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:

\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:

\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:

\begin{align} \quad B(z, \delta) \subseteq f^{-1}(B(f(z), \epsilon) \end{align}

- Hence:

\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 $A$. |