Proof of Convergence in the Complex Numbers by Real and Imaginary Parts, and Completeness in the Modulus Metric

Write $x_{n}=\operatorname{Re}z_{n}$ and $y_{n}=\operatorname{Im}z_{n}$ for $n\in\mathbb{N}$. By The Distance on the Complex Numbers §distance, $d_{\mathbb{C}}(z,w)=|z-w|$ for $z,w\in\mathbb{C}$, where $|\cdot|$ is the modulus; for a real number it is the absolute value by The Modulus of a Complex Number §agreement, so the absolute values in Convergent Sequences of Real Numbers §converges and Cauchy Sequences of Real Numbers §cauchy are moduli. $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, so Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares applies to it.

Step 1 (parts of a difference). Let $z,w\in\mathbb{C}$. Then $-w=(-1)w$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs, so Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums §parts gives $\operatorname{Re}(z-w)=\operatorname{Re}z+(-1)\operatorname{Re}w=\operatorname{Re}z-\operatorname{Re}w$, using Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs in $\mathbb{R}$ for the last step; likewise $\operatorname{Im}(z-w)=\operatorname{Im}z-\operatorname{Im}w$. Hence, by Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums §components,

$$|\operatorname{Re}z-\operatorname{Re}w|\le|z-w|,\qquad|\operatorname{Im}z-\operatorname{Im}w|\le|z-w|,\qquad|z-w|\le|\operatorname{Re}z-\operatorname{Re}w|+|\operatorname{Im}z-\operatorname{Im}w|.$$

Components, only if. Suppose $(z_{n})$ converges in $(\mathbb{C},d_{\mathbb{C}})$ to some $c\in\mathbb{C}$. Let $\varepsilon>0$ be real and choose $N\in\mathbb{N}$ with $d_{\mathbb{C}}(z_{n},c)<\varepsilon$ for all $n\ge N$, by Convergent Sequence in a Metric Space §converges. For $n\ge N$, Step 1 gives $|x_{n}-\operatorname{Re}c|\le|z_{n}-c|<\varepsilon$, so $|x_{n}-\operatorname{Re}c|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. Thus $x_{n}\to\operatorname{Re}c$ by Convergent Sequences of Real Numbers §converges; likewise $y_{n}\to\operatorname{Im}c$. Both are convergent.

Components, if. Suppose $x_{n}\to a$ and $y_{n}\to b$ for real numbers $a,b$, and put $c=a+bi$, so that $\operatorname{Re}c=a$ and $\operatorname{Im}c=b$ by Real and Imaginary Parts of a Complex Number §parts. Let $\varepsilon>0$ be real and $\eta=\varepsilon/2$, so $0<\eta$ and $\eta+\eta=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. By Convergent Sequences of Real Numbers §converges choose $N_{1}\in\mathbb{N}$ with $|x_{n}-a|<\eta$ for all $n\ge N_{1}$, then $N_{2}\in\mathbb{N}$ with $|y_{n}-b|<\eta$ for all $n\ge N_{2}$, and let $N=\max\{N_{1},N_{2}\}$, which exists and satisfies $N_{1}\le N$, $N_{2}\le N$ by The Maximum and Minimum of Two Elements of a Total Order §exists and The Maximum and Minimum of Two Elements of a Total Order §bounds for the total order of $\mathbb{N}$ (The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §order). For $n\ge N$ we have $n\ge N_{1}$ and $n\ge N_{2}$ by transitivity of the order of $\mathbb{N}$, so Step 1 and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum give

$$d_{\mathbb{C}}(z_{n},c)=|z_{n}-c|\le|x_{n}-a|+|y_{n}-b|<\eta+\eta=\varepsilon,$$

so $d_{\mathbb{C}}(z_{n},c)<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. Hence $(z_{n})$ converges to $c=a+bi$ in $(\mathbb{C},d_{\mathbb{C}})$ by Convergent Sequence in a Metric Space §converges, and is convergent there by Convergent Sequence in a Metric Space §convergent.

Limit. In that case $(x_{n})$ and $(y_{n})$ are convergent by the paragraph Components, only if; let $a=\lim_{n\to\infty}x_{n}$ and $b=\lim_{n\to\infty}y_{n}$, to which these sequences converge by The Limit of a Convergent Sequence §limit. By the paragraph Components, if, $(z_{n})$ converges to $a+bi$ in $(\mathbb{C},d_{\mathbb{C}})$; since its limit there is the point to which it converges, unique by The Limit of a Convergent Sequence in a Metric Space §limit and Uniqueness of Limits in a Metric Space §unique, that limit is $a+bi$.

Complete. Let $(z_{n})$ be a Cauchy sequence in $(\mathbb{C},d_{\mathbb{C}})$. Let $\varepsilon>0$ be real and choose $N\in\mathbb{N}$ with $d_{\mathbb{C}}(z_{m},z_{n})<\varepsilon$ for all $m,n\ge N$, by Cauchy Sequence in a Metric Space §cauchy. For such $m,n$, Step 1 and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed give $|x_{m}-x_{n}|<\varepsilon$ and $|y_{m}-y_{n}|<\varepsilon$. So $(x_{n})$ and $(y_{n})$ are Cauchy sequences by Cauchy Sequences of Real Numbers §cauchy, hence convergent by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences §cauchy, and $(z_{n})$ is convergent in $(\mathbb{C},d_{\mathbb{C}})$ by the paragraph Components, if. Thus $(\mathbb{C},d_{\mathbb{C}})$ is complete by Complete Metric Space §complete.