Show that $\cap _{M>0} \{z: -M<f(z)<M \}$ is connected. Suppose that $f$ is an entire function such that $f(0)=0$ and for every $M&gt;0$, the set $\{z\in \mathbb{C}: -M&lt;f(z)&lt;M\}$ is connected. Show that $\cap_{M&gt;0} \{z\in \mathbb{C}: -M&lt;f(z)&lt;M\}$ is connected.

I know that the intersection of nested, connected, compact sets is connected, but we do not have compactness here. A hint would be most welcome. Thank you.
Assume $f$ is not identically zero. Then the intersection (i.e. $f^{-1}(0)$) is discrete.
Then for sufficiently small $r&gt;0$, $f$ has no zero on $r\cdot S^1$. Let $h_r&gt;0$ be the minimum of the continuous function $|f(z)|$ on this compact set. We conclude that the single connected component of $\{\,z\in\mathbb C:|f|&lt;h_r\,\}$ is fully inside $rS 1$, i.e in $r\mathbb D$. Then 
$$\bigcap_{M&gt;0}\{\,z\in\mathbb C:|f|&lt;M\,\} \subseteq \bigcap _{r&gt;0}r\mathbb D=\{0\}.$$
If $f$ is constant, then we must have $f=0$ and the result follows, so assume $f$ is not constant.

Let $L_M = \{ z | |f(z)| &lt; M \}$. Then $z \in L_M$ for all $M&gt;0$ iff $|f(z)| &lt;M$ for all $M&gt;0$ iff $f(z) = 0$.

Let $L^* = \cap_{M&gt;0} L_M = f^{-1}( \{0\})$.
We are given that $f(0) = 0$, hence $0 \in L^*$

Suppose there is another zero $z \neq 0$. Since $L_M$ is open and connected, it is path connected, hence for $M={1 \over n}$ there is a path $\gamma_n$ from $0$ to $z$. 

Pick any $\delta&gt;0$ such that $\delta &lt; |z|$, and let $z_n$ be a point on $\gamma_n$ such that $|z_n| = \delta$. The sequence $z_n$ has an accumulation point $z_\delta$ and we must have $f(z_\delta) = 0$. Since $\delta&gt;0$ was arbitrary, it follows that $f=0$, a contradiction.

Hence we see that $L^* = \{0\}$.