Cauchy's Convergence Criterion/Complex Numbers/Proof 2

Theorem
Let $\left \langle {z_n} \right \rangle$ be a complex sequence.

Then $\left \langle {z_n} \right \rangle$ is a Cauchy sequence it is convergent.

Proof
Let $\left \langle {x_n} \right \rangle$ be a real sequence where
 * $x_n = Re\left( {z_n} \right)$ for every $n$


 * $Re\left( {z_n} \right)$ is the real part of $z_n$

Let $\left \langle {y_n} \right \rangle$ be a real sequence where
 * $y_n = Im\left( {z_n} \right)$ for every $n$


 * $Im\left( { z_n} \right)$ is the imaginary part of $z_n$

Lemma 1
$\left \langle {z_n} \right \rangle$ is a Cauchy sequence $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ are Cauchy sequences.

Necessary Condition
Let $\left \langle {z_n} \right \rangle$ be a Cauchy sequence.

This means that, for a given $\epsilon > 0$:


 * $\exists N: \forall m, n \in \N: m, n \ge N: \left|{z_n - z_m}\right| < \epsilon$

We have, for $m, n \ge N$:

Thus, by definition, $\left \langle {x_n} \right \rangle$ is a Cauchy sequence.

A similar argument shows that $\left \langle {y_n} \right \rangle$ is a Cauchy sequence.

Sufficient Condition
Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be Cauchy sequences.

This means for $\left \langle {x_n} \right \rangle$ that, for a given $\epsilon > 0$:
 * $\exists N_1: \forall m, n \in \N: m, n \ge N_1: \left|{x_n - x_m}\right| < \dfrac \epsilon 2$

Also, for $\left \langle {y_n} \right \rangle$:


 * $\exists N_2: \forall m, n \in \N: m, n \ge N_2: \left|{y_n - y_m}\right| < \dfrac \epsilon 2$

Let $N = max(N_1, N_2)$.

We have, for $m, n \ge N$:

Thus, by definition, $\left \langle {z_n} \right \rangle$ is a Cauchy sequence.

Lemma 2
$\left \langle {z_n} \right \rangle$ is a convergent $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ are convergent.

Necessary Condition
Let $\left \langle {z_n} \right \rangle$ converge to the complex number $l$.

This means that, for a given $\epsilon > 0$:
 * $\exists N: n > N \implies \left|{z_n - l}\right| < \epsilon$

We have, for $n > N$:

Thus, by definition, $\left \langle {x_n} \right \rangle$ is convergent.

A similar argument shows that $\left \langle {y_n} \right \rangle$ is convergent.

Sufficient Condition
Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be convergent.

This means for $\left \langle {x_n} \right \rangle$ that, for a given $\epsilon > 0$:


 * $\exists a \in \R, N_1 \in \N: n > N_1 \implies \left|{x_n - a}\right| < \dfrac \epsilon 2$

Also, for $\left \langle {y_n} \right \rangle$:


 * $\exists b \in \R, N_2 \in \N: n > N_2 \implies \left|{y_n - b}\right| < \dfrac \epsilon 2$

Let $N = max\left( {N_1, N_2} \right)$.

Let $i = \sqrt {-1}$.

Then, for $n > N$:

Thus, by definition, $\left \langle {z_n} \right \rangle$ is convergent.

We aim to prove that $\left \langle {z_n} \right \rangle$ being a Cauchy sequence is equivalent to $\left \langle {z_n} \right \rangle$ being convergent.

We have:


 * $\left \langle {z_n} \right \rangle$ is a Cauchy sequence


 * $\Longleftrightarrow$ $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ are Cauchy sequences (by Lemma 1)
 * $\Longleftrightarrow$ $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ are convergent (by Sequence is Cauchy Sequence iff it is Convergent/Real Case)
 * $\Longleftrightarrow$ $\left \langle {z_n} \right \rangle$ is convergent (by Lemma 2)