Equivalence of Definitions of Continuous Complex Function

== Theorem ==
Let $A_1, A_2 \subseteq \C$ be subsets of the complex plane.
Let $f: A_1 \to A_2$ be a complex function from $A_1$ to $A_2$.
Let $a \in A_1$.
=== Definition using Limit ===
=== Epsilon-Delta Definition ===
=== Epsilon-Neighborhood Definition ===
=== Open Sets Definition ===
== Proof ==
=== Definition using Limit implies Epsilon-Delta Definition ===
Let $\epsilon > 0$.
By the Epsilon-Delta condition of , there exists $\delta > 0$, so for all $z \in A_1$:
:$ 0 < \cmod {z - a} < \delta \implies \cmod {\map f z - \map f a} < \epsilon $
In the case of $0 = \cmod{ z - a }$, we must have $z = a$, so
:$\cmod {\map f z - \map f a} = 0 < \epsilon$
=== Epsilon-Delta Definition implies Definition using Limit ===
Let $\epsilon > 0$.
By Epsilon-Delta Definition, there exists $\delta > 0$, so for all $z \in A_1$:
:$\cmod {z - a} < \delta \implies \cmod {\map f z - \map f a} < \epsilon $
By the Epsilon-Delta condition of the , it follows that
:$\ds \lim_{z \mathop \to a} \map f z = \map f a$
=== Epsilon-Delta Definition implies Epsilon-Neighborhood Definition ===
Let $A_1$ be open in $\C$.
Let $\epsilon > 0$.
By Epsilon-Delta Definition, there exists $\delta_0 > 0$, so for all $z \in A_1$:
:$\cmod {z - a} < \delta_0 \implies \cmod {\map f z - \map f a} < \epsilon$
By definition of $\epsilon$-neighborhood, $\cmod {z - a} < \delta_0$ implies $z \in \map { N_{\delta_0} } {a}$.
By definition of open set, there exists $\delta_1 > 0$ so $\map { N_{\delta_1} } {a} \subseteq A_1$.
Set $\delta = \min \paren{ \delta_0 , \delta_1 }$.
Now $\map { N_\delta } {a} \subseteq \map { N_{\delta_0} } {a}$, and $\map { N_\delta } {a} \subseteq \map { N_{\delta_1} } {a} $.
For all $z \in \map { N_\delta } {a}$, we have: 
:$\cmod {\map f z - \map f a} < \epsilon$
which implies $\map f z \in \map { N_\epsilon } { \map f a }$.
It follows that $ f \sqbrk{ \map {N_\delta} a } \subseteq \map { N_\epsilon } { \map f a }$.
=== Epsilon-Neighborhood Definition implies Epsilon-Delta Definition ===
Let $A_1$ be open in $\C$.
Let $\epsilon > 0$.
By Epsilon-Neighborhood Definition, there exists $\delta > 0$ such that:
:$ f \sqbrk{\map {N_\delta} a} \subseteq \map { N_\epsilon } { \map f a }$.
Let $z \in A_1$ with $\cmod{ z - a } < \delta$.
By definition of $\epsilon$-neighborhood, we have $z \in \map { N_\delta } {a}$.
Then $\map f z \in \map { N_\epsilon } { \map f a }$.
By definition of $\epsilon$-neighborhood, we have $\cmod {\map f z - \map f a} < \epsilon$.
=== Epsilon-Neighborhood Definition implies Open Sets Definition ===
Let $A_1$ be open in $\C$.
Let $U \subseteq \C$ be open in $\C$.
Suppose $f^{-1} \sqbrk{ U } = \emptyset$.
Empty Set is Open in Metric Space shows that $f^{-1} \sqbrk{ U }$ is open in $\C$.
Otherwise, let $a \in f^{-1} \sqbrk{ U }$.
Then $\map f a \in U$.
As $U$ is open, there exists $\epsilon > 0$ so $\map { N_\epsilon } { \map f a } \subseteq U$.
By Epsilon-Neighborhood Definition, there exists $\delta > 0$ such that:
:$ f \sqbrk{ \map { N_\delta } a } \subseteq \map { N_\epsilon } { \map f a }$
For $z \in \map {N_\delta} a$, we have $\map f z \in \map { N_\epsilon } { \map f a } \subseteq U$.
It follows that $\map {N_\delta} a \subseteq f^{-1} \sqbrk{ U }$.
By definition of open sets, it follows that $f^{-1} \sqbrk{ U }$ is open.
=== Open Sets Definition implies Epsilon-Neighborhood Definition ===
Let $A_1$ be open in $\C$.
Let $a \in A_1$.
Let $\epsilon > 0$.
Open Ball of Metric Space is Open Set shows that $\map{ N_{ \epsilon} }{ \map f a }$ is open in $\C$.
By Open Sets Definition, it follows that $f^{-1} \sqbrk{ \map{ N_{ \epsilon} }{ \map f a } }$ is open in $\C$.
We have $ a \in f^{-1} \sqbrk{ \map{ N_{ \epsilon} }{ \map f a } }$.
As $f^{-1} \sqbrk{ \map{ N_{ \epsilon} }{ \map f a } }$ is open, there exists $\delta > 0$ such that:
Category:Complex Analysis