Open and Closed Set Criterion for the Continuity of Complex Functions

Open and Closed Set Criterion for the Continuity of Complex Functions

On the Continuity of Complex Functions page we said that a complex function $f : A \to \mathbb{C}$ is continuous at $z_0 \in A$ if:

(1)
\begin{align} \quad \lim_{z_0 \to z} f(z) = f(z_0) \end{align}

We then proved a lemma that said $f$ is continuous at $z_0$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that:

(2)
\begin{align} \quad f(B(z_0, \delta)) \subseteq B(f(z_0), \epsilon) \end{align}

We will now look at a very important criterion for determining if a complex function is continuous. We will see that a complex function $f$ is continuous if and only if the inverse image of any open set in $\mathbb{C}$ is also open in $A$ (and similarly, $f$ is continuous if and only if the inverse image of any closed set in $\mathbb{C}$ is also closed in $A$).

 Theorem 1: Let $f : A \to \mathbb{C}$. Then $f$ is continuous if and only if the inverse image of every open set in $\mathbb{C}$ is open in $A$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous on $A$. Let $V \subseteq \mathbb{C}$ be an open set in $\mathbb{C}$.
• Let $z \in f^{-1}(V)$. Then $f(z) \in V$. Since $V$ is open in $\mathbb{C}$ and $f(z) \in V$ there exists an $\epsilon > 0$ such that the ball centered at $f(z)$ with radius $\epsilon$ is fully contained in $V$, that is:
(3)
\begin{align} \quad B(f(z), \epsilon) \subseteq V \end{align}
• Since $f$ is continuous on $A$ it is continuous at $z$, from above, we have that there exists a $\delta > 0$ such that:
(4)
\begin{align} \quad f(B(z, \delta)) \subseteq B(f(z), \epsilon) \end{align}
• So $f(B(z, \delta)) \subseteq V$. So $z \in B(z, \delta) \subseteq f^{-1}(V)$. Hence $z \in \mathrm{int} f^{-1}(V)$. Since $z$ was arbitrary, $f^{-1}(V) = \mathrm{int} f^{-1}(V)$ and so $f^{-1}(V)$ is open.
• $\Leftarrow$ Suppose that for every open set $V$ in $\mathbb{C}$ we have that $f^{-1}(V)$ is open.
• Let $z \in A$. Let $\epsilon > 0$ be given. Consider $B(f(z), \epsilon)$. Observe this is an open set in $\mathbb{C}$. So $f^{-1}(B(f(z), \epsilon)$ is open in $A$. Since $f(z) \in B(f(z), \epsilon)$ we have that $z \in f^{-1}(B(f(z), \epsilon)$. So there exists a $\delta > 0$ such that:
(5)
\begin{align} \quad B(z, \delta) \subseteq f^{-1}(B(f(z), \epsilon) \end{align}
• Hence:
(6)
\begin{align} \quad f(B(z, \delta)) \subseteq B(f(z), \epsilon) \end{align}
• From the lemma stated at the top of the page, this implies that $f$ is continuous at $z$ and hence $f$ is continuous on $A$. $\blacksquare$
 Theorem 2: Let $f : A \to \mathbb{C}$. Then $f$ is continuous if and only if the inverse image of every closed set in $\mathbb{C}$ is open in $A$.