Cauchy's Convergence Criterion/Complex Numbers/Proof 1
Theorem
Let $\sequence {z_n}$ be a complex sequence.
Then $\sequence {z_n}$ is a Cauchy sequence if and only if it is convergent.
Proof
Lemma
Let $\sequence {z_n}$ be a complex sequence.
Let $\NN$ be the domain of $\sequence {z_n}$.
Let $x_n = \map \Re {z_n}$ for every $n \in \NN$.
Let $y_n = \map \Im {z_n}$ for every $n \in \NN$.
Then $\sequence {z_n}$ is a (complex) Cauchy sequence if and only if $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences.
$\Box$
Let $\sequence {x_n}$ be a real sequence where:
- $x_n = \map \Re {z_n}$ for every $n$
- $\map \Re {z_n}$ is the real part of $z_n$
Let $\sequence {y_n}$ be a real sequence where:
- $y_n = \map \Im {z_n}$ for every $n$
- $\map \Im {z_n}$ is the imaginary part of $z_n$
Necessary Condition
Let $\sequence {z_n}$ be a Cauchy sequence.
We aim to prove that $\sequence {z_n}$ is convergent.
We find:
- $\sequence {z_n}$ is a Cauchy sequence
- $\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma
- $\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Cauchy's Convergence Criterion on Real Numbers
- $\leadsto \sequence {z_n}$ is convergent by definition of convergent complex sequence.
$\Box$
Sufficient Condition
Let $\sequence {z_n}$ be convergent.
We aim to prove that $\sequence {z_n}$ is a Cauchy sequence.
We find:
- $\sequence {z_n}$ is convergent
- $\leadsto \sequence {x_n}$ and $\sequence {y_n} $ are convergent by definition of convergent complex sequence
- $\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Cauchy's Convergence Criterion on Real Numbers
- $\leadsto \sequence {z_n}$ is a Cauchy sequence by Lemma
$\blacksquare$