The Metric Structure and Sequences of Complex Numbers 2

Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost, W.S. Anglin.

He who fights with monsters might take care lest he thereby become a monster. And if you gaze for long into an abyss, the abyss gazes also into you, Friedrich Nietzsche.

Recall

Definition. A sequence of complex numbers is a function a $\mathbb{N} \to \mathbb{C}$, mapping natural numbers to complex numbers.

Definition. A sequence $\{ a_n \}_{n=1}^∞$ is said to have a limit $L \in \mathbb{C}$ (or converges to a limit L) if $\forall \varepsilon > 0, \; \exists N \in \mathbb{N} \text{ such that } \forall n \geq N \Rightarrow |a_n - L| < \varepsilon$.

The following are all equivalent definitions of $a_n \to L$:

1. ε-N. $\forall \epsilon > 0, \exist N \in ℕ \text{ such that } |a_n - L| < \epsilon, \forall n ≥ N$.
2. Neighborhoods. Every neighborhood of L contains all but finitely many terms.
3. Distance. $d(a_n, L) \to 0$ as $n \to \infty$. The distance $d(a_n, L) = |a_n - L|$.
4. Component-wise Criterion. $a_n \to L \text{ in } \mathbb{C} \iff \begin{cases} x_n \to a \text{ in } \mathbb{R} \\ y_n \to b \text{ in } \mathbb{R} \end{cases}$.
5. Modulus & Argument . $|a_n| \to |L|$ and (if L ≠ 0) $arg(a_n) \to arg(L)$

Subsequences

Definition. A subsequence of {aₙ} is a sequence $\{ a_{n_k} \}_{k=1}^{\infty}$ where $n_1 < n_2 < n_3 < \cdots$ is a strictly increasing sequence of natural numbers.

Examples: If {aₙ} = {a₁, a₂, a₃, a₄, a₅, …}: {a₂, a₄, a₆, …} = $\{a_{2k}\}$ — even-indexed terms; {a₁, a₃, a₅, …} = $\{a_{2k-1}\}$ — odd-indexed terms; {a₁, a₄, a₉, a₁₆, …} = $\{a_{k^2}\}$ — square-indexed terms; {a₂, a₃, a₅, a₇, a₁₁, …} = $\{a_{p_k}\}$ — prime-indexed terms

Properties

Theorem. If $a_n \to L$, then every subsequence $a_{n_k} \to L$.

Proof. Let ε > 0. There exists N such that n ≥ N ⟹ |aₙ - L| < ε.

Since $n_k \to \infty$ (as n₁ < n₂ < ⋯, a strictly increasing sequence), there exists K such that $\forall k \ge K \implies n_k \ge N$

$\forall k ≥ K: |a_{n_k} - L| < \varepsilon$ (since $n_k \ge N$). $\blacksquare$

Corollary. If a sequence has two subsequences with different limits, the sequence diverges.

Example: For $a_n = i^n$ we have two subsequences with different limits: (i) subsequence $a_{4k} = 1 \to 1$, (ii) subsequence $a_{4k+1} = i \to i$. Different limits, so (iⁿ) diverges.

Cauchy Sequences and Completeness

Cauchy’s criterion

The standard definition of convergence ($|a_n - L| < \epsilon$) requires you to know the limit L before you can prove convergence.

In many situations it’s easier to check whether the terms of a sequence get arbitrarily close to each other —rather than trying to guess the limit and verify convergence directly. Cauchy’s criterion gives exactly that test.

Definition. A sequence $\{ a_n \}_{n=1}^∞$ is called a Cauchy sequence if $\forall \varepsilon > 0, \; \exists N \in \mathbb{N} \text{ such that } m, n \geq N \Rightarrow |a_m - a_n| < \varepsilon$.

Intuition. Beyond some index N, every pair of terms lies within an $\varepsilon$-neighborhood of each other (the gap between a_m and a_n is less than ε).

• ε represents an arbitrarily small positive number. It defines the “closeness” or “tolerance” we want between terms of the sequence.
• N is a point in the sequence beyond which all terms are “close enough” to each other.
• |a_m - a_n| < ε states that the absolute difference between any two terms a_m and a_n beyond N is smaller than our chosen tolerance ε. In other words, a sequence is Cauchy (and therefore convergent) if its terms eventually become arbitrarily close to each other (not necessarily to a specific limit).

The most important property of the complex numbers is that ℂ is a complete metric space. Completeness Theorem. A sequence in ℂ converges if and only if it is a Cauchy sequence.

Proof

(⟹) Convergent ⟹ Cauchy:

Suppose $a_n \to L$. For any ε > 0, there exists N such that n ≥ N implies |aₙ - L| < ε/2.

For m, $\forall n ≥ N: |a_m - a_n| \leq |a_m - L| + |L - a_n| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

(⟸) Cauchy ⟹ Convergent:

1. Cauchy sequences are bounded. Taking ε = 1, there exists N such that m, n ≥ N ⟹ |aₘ - aₙ| < 1.
   In particular, $|a_n - a_N| < 1$ for all n ≥ N, so $|aₙ| \le[\text{By the triangle inequality}] |a_n - a_N| + |a_N| \lt 1 + |a_N|$.
   Let $M = max\{|a_1|, ..., |a_{N-1}|, |a_N| + 1\}$. Then, $|a_n| \le M$ and {aₙ} is bounded.
   

   A Cauchy sequence “clumps together” as n increases. The bound M confines the sequence to a disk of radius M in C, preventing divergence to infinity.

2. $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$ (via $a+ib \to (a, b)$). The Bolzano-Weierstrass theorem guarantees that every bounded sequence in $\mathbb{R}^2$ has a convergent subsequence. Since {aₙ} is bounded, it has a convergent subsequence $a_{n_k} \to L$.
3. The full sequence converges to L. Let ε > 0.
   Since {aₙ} is Cauchy: ∃N₁ such that m, n ≥ N₁ ⟹ |aₘ - aₙ| < ε/2.
   Since $a_{n_k} \to L$: ∃K such that k ≥ K ⟹ $|a_{n_k} - L| < \varepsilon/2$.
   Choose nₖ ≥ N₁ (possible since nₖ → ∞). For n ≥ N₁: $|a_n - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$. Therefore, $aₙ \to L$. ∎

Why Completeness Matters

Completeness is what distinguishes ℂ (and ℝ) from incomplete spaces like ℚ.

Example in ℚ (incomplete space). The sequence of rational approximations to $\sqrt{2}$: $a_1 = 1, \quad a_2 = 1.4, \quad a_3 = 1.41, \quad a_4 = 1.414, \quad \ldots$

This is Cauchy in ℚ (terms get arbitrarily close), but does not converge in ℚ because $\sqrt{2} \notin \mathbb{Q}$. In ℂ (or ℝ), Cauchy sequences always have a limit within the space.

Examples

• A Real Sequence. Consider the sequence $a_n = \frac{1}{n}$ in ℝ. Cauchy check: $|\frac{1}{m}-\frac{1}{n}| \le max(\frac{1}{m}, \frac{1}{n}) \le \frac{1}{N}, \forall m, n \ge N$. So choosing $N \gt \frac{1}{\varepsilon}$ suffices. Therefore, this sequence {1/n} is Cauchy, hence converges (in fact to 0).

• A Complex Sequence. $z_n = \frac{1+i}{n}$. Cauchy check: $|z_m - z_n| = \left| \frac{1+i}{m} - \frac{1+i}{n} \right| = |1+i|\cdot\left|\frac{1}{m}-\frac{1}{n}\right| = \sqrt{2} \cdot \left|\frac{1}{m}-\frac{1}{n}\right| =[\text{Using the previous argument}] \le \frac{\sqrt{2}}{N} \lt \epsilon$. For any arbitrary ε > 0, choose N so that $N \gt \frac{\sqrt{2}}{\epsilon}$. Hence, {z_n} is Cauchy. Because ℂ is a complete metric space, every Cauchy sequence in ℂ converges. So, $z_n \to 0 \in \mathbb{C}$.

  This sequence is shrinking since the denominator grows without bound. It’s essentially a scaled-down version of $\frac{1}{n}$, stretched by the complex constant 1+i.

• A Non­Cauchy Oscillating Sequence. Consider the sequence a_n = (-1)ⁿ in ℝ. The terms oscillate or alternate between -1 and 1. No matter how large we choose N, we can always find m and n ≥ N such that |a_m - a_n| = |-1 - 1| = $2 \nleq \epsilon$ (if m is odd and n is even) for, say, ε = 1. Since we can’t make the difference arbitrarily small, this sequence is not Cauchy and indeed does not converge.

• Partial Sums: $a_n = \sum_{k=1}^{n} \frac{i^k}{k!}$ (partial sums of the series $\sum \frac{i^k}{k!}$).
  For m > n: $|a_m - a_n| = \left|\sum_{k=n+1}^{m} \frac{i^k}{k!}\right| \leq \sum_{k=n+1}^{m} \frac{1}{k!} \leq \sum_{k=n+1}^{\infty} \frac{1}{k!}$
  The series $\sum _{k=0}^{\infty }\frac{1}{k!}$ converges (to e), so its tail goes to 0 as $n \rightarrow \infty$, so for every $\varepsilon >0$, there exists N such that for all $m > n \geq N, |a_m-a_n|<\varepsilon$.
  Therefore, {aₙ} is Cauchy, hence convergent.
  We know $\sum _{k=0}^{\infty }\frac{i^k}{k!}=e^i=\cos(1) + i\sin(1)$. The sequence omits the k = 0 term, so $\sum _{k=1}^{\infty }\frac{i^k}{k!}=e^i-1$. Therefore, $\lim _{n\rightarrow \infty }a_n=\sum _{k=1}^{\infty }\frac{i^k}{k!}=e^i-1=(\cos(1)+i\sin(1))-1$.

• The sequence aₙ = iⁿ is NOT Cauchy.
  The cycle iⁿ ∈ {1, i, −1, −i} prevents convergence.
  Take ε = 1. For any N, let m = 4N and n = 4N + 2.
  Then aₘ = i⁴ᴺ = 1 and aₙ = i⁴ᴺ⁺² = -1.
  |aₘ - aₙ| = |1 - (-1)| = 2 > 1 = ε. This cleanly violates the Cauchy condition.

• Geometric series. Consider $S_n = \sum_{k=0}^{n} (\frac{i}{2})^k$. Does this sequence of partial sums converge?
  For a geometric series $\sum_{k=0}^{\infty} r^k$, it converges if $|r| \lt 1$.
  In our particular case, $r = |\frac{i}{2}| = \frac{1}{2} \lt 1$, so the series converges.
  The sum is $\frac{1}{1-r}=\frac{1}{1-\frac{i}{2}} = \frac{2}{2-i} =[\text{Rationalizing the denominator}] \frac{2}{2-i}\frac{2+i}{2+i} = \frac{4+2i}{5} = \frac{4}{5}+\frac{2}{5}i$.
  Cauchy Criterion Verification: $|S_m - S_n|$ for $m > n$:
  $|S_m - S_n| = \left| \sum_{k=n+1}^m \left(\frac{i}{2}\right)^k \right| \le \sum_{k=n+1}^m \left|\frac{i}{2}\right|^k = \sum_{k=n+1}^m \frac{1}{2^k} \le \sum_{k=n+1}^{\infty} \frac{1}{2^k}$
  As $n \to \infty$, the tail sum $\sum_{k=n+1}^{\infty} \frac{1}{2^k}$ is a geometric series with first term $a = \frac{1}{2^{n+1}}$ and ratio $\frac{1}{2}$.
  Using the formula for the sum of an infinite geometric series (valid since ∣r∣ < 1): $\sum_{k=n+1}^{\infty} \frac{1}{2^k} = \frac{a}{1-r} = \frac{\frac{1}{2^{n+1}}}{1-\frac{1}{2}} = \frac{\frac{1}{2^{n+1}}}{\frac{1}{2}} = \frac{1}{2^n}$
  It goes to 0 ($\frac{1}{2^n} \to 0$) as $n \to \infty$. This result is characteristic of convergent geometric series where the terms decay exponentially.
  Since the terms get arbitrarily close, the sequence is Cauchy, and therefore must converge to some complex number.

Bolzano-Weierstrass theorem

The Bolzano–Weierstrass theorem is a cornerstone of analysis and topology in both ℝⁿ and ℂ. This theorem captures the essence of compactness in finite-dimensional spaces.

Bolzano-Weierstrass theorem. Every bounded sequence in ℝⁿ (or ℂ) admits a convergent subsequence. Equivalently, every bounded infinite subset of ℝⁿ (or ℂ) has at least one limit point (accumulation point).

Deconstruction:

1. Bounded: The sequence is contained entirely inside some large disk B(0; R).
   

   A sequence {x_n} in ℝⁿ is bounded if there exists a number M > 0 such that ||x_n|| ≤ M for all n, where ||.|| denotes the Euclidean norm. In ℂ, a sequence {x_n} is bounded if there exists a number M > 0 such that |z_n| ≤ M for all n.

2. Subsequence: We can pick infinitely many terms $a_{n_1}, a_{n_2}, \dots$ (preserving order $n_1 < n_2 < \dots$. In other words, this is an increasing sequence of indices) that form a new sequence.
3. Convergent: This selection settles down to a limit point. A point x is a limit (accumulation) point of a set S if every open neighborhood of x contains infinitely many points of S.
4. Intuition. 💭🧠 If you confine (cram) infinitely many points into a finite region (a bounded set), they must “pile up” somewhere (there must be at least one point where they accumulate or cluster together).
5. Limit Points vs. Limits. It is crucial to distinguish these two concepts:
   Limit of a Sequence $L$: L is a limit point of {aₙ} if and only if there exists a subsequence converging to L. All terms (eventually) go there. It is unique.
   Limit Point of a Set (Accumulation Point): L is a limit point of {aₙ} if and only if there exists a subsequence converging to L. Some subsequence goes there. Not necessarily unique.

Proof (Bisection Method)

Cantor’s Intersection Theorem. Let (X, d) be a complete metric space (such as $\mathbb{R}^n$ or $\mathbb{C}$). Let $\{ K_j \}_{j \in \mathbb{N}}$ be a nested sequence of non-empty compact sets: $K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots$ (i.e., the sets are nested and decreasing). Then, the intersection is non-empty: $\bigcap_{j=1}^{\infty} K_j \neq \emptyset$.

Corollary. If additionally $diam(K_j) \to 0$, then $\bigcap_{j=1}^{\infty} K_j$ contains exactly one point.

The pigeonhole principle states that if n items (pigeons) are placed into m containers (holes), and n > m, then at least one container must hold more than one item.

Proof.

1. Let {aₙ} be a bounded sequence with |aₙ| ≤ M for all n. All terms lie in the square $Q_0 = [-M, M] \times [-M, M]$.
2. Bisection.
   Divide Q₀ into four equal quadrants. At least one quadrant contains infinitely many terms of the sequence (by pigeonhole).
   We have infinitely many terms of the sequence (points) contained in the square Q₀. When Q₀ is divided into four congruent smaller squares (quadrants), each point falls into exactly one quadrant.
   Since there are infinitely many points and only four quadrants, at least one quadrant must contain infinitely many points.
   Otherwise, if all quadrants had only finitely many points, their union would be finite —a contradiction.
   This is an application of the infinite pigeonhole principle: if an infinite set is partitioned into finitely many subsets, at least one subset is infinite.
   Call it Q₁ and pick a term $a_{n_1}$ from Q₁.
   Divide Q₁ into four quadrants. At least one contains infinitely many terms. Call it Q₂ and pick $a_{n_2}$ with n₂ > n₁.
   Continue: we get nested squares Q₀ ⊃ Q₁ ⊃ Q₂ ⊃ ⋯ with:
   (i) $diameter(Qₖ) = M \cdot 2^{-k-1/2} \to 0$; (ii) Each Qₖ contains infinitely many terms; (iii) A subsequence $a_{n_k} \in Q_k$.

   $Q_0$ has side length 2M, so its diameter is $2M\sqrt{2}$. Each division reduces the side length by half. For $Q_k$, its side length is $\frac{2M}{2^k} = \frac{M}{2^{k+1}}$. The diameter formula is as follows: $diameter(Qₖ) = (\frac{M}{2^{k+1}}) \cdot \sqrt{2} = \frac{M\sqrt{2}}{2^{k+1}} = M \cdot 2^{-k-1/2}$.

3. Apply Cantor’s Intersection Theorem. The Qₖ are nested, non-empty, compact sets with $diameter \to 0$. By Cantor: $\bigcap_{k=0}^{\infty} Q_k = \{L\}$ for some L ∈ ℂ.
4. The subsequence converges. Since $a_{n_k} \in Q_k$ and L ∈ Qₖ: $|a_{n_k} - L| \leq \text{diam}(Q_k) \to 0$. Therefore, $a_{n_k} \to L$ ∎

Examples

1. Sequence with one limit point: $a_n = \frac{1}{n} + i\frac{(-1)^n}{n}$. Only limit point: 0 (and the sequence converges to 0).

2. Oscillating Real Sequence. Consider the bounded sequence a_n = (-1)ⁿ, ∣a_n∣ = 1 ≤ 1. This sequence itself does not converge (no single limit). However, it has two convergent subsequences. Sequence with two limit points: {1, -1}.
   The subsequence {a₂ₙ} = 1, 1, 1, … converges to 1.
   The subsequence {a₂ₙ₊₁} = -1, -1, -1, … converges to -1.

3. Sequence with infinitely many limit points: $a_n = e^{i\theta_n} \text{ where } \theta_n \text{ is a rational in } [0, 2\pi)$ (and the set $\{ \theta_n \}$ is dense in $[0, 2\pi)$).
   Limit set = entire unit circle {z : |z| = 1}.
   Take any point on the unit circle: $z_0=e^{i\alpha }, \alpha \in [0, 2\pi)$.
   Because $\{ \theta _n\}$ is dense in $[0, 2\pi)$, for every $\varepsilon > 0$ there exists some n such that $|\theta _n-\alpha | < \varepsilon$.
   The map $\theta \mapsto e^{i\theta}$ is continuous, so $|\theta _n-\alpha |\rightarrow 0\quad \Rightarrow \quad |e^{i\theta _n}-e^{i\alpha }|\rightarrow 0$.
   Thus, for any neighborhood of $z_0$, there are infinitely many $a_n$ inside it (because density gives us infinitely many rationals approaching $\alpha$ and continuity), so $z_0$ is a limit point of the sequence.
   Since $z_0$ was arbitrary on the unit circle, every point with |z| = 1 is a limit point.

4. Dense sequence: $a_n = \text{n-th point in some enumeration of } \mathbb{Q} + i\mathbb{Q}$ (since $\mathbb{Q}+i\mathbb{Q}$ is countable, we can list its elements as $a_1, a_2, a_3, \dots$.) A subset $S \subseteq \mathbb{C}$ is dense if every complex number $z \in \mathbb{C}$ can be approximated arbitrarily closely by elements of $S$. Limit set = ℂ (every complex number is a limit point).
   The set $\mathbb{Q}+i\mathbb{Q}=\{ q_1+iq_2:q_1,q_2\in \mathbb{Q}\}$ is countable and dense in $\mathbb{C}$ (so its closure is all of $\mathbb{C}$. That is, $\overline{\mathbb{Q}+i\mathbb{Q}} = \mathbb{C}$):
   for any $z=x+iy \in \mathbb{C}$ and any $\varepsilon >0$, we can choose rationals $q_1,q_2$ with $|q_1-x|<\frac{\varepsilon }{\sqrt{2}},\quad |q_2-y|<\frac{\varepsilon }{\sqrt{2}},$
   so $|(q_1+iq_2)-(x+iy)|\leq \sqrt{(q_1-x)^2+(q_2-y)^2}<\varepsilon$.
   Because $\{ a_n\}$ enumerates a dense subset of $\mathbb{C}$, its closure is all of $\mathbb{C}$.
   Since any neighbourhood of $z$ contains infinitely many terms of the sequence (because the set is dense and each term is listed exactly once, but other terms crowd around), we can indeed pick a subsequence converging to $z$. More explicitly: for each $k \geq 1$, choose $n_k$ such that $|a_{n_k} - z| < 1/k$ and $n_k$ increasing.
   This sequence has every complex number as a limit point, illustrating that a countable dense set can fill an entire non-compact space (though $\mathbb{C}$ is not compact, its one-point compactification is the Riemann sphere).
   The sequence does not converge to any one point; it just “fills” the whole complex plane.

5. Grid in the Unit Square. Consider the set S = {$(\frac{1}{n}, \frac{1}{m}) : n, m \in \mathbb{N}$}. This is an infinite subset of the compact set [0, 1] × [0, 1] in ℝ².
   Limit Points: Axes and Origin. Points like (0, 0),(0, 1/m), and (1/n, 0) are limit points (e.g., $(\frac{1}{n}, \frac{1}{m}) \to (0, 0) \text{ as } n \to \infty$).
   Interior Points: $\forall (a, b) \in [0, 1]^2$, choose n > 1/a, m > 1/b (if a, b > 0) to get (1/n, 1/m) close to (a, b).
   The closure of S is [0, 1] ×[0, 1], which is compact (closed and bounded in $\mathbb{R}^2$). S is not closed (it misses limit points like (0, 0)), but its closure is compact.

In ℝⁿ (and ℂ), the Heine–Borel theorem theorem characterizes compact subsets as precisely those that are closed and bounded.

The Bolzano–Weierstrass theorem states that every bounded sequence in ℝⁿ (or ℂ) has a convergent subsequence. This allows extraction of convergent subsequences when we only know boundedness.

Limits Involving Infinity

The extended complex plane (or Riemann sphere) is $\hat{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$. This one-point compactification transforms the unbounded complex plane into a compact, connected surface topologically equivalent to a sphere.

Definition. A sequence $\{ a_n \}$ converges to infinity, written $a_n \to \infty$, if $\lim_{n \to \infty} |a_n| = \infin$. Equivalently, for every M > 0, there exists $N \in \mathbb{N}$ such that $\forall n \ge N, |aₙ| > M$.

The sequence eventually escapes every finite disk B(0;M), no matter how large. On the Riemann sphere, this is ordinary convergence to the north pole.

On the Riemann sphere, each point $z \in \mathbb{C}$, corresponds to a point on the unit sphere (excluding the north pole). As $|z| \to \infty$, the projection approaches the north pole — this is ∞.

Characterization. The following are equivalent:

1. $a_n \to \infty$
2. $|a_n| \to \infty$ in ℝ
3. For every M > 0, ∃N such that n ≥ N ⟹ |aₙ| > M
4. $\frac{1}{a_n} \to 0$ (when aₙ ≠ 0)