someone please guide me.

risefrominfinite
  • 840
  • You should look at these posts: 1) https://math.stackexchange.com/questions/1334557/entire-function-with-uncountably-many-zeros 2) https://math.stackexchange.com/questions/27546/zeroes-of-holomorphic-function 3) https://math.stackexchange.com/questions/1664066/analytic-functions-with-uncountable-zeros 4) https://math.stackexchange.com/questions/892864/can-a-holomorphic-function-have-uncountable-many-zeros – MMASRP63 Sep 27 '17 at 03:35

1 Answers1

1

Let $K_n:=\{z \in \mathbb C: |z| \le n\}$ for $n \in \mathbb N$ and suppose that $Z(f)$ has no limit point in $ \mathbb C$. Since $f$ is continuous and $K_n$ is compact, $Z(f) \cap K_n$ must be finite or empty. But then

$$Z(f)= \bigcup _{n=1}^{\infty}Z(f) \cap K_n$$

is at most countable.

Fred
  • 77,394