$\psi$ is upper semicontinous $\Longleftrightarrow\{z:\psi(z)Let $\Omega\subseteq\Bbb C$ be open. A function $\psi:\Omega\to[-\infty,+\infty[$ is called upper semicontinous if $\psi(z_0)\ge\limsup_{z\to z_0}\psi(z)\;\;\forall z_0\in\Omega$.
How can I show that $\psi$ is upper semicontinous IFF $\{z:\psi(z)<c\}$ is open $\forall c\in\Bbb R$?
I have no ideas! Can someone help me?
Thanks a lot!
 A: Suppose $\psi$ is upper semicontinuous. Fix $c\in \Bbb R$ and let $X_c := \{z : \psi(z) < c\}$. Let $z$ be a limit point of $\Bbb C \setminus X_c$. Then $z = \lim_{n\to \infty} z_n$ for some sequence $z_n$ in $\Bbb C \setminus X_c$. Since $\psi(z_n) \ge c$ for all $n \in \Bbb N$, $\limsup_{n\to \infty} \psi(z_n) \ge c$. Since $\psi$ is upper semicontinuous and $z_n \to z$, $\psi(z) \ge \limsup_{n\to \infty} \psi(z_n) \ge c$. Therefore $z \in \Bbb C \setminus X_c$, and consequently $\Bbb C\setminus X_c$ is closed. Therefore, $X_c$ is open.
Conversely, suppose $X_c$ is open for all $c\in \Bbb R$. Given $\epsilon > 0$ and $z_0 \in \Bbb C$, $X_{\psi(z_0) + \epsilon}$ is an open set. So since $z_0 \in X_{\psi(z_0) + \epsilon}$, there exists $\delta > 0$ such that for all $z$, $|z - z_0| < \delta$ implies $z\in X_{\psi(z_0) + \epsilon}$. That is, there is $\delta$-neighborhood of $z_0$ such that $\psi(z) < \psi(z_0) + \epsilon$ for all $z$ in the neighborhood. Hence, $\limsup_{z\to z_0} \psi(z) \le \psi(z_0) + \epsilon$. Since $\epsilon$ was arbitrary, $\limsup_{z\to z_0} \psi(z) \le \psi(z_0)$. Hence, $\psi$ is upper semicontinuous.
