User:Anghel/Sandbox

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 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 $A_1, A_2$ 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 $A_1, A_2$ 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 $A_1, A_2$ 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 $\epsilon_0 > 0$ so $\map { N_{ \epsilon_ 0 } } { \map f a } \subseteq U$.

As $A_2$ is open, there exists $\epsilon_1 > 0$ so $\map { N_{ \epsilon_ 1} } { \map f a } \subseteq A_2$.

Set $\epsilon = \min \paren { \epsilon_0, \epsilon_1 }$.

Now $\map { N_\epsilon } {\map f a} \subseteq \map { N_{\epsilon_0} } {\map f a}$, and $\map { N_\epsilon } {\map f a} \subseteq \map { N_{\epsilon_1} } {\map f a} $.

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 $A_1, A_2$ 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:

RemCategory:Complex Analysis