Boundedness, Compactness & Connectivity in the Complex Plane

To err is human, to blame it on someone else is even more human, Jacob’s Law

Recall

A complex number is specified by an ordered pair of real numbers (a, b) ∈ ℝ² and expressed or written in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit, defined by the property i² = −1 ⇔ i = $\sqrt{-1}$, e.g., 2 + 5i, $7\pi + i\sqrt{2}.$ ℂ= { a + bi ∣a, b ∈ ℝ}.

Definition. Let D ⊆ ℂ be a set of complex numbers. A complex-valued function f of a complex variable, defined on D, is a rule that assigns to each complex number z belonging to the set D a unique complex number w, f: D ➞ ℂ.

• The set D is called the domain of the complex function f, D = Dom(f).
• The set of all actual outputs {f(z): z ∈ D} is called the range (or image) of the complex function f, also denoted f(D) = {f(z): z ∈ D}.

We often call the elements of D as points. If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. f: D ➞ ℂ means that f is a complex function with domain D. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts.

Definition. A set S ⊆ ℂ is said to be bounded if there exists an M > 0 such that |z| < M for every z ∈ S. This means all points in S lie inside the open disk of radius M centered at the origin, B_M(0) = {z: |z| < M}

Definition. A set S ⊆ ℂ is compact if it is both closed and bounded. This is a direct consequence of the Heine-Borel theorem in ℝ², since ℂ ≅ ℝ².

Definition. Let S be an arbitrary set in the complex plane, S ⊆ ℂ. An open cover of a set S is a collection of open sets $\mathbf{U}$ = { $\mathbf{U_{i \in I}}$} such that $S ⊂ \cup_{i ∈I}{\mathbf{U_i}}$. A finite subcover is a finite subcollection {$ {\mathbf{U_{i_1}}}, {\mathbf{U_{i_2}}}, ···, {\mathbf{U_{i_n}}}$ } $⊂ \mathbf{U}$ that still covers S.

Definition. A set S in the complex plane (S ⊆ ℂ) is compact if every open cover of S admits a finite subcover ↭[Heine-Borel Theorem] S is both closed and bounded in ℂ.

Counter-examples (Noncompact Sets)

Open Unit Disk B(0, 1) is not compact because:

1. It is not closed (it does not contain its boundary circle { z ∈ ℂ : ∣z∣ = 1}.
2. $\mathbf{U}$ = {B(0,r) : 0 < r < 1}, the open cover $∪_{0 < r < 1}B(0, r) = B(0, 1)$, but no finite subcollection covers the whole disk (any finite max-radius r_max = max{r₁,r₂, ···, r_n}, r_max < 1 misses points with ∣z∣ near 1).

Infinite Discrete Set. S = {n + i·0: n ∈ ℕ} ⊆ ℂ is not compact because

1. It is not bounded. Its points extend infinitely along the real axis.
2. The opening cover $\mathbf{U}$ = {B(n; 1/2): n ∈ ℕ} ⊇ S has no finite subcollection that covers all natural indices.

Heine-Borel Theorem in ℂ

Heine-Borel Theorem (in ℂ). A subset S of the complex plane S ⊆ ℂ is compact, that is, every open cover of S has a finite subcover if and only if the set S is both closed and bounded in ℂ.

1. S is closed ↭ The complement of S, ℂ∖C ≋ S^c is an open set ↭(if and only if) $S =A uyoip[]\-\bar{S} = S ∪ ∂S$ ↭ S contains all of its limits points ↭ S contains its boundary (all of its boundary points).
2. S is bounded: This means that S can be contained within an open disk of finite radius about the origin. In other words, there exists some real number M > 0 such that |z| ≤ M for all z in S.

Examples

• The unit circle in ℂ, defined as {z ∈ ℂ : |z| = 1}, is compact because:
  Closed: ℂ∖{z ∈ ℂ : |z| = 1} is open.
  It is obviously bounded, e.g., by a disk of radius 2 centered at the origin. \56f4ytg3sd2a`3456H7G 89JKL-0;* Closed Disk: The set $\overline{B_r(0)} = \{z ∈ ℂ : |z| ≤ r\}$ (a disk of radius r including the boundary) is compact. It’s closed because it contains its boundary (the circle |z| = r), and it’s bounded by definition (by r).

• Closed Rectangle: Consider the set {z = x + iy : a ≤ x ≤ b, c ≤ y ≤ d}, where a, b, c, and d are real numbers. This is a rectangle in the complex plane, including its edges. It is compact because it is both bounded (bounded by max{∣a∣, ∣b∣, ∣c∣, ∣d∣}) and closed as a Cartesian product of closed intervals [a, b] x [c, d].

• Finite Set of Points: Any finite set of points S = {z₁, z₂, ···, z_n} in ℂ is compact. It is trivially bounded (take M = max_i(z_i), |z| ≤ M for every z ∈ S). Furthermore, it is also closed because it contains all its limit points since finite sets have no limit points (there’s no way for a small enough disk or neighborhood around any of these points to contain another distinct point from the set.).

• The set {z ∈ ℂ : Re(z) ≥ 0 and |z| ≤ 1} = {z: ℜ(z) ≥ 0} ∩ {z: ∣z∣ ≤ 1}. The half-plane {z: ℜ(z) ≥ 0} is closed. The closed disk {z: ∣z∣ ≤ 1} is closed and bounded. An intersection of any family of closed sets is closed, and T ⊂ {z: ∣z∣ ≤ 1} is bounded, too. Hence, T is compact.

• A concrete finite example. The set S = {1, i, −1, −i} ⊂ ℂ is compact because it is both bounded (every point in S satisfies ∣z∣ ≤ 1) and closed (it is a finite set ⇒ it has no limit points).

Counterexamples (Non-Compact Sets)

• Open Disk: The set B_r(0) = {z ∈ ℂ : |z| < r} (a disk without the boundary) is not compact. It is bounded but not closed (it does not contain its boundary).

  The open cover {$\mathbf{B}_{r-\frac{1}{n}}(0): n \in \mathbf{N}$} admits no finite subcover, so B_r(0) fails compactness.

• The Entire Complex Plane (ℂ): ℂ is closed (it contains all its limit points) but unbounded. Therefore, it is not compact.

• Right half-plane {z ∈ ℂ : ℜ(z) ≥ 0} This set is closed but not bounded, therefore it is not compact.

These illustrate the two pillars of Heine–Borel:no escape to infinity (boundedness) and no leaks at the edge (closedness).

Cantor Theorem

Cantor Theorem. Let {K_j}_{j∈ ℕ} be a nested sequence of non‐empty compact (closed and bounded) subsets of ℂ: K₁ ⊃ K₂ ⊃··· ⊃ K_n ⊃ ⋯ (i.e., the sets are nested and decreasing). Then, the intersection is not empty: $\cap_{j=1}^{\infty} K_j \ne \emptyset$.

Proof (by contradiction).

Let’s assume for the sake of contradiction that $\cap_{i=1}^{\infty} K_i = \empty$, then its complement $\Complex = (\cap_{i=1}^{\infty} K_i )^c = \cup _{i=1}^{\infty} K_i^c$. In particular, $K_1 ⊆ \cup _{i=1}^{\infty} K_i^c$.

Since K₁ is compact and {$K_j^c$} is an open cover of K₁, there is a finite subcover ⇒ ∃j₁, j₂, ···, j_n and (this is somehow more restricted, but it still applies), there exist indices j₁ < j₂ < ··· < j_n such that $K_1 ⊂ \cup _{m=1}^{n}$ K_jₘ^c ⊂ K_jₙ+1^c because by assumption K_jₙ+1 ⊂ K_jₙ ⊂ ··· ⊂ K_j₁

This leds to K₁ ∩ K_jₙ+1 = Ø, this contradicts the given assumption of a nested subsets of ℂ ⊥

A common corollary is that a nested sequence of closed intervals [a_n, b_n] in ℝ with a_n ≤ a_n+1 ≤ b_n+1 ≤ b_n also has non‐empty intersection.

Proposition. Let K be a compact set in a topological space (which could be ℝⁿ, ℂ, or any metric space). Let C be a closed subset of K, C ⊆ K. Then, C is compact.

Proof.

To show that C is compact, we must show that every open cover of C has a finite subcover.

Consider an arbitrary open cover of C. Let {U_α}_α∈A be an open cover of C, where α belongs to some index set A. This means that: C ⊆ ⋃_α∈A U_α

Construct an open cover of K: Since C is a subset of K, we can extend the open cover of C to an open cover of K. Because C is closed, its complement in K, C^c (the set of all points that are in K but not in C), is open. Now, consider the collection of open sets: $\mathbf{U}$ = {U_α}_α∈A ∪ {C^c}. This collection forms an open cover of K

If x ∈ C, then it must be in one of the U_α sets (because {U_α} covers C).

Otherwise, if a point x is in K but not in C (i.e. x∈K and x∉C), then it must be in C^c.

Therefore, K ⊆ (⋃_α∈A U_α) ∪ C^c.

Since K is compact, the open cover {U_α}_α∈A ∪ {C^c} has a finite subcover. Let’s denote this finite subcover as ( there exist finitely many indices): {U_α1, U_α2, …, U_αn, C^c}. This finite subcover covers K: K ⊆ (U_α1 ∪ U_α2 ∪ … ∪ U_αn) ∪ C^c

Now, consider the sets U_α1, U_α2, …, U_αn. These are a subset of the original open cover {U_α} of C because C ⊆ K and a finite subcover of C (notice that C^c does not meet C) ∎

Definition. A set S ⊆ ℂ is connected if it cannot be written as the union of two disjoint non-empty open subsets of ℂ which have a non-trivial intersection with S (both meet S) Mathematically, ∄G₁, G₂ such that G₁∪G₂ ⊃ S, G₁ ≠ ∅, G2 ≠ ∅, G₁∩G2 = ∅, G₁ and G₂ are open, G₁∩S ≠ ∅, G₂∩S ≠ ∅ (S is disconnected Fig b).

Definition. An open, non-empty, connected subset of ℂ is called a region or a domain.

• Open: for every z ∈ G, there is an ε > 0 so that the disk B(z; ε) ⊂ G.
• Connected: G cannot be split as G = G₁ ∪ G₂ where G₁, G₂ are disjoint, non-empty, open in ℂ.

Proposition. Let G a non-empty, open subset of ℂ. Then, G is a region if and only if any two points of G can be connected by a polygonal path lying in G.

A polygonal path from p to q is a finite union of line segments [z₀, z₁] ∪ [z₁, z₂] ∪ ⋯ ∪ [z_n-1, z_n] with z₀ = p, z_n = q.

Proof

Part 1. Region ⇒ Polygonal Connectivity.

Suppose G is a region, fix a point a ∈ G. Let G₁ = {z ∈ G: there is a polygonal path from a to z in G} and let G₂ = G \ G₁.

Claim: G₁ and G₂ are open sets.

If z ∈ G₁, then since G₁ ⊂ G and G is open, there is an r>0 such that B(z; r) ⊂ G. Since a and z are connected by a polygonal path, we can extend this path to any point w∈ B(z; r) by the straight segment [z, w] ⊂ B(z; ε) (Figure C), hence w ∈ G₁ ⇒ B(z; r) ⊂ G₁ ⇒ G₁ is open.

Suppose z ∈ G₂. This means z is in G, but there is no polygonal path from a to z within G. Since G is open, there exists an r > 0 such that the open ball B(z; r) = {w ∈ ℂ : |w - z| < r} is entirely contained in G (i.e., B(z; r) ⊂ G).

Now, we will show that B(z; r) is also contained in G₂. Suppose, for the sake of contradiction, that there exists a point w ∈ B(z; r) such that w ∈ G₁. This would mean there is a polygonal path from a to w within G.

Since w is in B(z; r), we can connect z and w by a straight line segment (which is a simple polygonal path) that lies entirely within B(z; r), [w, z] ⊂ B(z;ε) and therefore within G. If there’s a polygonal path from a to w and a polygonal path (the line segment) from w to z, then by concatenating these paths, we can construct a polygonal path from a to z within G ⊥ (z ∉ G₁)

Therefore, G₁ and G₂ are disjoint open subsets covering G and G is connected, one of them must be empty ⇒ either G₁ = ∅ or G₂ = ∅, but we know that G₁ ≠ ∅ (a ∈ G), then G = G₁. In other words every z ∈ G is reachable from a by a polygonal path.

Finally, given any two points p, q ∈ G, first go from a to p, then reverse the path from a to q. Concatenation gives a polygonal path p→q all insideG.

Part II: Polygonal Connectivity ⇒ Connected

Suppose for the sake of contradiction that G can be split into two non-empty disjoint open sets: G = U ∪ V, U ∩ V = ∅, U ≠ ∅, V ≠ ∅.

Choose two arbitrary points p ∈ U, q ∈ V. By assumption, G has polygonal connectivity, hence there is a finite chain of segments joining p and q, p → q, say [z₀, z₁] ∪ ⋯ ∪ [z_n-1 , z_n], z₀ = p, z_n = q.

Each closed segment ℓ = [z_k-1, z_k] = {(1 - t)·z_k-1 + t·z_k : t ∈ [0, 1]} is connected in the classical sense because we can define the continuous map f: [0, 1] ⟶ℓ, f(t) = (1 - t)·z_k-1 + t·z_k. The domain [0, 1] is a connected topological space. A continuous image of a connected space is connected. Hence, ℓ = f([0,1]) is a connected subset of ℂ (it is homeomorphic to [0,1], a connected set in R).

Since z₀ = p, z_n = q, there must exist a segment [z_j-1, z_j] such that: z_j-1 ∈ U and z_j ∈ V. The segment [z_j-1, z_j] is connected and met both U and V (as it contains points from both), [z_j-1, z_j] ⊆ U ∪ V

If ℓ met both U and V, but U and V are disjoint, that means ℓ would be contained in the union U ∪ V yet intersect both pieces. That is exactly a separation of ℓ into two non‐empty, disjoint, open in the subspace topology of [z_j-1, z_j], and that is impossible because ℓ is connected. A similar argument is that ℓ must intersect the boundary between U and V. However, since U and V are disjoint and open in G, their boundary in G is empty. This is a contradiction, too.

Sequences of complex numbers

The complex plane ℂ is equipped with the standard metric topology, where the distance between two complex numbers z₁​ and z₂​ is given by the modulus: d(z₁, z₂) = ∣z₁ − z₂∣. This metric induces a topology on ℂ, where open sets are formed as unions of open disks: B_ϵ(z₀) = { z ∈ ℂ : ∣z − z₀∣ < ϵ}.

A set U ⊆ ℂ is considered open if, for every point z₀ ∈ U, there exists some radius ϵ > 0 such that the entire disk B_ϵ(z₀) is contained in U.

Definition. A sequence of complex numbers, denoted by (a_n) or $\{ a_n \}_{n=1}^∞$ is an ordered list of complex numbers such that for each natural number n, there is a corresponding n-th complex number a_n​ in the list.

For example:

• a_n := 12 + ¹⁄_n for n ≥ 1.
• a₀ := 1, a₁ = 1, a_n := a_n-1 + 2a_n-2 + i(a_n-1 -7a_n-2).

Definition. A sequence $\{ a_n \}_{n=1}^∞$ is said to have a limit L ∈ ℂ (or converges to a limit L) if for every epsilon (∀ε > 0), there exist a natural number (∃ n₀∈ ℕ) such that for each ∀n ≥ n₀, |a_n -L| < ε. A convergent sequence approaches a specific complex number L as n grows larger. Equivalently, eventually all terms lie inside every disc B_ε(L).

The limit L (if it exists) is unique for a convergent sequence.

Example. Consider the sequence $a_:=\frac{1}{n}+i(1−\frac{1}{n})$. As n→∞, a_n→0+i(1−0)=i. Thus, the limit L = i, and the sequence is convergent.

Interpretation: {a_n} converges to L in ℂ, means that the terms a_n cluster around L as n grows larger.

A sequence of complex numbers can be visualized as a set of points in the complex plane. A sequence {a_n} converges to a limit L ∈ ℂ if, for every open neighborhood B_ϵ(L) of L, there exists a natural number n₀ ∈ ℕ such that for all n ≥ n₀​, the terms a_n lie within B_ϵ(L). In other words: ∀ ϵ > 0,  ∃ n₀ ∈ ℕ such that n ≥ n₀ ⇒  a_n ∈ B_ϵ(L). This is equivalent to the standard definition $\lim_{n \to \infty}a_n = L \iff \forall \epsilon > 0, \exist n_0 \in ℕ \text{ such that } |a_n - L| < \epsilon, \forall n ≥ n_0$.

Definition. A sequence is said to be convergent if it has a limit. A sequence is said to diverge, diverge to infinite or be no convergent (written a_n → ∞) if for every M > 0, there exists a natural number n₀∈ ℕ such that for all n ≥ n₀, |a_n| > M. A divergent sequence grows without bound (in magnitude) as n increases. Equivalently, its terms eventually leave every bounded disk.

Example: Polynomial growth. Consider the sequence a_n: = n + in². As n→∞, ∣a_n∣ = $\sqrt{n²+n⁴}$→∞. Thus, the sequence diverges.

A sequence (a_n) is divergent if it does not converge to any finite limit L ∈ ℂ, that is, the sequence “escapes” every bounded region of the complex plane. Formally: ∀M > 0,  ∃ n₀ ∈ ℕ such that ∀n ≥ n₀ ⇒ ∣a_n∣ > M. This is equivalent to saying that the sequence (a_n​) “goes to infinity” in the extended complex plane ℂ ∪ {∞}.

In complex analysis, a set is said to be compact if it is both closed and bounded. The Bolzano-Weierstrass theorem states that every bounded sequence in ℂ has a convergent subsequence. This is a key result in complex topology and analysis.

Cauchy’s criterion

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.

Cauchy’s criterion in ℂ (or ℝ) A sequence $\{ a_n \}_{n=1}^∞$ of complex numbers is convergent in ℂ if and only if it is a Cauchy sequence, meaning: for every given epsilon ∀ε > 0 (no matter how small) there exist a natural number ∃N ∈ ℕ such that for all ∀n, m ≥ N, we have |a_m - a_n| < ε. In words, 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.

Why it works

1. If $\{ a_n \}_{n=1}^∞$ converges to some limit L, then eventually all a_n lie in B_ε/2(L). Hence for m, n large enough, |a_m - a_n| ≤ |a_m - L| + |L - a_n| ≤ ε/2 + ε/2 = ε. Thus, every convergent sequence is Cauchy.
2. Conversely, ℂ ≅ R² (a complete metric space), hence ℂ inherits completeness. Every Cauchy sequence is bounded and [By the Bolzano-Weierstrass theorem, every bounded sequence in ℝⁿ (or ℂⁿ) has a convergent subsequence] it has a convergent subsequence $a_{n_k} \to L$. Since ${a_n}$ is Cauchy, the entire sequence converges to $L$.

Examples

• A Real Sequence. Consider the sequence a_n = 1/n in ℝ. Cauchy check: For any arbitrary ε > 0, choose N so that 1/N < ε/2. Then, for all n, m ≥ N, |1/m - 1/n| ≤ 1/m + 1/n ≤ 1/N + 1/N < ε. This sequence {1/n} is Cauchy, hence converges (in fact to 0).

• A Complex Sequence. Cauchy check: z_n = (1 + i)/n. $|z_m - z_n| = \left| \frac{1+i}{m} - \frac{1+i}{n} \right| = |1+i|\cdot\left|\frac{1}{m}-\frac{1}{n}\right| = \sqrt{2} \cdot \left|\frac{1}{m}-\frac{1}{n}\right| =[\text{Using the previous argument}] \le \sqrt{2}·(\frac{1}{N}+\frac{1}{N}) = \frac{2·\sqrt{2}}{N} \lt \epsilon$. For any arbitrary ε > 0, choose N so that $N \gt \frac{2·\sqrt{2}}{\epsilon}$. 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.

Summary

1. Cauchy test lets you verify convergence without knowing the limit in advance.
2. In any complete metric space (including ℂ), Cauchy ⇔ convergent.
3. If you can show ∣a_m - a_n∣ gets arbitrarily small as m, n→∞, you’ve proven convergence.

Bolzano-Weierstrass theorem

The Bolzano–Weierstrass theorem is a cornerstone of analysis and topology in both ℝⁿ and ℂ.

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

Recall:

• 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.
• A subsequence of {x_n} is a sequence formed by selecting some of the terms of {x_nk} in their original order, that is, n₁ < n₂ <⋯ is an increasing sequence of indices.
• A point x is a limit (accumulation) point of a set S if every open neighborhood of x contains infinitely many points of S.

Intuitive idea 💭: If you cram infinitely many points into a finite region (a bounded set), they can’t all stay isolated, then there must be at least one point where they accumulate or cluster together.

Examples

1. 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:

   • The subsequence a_2n = 1, 1, 1, … converges to 1.
   • The subsequence a_2n+1 = -1, -1, -1, … converges to -1.

2. Grid in the Unit Square. Consider the set S = {(1/n, 1/m) : n, m ∈ ℕ}. This is an infinite subset of the compact set [0, 1] × [0, 1] in ℝ². The point (0, 0) is a limit point of S since arbitrarily close to (0, 0) you can find (1/n, 1/m) for large values of n, m.

In ℝⁿ (and ℂ), the Heine–Borel theorem theorem states: a set is compact if and only if it is closed and bounded. Hence the Bolzano–Weierstrass theorem is exactly the statement that every infinite subset of a compact set has an accumulation point. This allows extraction of convergent subsequences when we only know boundedness.