Bounded Sets, Compactness, and Connectedness in the Complex Plane

We shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender, Winston Churchill

Introduction

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, $f: D \to \mathbb{C}$.

• Domain D: A subset of $\mathbb{C}$ on which the function is defined. Every $z \in D$ is called a point of the domain.
• Function $f: D \to \mathbb{C}$: A rule that assigns to each $z \in D$ a unique complex number w = f(z).
• Range (or image) f(D): The set of all actual outputs {f(z): z ∈ D}.
• Decomposition into real and imaginary parts: If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts (real‑valued functions of two real variables).

Bounded Sets

Definition. A subset $S \subseteq \mathbb{C}$ is bounded if it fits entirely inside some sufficiently large open disk. Formally, S is bounded if there exists $R > 0$ and $z_0 \in \mathbb{C}$ such that: $S \subseteq D_R(z_0) = \{z \in \mathbb{C} : |z - z_0| < R\}$. This means all points of S lie within a distance R of some fixed point $z_0$.

Equivalent formulation: (1) $S$ is bounded if $\sup\{|z| : z \in S\} < \infty$. (2) S is bounded if there exists an M > 0 such that $|z| < M, \forall z \in S$ (all points in S lie inside the open disk of radius M centered at the origin).

Proof (1) $S \subseteq D_R(z_0) = \{z \in \mathbb{C} : |z - z_0| < R\}$ and (2) $\sup\{|z| : z \in S\} < \infty$ are equivalent.

(1) ⇒ (2):

Suppose $S \subseteq D_R(z_0)$, then $\forall z \in S, |z - z_0| < R$. By the triangle inequality, $∣z∣ \le |z - z_0| + |z_0| < R + |z_0|$. Thus, $sup\{ ∣z∣ : z \in S \} \le R + |z_0| \lt \infty$.

(2) ⇒ (1):

Suppose $sup\{ ∣z∣ : z \in S \} = M \lt \infty$. Let $z_0 = 0$ (the origin) and R = M + 1. Then, for all $z \in S$, $|z - 0| = |z| \le M \lt M$, so $S \subseteq D_R(0)$

Examples: Open Unit Disk $B(0; 1)$ (fits inside $D_1(0)$); Unit circle $\{z : |z| = 1\}$ (fits inside $D_2(0)$); Sequence $\{1/n : n \in \mathbb{Z}^+\}$ (all points lie inside $D_1(0)$); $\{1, i, -1, -i\}$ (finite set fits inside $D_2(0)$).

The two definitions are equivalent because both express the idea that all points of S lie within some finite distance from a fixed point (either $z_0$ or the origin).

Counterexamples: Entire Real Line $\mathbb{R}$ (extends infinitely in both directions); Upper Half-Plane $\mathbb{H} = \{ z \in \mathbb{C}: Im(z) \gt 0 \}$ (extends infinitely upward); Infinite Strip $\{0 < \text{Re}(z) < 1\}$ (extends infinitely in the imaginary direction).

Why Are Finite Sets Compact?

• Finite sets are bounded. If S = $\{ z_1, z_2, \cdots, c_n \}$ is finite, let M = $max\{ |z_1|, |z_2|, \cdots, |c_n| \} + 1$. Then, $|z| \lt M, \forall z \in S$, so S is bounded.
• A set is closed if it contains all its limit points. A limit point is a point $z_0$ such that every open disk $D_{\varepsilon}(z_0)$ contains at least one point of S other than $z_0$ itself.

1. If $z_0 \in S$, say $z_0 = z_k$, then any open disk $D_{\varepsilon}(z_0)$ with $\varepsilon$ small enough will not contain any other points of S (the distances $|z_k-z_i|\quad (i\neq k)$ form a finite set of positive numbers. Let $\delta =\min _{i\neq k}|z_k-z_i|>0$. Then, the disk $D_{\delta /2}(z_k)$ contains no other points of S). Thus, $z_0$ is not a limit point.
2. If $z_0 \notin S$, there exists a minimum distance $\delta = min\{ |z_0 -z_i|: z_i \in S \}$. Taking $\varepsilon = \delta/2$, the disk $D_{\varepsilon}(z_0)$ contains no point of S. Thus, $z_0$ is not a limit point.
3. A finite set has no limit points at all, so it contains all its limit points vacuously. Therefore, it is closed.
4. In $\mathbb{R^{\mathnormal{n}}}$ or $\mathbb{C}$, the Heine–Borel theorem states: A set is compact iff it is closed and bounded. We’ve shown both properties, so the conclusion follows immediately.

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

• Key Properties of Compact Sets:
  Closed: Contains all its limit points (no “holes” or “missing boundaries”).
  Bounded: Fits inside some large disk (does not “stretch to infinity”).
• In analysis, compact sets behave nicely (e.g., every continuous function on a compact set attains its maximum and minimum). In topology, compactness is widely considered a generalization of the notion of finiteness to infinite sets. A topological space X is compact if every open cover of X has a finite subcover. This means that for any collection of open sets covering the space, you can always pick a finite number of them that still cover the entire space.
• Examples: Closed Unit Disk $\{ z \in \mathbb{C}: |z| \le 1 \}$; Unit circle $\{z : |z| = 1\}$; Finite sets; Closed and Bounded Sequence $\{1/n : n \in \mathbb{Z}^+\} \cup \{ 0 \}$
• Counterexamples: Open Unit Disk $B(0; 1)$ (bounded but not closed); Unbounded Closed Sets: $\mathbb{R}$ and $\mathbb{H} = \{ z \in \mathbb{C}: Im(z) \gt 0 \}$; Half-Open Interval $\{ z\in \mathbb{C}: 0 \le Re(z) \lt 1 \}$ (it is not compact).
• Important Theorems Related to Compactness: Heine-Borel Theorem: In $\mathbb{C}$, a set is compact if and only if it is closed and bounded.
  Bolzano-Weierstrass Theorem: Every bounded sequence in $\mathbb{C}$ has a convergent subsequence.
  Extreme Value Theorem: Every continuous function on a compact set attains its maximum and minimum.

Definition. For $a, b \in \mathbb{C}$, the closed line segment (or interval) from $a$ to $b$ is: $[a, b] = \{a + t(b - a) : t \in [0, 1]\} = \{(1-t)a + tb : t \in [0, 1]\}$. This is the straight line joining a and b, including both endpoints.

The segment $[a, b]$ traces from $a$ (at $t = 0$) to $b$ (at $t = 1$). Besides, $\frac{a + b}{2}$ is the midpoint at t = 0.5.

Definition. A polygonal path (or broken line) connecting two points $p$ and $q$ is defined as the union of finitely many consecutive line segments: $\gamma = [p = z_0, z_1] \cup [z_1, z_2] \cup \cdots \cup [z_{n-1}, z_n = q]$ where each closed line segment or interval $[z_i, z_{i+1}]$ represent the straight line joining the consecutive pair of points z_i​ and z_i+1.

The points $z_0, z_1, \ldots, z_n$ are called vertices of the path.

Definition. Connected Set (Polygonally) A set $S \subseteq \mathbb{C}$ is connected if for any two points $p, q \in S$, there exists a polygonal path $\gamma$ connecting $p$ and $q$ that lies entirely within $S$ (every pair of points of S can be joined by a polygonal path lying entirely in S): $\gamma = \bigcup_{i=0}^{n-1} [z_i, z_{i+1}] \subseteq S$ where $z_0 = p$ and $z_n = q$.

This indicates that you can join any two points in S with a finite sequence of straight line segments, all remaining within the set S. It means there are no “gaps” or separations within the set — you can walk from any point to any other without leaving the set.

Examples: $B(0; 1)$ and $\overline{B(0; 1)}$ (any two points joined by straight line in disk), $\mathbb{C}$, $\mathbb{H}$, $A(0; 1, 2)$ (annulus), $\mathbb{C} \setminus \{0\}$ (even though $0$ is removed, any two points can be connected by going around the origin).

Counterexamples: $B(0; 1) \cup B(5; 1)$ (no, it is not possible to cross the gap), $\{z : |z| \neq 1\}$ (inside and outside the circle of radius 1 are separated), $\mathbb{R} \setminus \{0\}$ (positive and negative reals are separated).

Definition: A set $S$ is convex if for any $a, b \in S$, the entire segment $[a, b] \subseteq S$.

Proposition: Every convex set is connected.

Proof: For any $p, q \in S$, the path $\gamma = [p, q]$ is a single segment contained in $S$ (by convexity). $\blacksquare$

Domains

Definition. An open, connected subset of $\mathbb{C}$ is called a domain. This means a domain has two important properties:

1. Open: For every point z in the domain, z ∈ D there exists an open disk centered at z (ε-disk around it) that is entirely contained within the domain.
2. Connected: As defined above, any two points in the domain can be joined by a polygonal path within the domain.

Definition. A region is a domain together with some, none, or all of its boundary. In other words: Start with a domain $D$; a region is any set $R$ with: $D \subseteq R \subseteq \bar{D}$

This means a region can be:

• An open region (if it contains none of its boundary points), e.g., $B(0; 1)$.
• A closed set (if it contains all of its boundary points), e.g., $\overline{B(0; 1)}$.
• Something in between (if it contains some, but not all, of its boundary points), e.g., $\{z : |z| \leq 1, z \neq 1\}$.

Examples

1. $D = \{z : |z| < 1\}$. D is a domain (it’s open and connected) and also a region (it’s a domain with none of its boundary points).
2. $\bar{D}$ = {z | |z| ≤ 1} is not a domain (it’s not open), but it is a region (it’s a domain D plus all of its boundary points).
3. $R = \{z : |z| \leq 1, z \neq 1\}$ (the closed disk excluding the point 1) is not a domain (it’s not open). However, R is a region (it’s the domain D plus some of its boundary points, but not the point 1).

Star-Shaped Sets

Definition. A set S in the complex plane is star-shaped if there exists a point $z_0 \in S$ (called a star center) such that for every $z \in S$, the entire line segment connecting z₀ and z $[z_0, z]$ is entirely contained in $S$. Formally, $\exists z_0 \in S : \forall z \in S, \forall t \in [0, 1], \quad (1-t)z_0 + tz \in S$

Formal Characterization. A set S is star-shaped if and only if there exists $z_0 \in S$ such that the kernel of S (the set of all points that can serve as star centers) is non-empty. The kernel is defined as: $Kernel(S) = \{ z_0 \in S: \forall z \in S, [z_0, z] \subset S \}$

• If S is convex, Kernel(S) = S.
• If S is star-shaped but not convex, Kernel(S) is a proper subset of S.

Intuition. This means you can see every point in S from $z_0$ without any obstructions. The star center acts like a vantage point from which the entire set is visible.

Key Properties and Examples

1. Convex Sets Are Star-Shaped. In a convex set, any two points can be connected by a line segment entirely within the set. Thus, choosing any $z_0 \in S$ makes S star-shaped (in fact, every point in a convex set is a star center), e.g., B(0; 1)$ is convex, so every point inside it is a star center.
2. Star-Shaped Sets Need Not Be Convex.
   Counterexample: A star polygon (e.g., a 5-pointed star) is star-shaped (with the center as the star center) but not convex because some line segments between points lie outside the set (e.g., take two points at the tips of adjacent “arms” of the star. The straight line segment between them will pass outside the body of the star).
   Another Example: An L-shaped region is star-shaped if the star center is at the corner where the two legs of the “L” meet, but it is not convex.
   Take a point at the end of the top leg and a point at the end of the right leg. The line segment connecting them will pass through the “empty” square region outside the L-shape.
   A crescent moon shape is star-shaped if you pick a point in its thickest part, but it is clearly not convex.
3. Star-shaped sets are always connected (since you can traverse from any point to $z_0$ and then to any other point via $z_0$).
4. A connected set need not be star-shaped.
   Counterexample: An annulus A(0; 1, 2) = $\{ z \in \mathbb{C}: 1 \lt ∣z∣ \lt 2 \}$ is connected but not star-shaped (no single point can “see” all others due to the hole in the middle).
   A horseshoe shape (or a U-shaped region, two arms facing each other, with a bend at the buttom) is connected but not star-shaped.
   Why it’s connected: You can travel continuously between any two points within the horseshoe without leaving the set, because it’s all one piece.
   Why it’s not star-shaped: Take any candidate star center p: (i) If p is in the left arm, then a point q on the right arm will have the segment $\bar{pq}$ cut across the “gap” in the middle, which is outside the set. (ii) If p is in the bend, then points on the tips of the arms may still have segments $\bar{pq}$ that leave the set. In other words, the narrow “neck” blocks visibility.

Exercise: An annulus A(0; 1, 2) = $\{ z \in \mathbb{C}: 1 \lt ∣z∣ \lt 2 \}$ is not star-shaped (no single point can “see” all others due to the hole in the middle).

Let $p \in A$ be any point. Write $p = re^{i\theta}$ with 1 < r < 2.

Consider the point diametrically opposite to p through the origin: $z = -p = re^{i(\theta + \pi)}$. Obviously, |z| = r, so 1 < |z| < 2, hence $z \in A$.

The segment from p to −p can be parameterized as: $\gamma(t) = (1 −t)p + t(−p) = (1−2t)p, t \in [0, 1]$. The modulus is: $|\gamma(t)| = |(1−2t)p| = |1-2t||p| = |1-2t|r$

As t goes from 0 to 1, ∣1−2t∣ decreases from 1 to 0 at t = 1/2, then increases back to 1. In particular:

• At t = 1/2, $\gamma(\frac{1}{2}) = 0$, which is not in A (since |0| = 0 < 1).
• For t such that $|1-2t|r \le 1 \implies |\gamma(t)| \le 1$, the point lies in {w: ∣w∣ ≤ 1}, which is outside A.
• Because $r \gt 1$, there exists an interval of t around 1/2 where ∣γ(t)∣≤1. Thus, the entire segment $\overline{pz}$ is not contained in A $\blacksquare$.

Note:

Because $r \gt 1$, there exists an interval of t around 1/2 where ∣γ(t)∣≤1: $|1-2t|r \le 1 \iff |1-2t| \le \frac{1}{r} \implies \frac{-1}{r} \le 1 -2t \le \frac{1}{r}$.

$1 -2t \le \frac{1}{r} \implies 2t \ge 1 - \frac{1}{r} \implies t \ge \frac{1}{2}-\frac{1}{2r}$. Similarly, $1 -2t \le \frac{-1}{r} \implies t \le \frac{1}{2}+\frac{1}{2r}$.

Since $r \gt 1$, we have $0 \le \frac{1}{r} \lt 1$, so $t \in [\frac{1}{2}-\frac{1}{2r}, \frac{1}{2}+\frac{1}{2r}]$ and the interval has positive length and is centered at t = 1/2.

Important Relationships

$\boxed{\text{Convex} \Rightarrow \text{Star-shaped} \Rightarrow \text{Connected}}$

Simply Connected Domains

Definition: A domain $D$ is called simply connected if it is path-connected and every closed curve (loop) in D can be continuously shrunk to a point without ever leaving $D$.

Intuition 🔮.Simply connected domains have no holes. If you were to stretch a rubber band around any part of the domain, you could pull the rubber band tight to a single point without getting it snagged on a missing point or an obstacle.

Examples

• Simply Connected Domains (No Holes): $\mathbb{C}$ (the whole plane); $B(0; 1)$ (the open unit disk); $\mathbb{H}$ (the upper half-plane), any convex set.

  A set D is convex if for any two points $z_1, z_2 \in D$, the line segment connecting them is also in D. All convex sets are simply connected.

• Non-Simply Connected Domains (Multiply Connected): $\mathbb{C} \setminus \{0\}$ (the punctured plane has a “hole” at 0. A loop encircling the origin cannot be shrunk to a point without crossing the missing 0); the annulus $A(0; 1, 2)$; the plane with all integers removed $\mathbb{C} \setminus \mathbb{Z}$ has infinitely many holes.

  Why the Annulus Fails: Consider a circle around the origin with radius r (where 1 < r < 2). This closed curve cannot be contracted to a point within the annulus because the “hole” (the missing inner disk |z| ≤ 1) obstructs the contraction —the curve would need to cross the hole, which is not part of the domain

Significance in Complex Analysis

The distinction between simply connected and multiply connected domains is fundamental because key theorems behave differently depending on the topology of the region:

1. Cauchy’s Theorem: If f is analytic on a simply connected domain D, then $\int_{\gamma} f(z)dz = 0$ for every closed contour γ in D. This is not generally true for domains with holes (e.g., $\int_{|z| = 1}\frac{1}{z}dz = 2\pi i \ne 0$ in $\mathbb{C} \setminus \{ 0 \}$).
2. Existence of Antiderivatives: An analytic function f on a simply connected domain D always has an antiderivative F (such that F′ = f) on D. This fails in punctured domains (e.g., 1/z does not have a global antiderivative on $\mathbb{C} \setminus \{ 0 \}$, though the logarithm function attempts to serve this role).
3. Riemann Mapping Theorem: Let $\Omega$ be a simply connected domain in the complex plane $\mathbb{C}$ such that $\Omega \ne \mathbb{C}$. Then, there exists a conformal (bijective analytic) map $f: \Omega \to \mathcal{D} = \{ z \in \mathbb{C}: |z| \lt 1 \}$. Moreover, given $z_0 \in \Omega$, we can choose f so that $f(z_0) = 0$ and $f'(z_0) \ge 0$; under these conditions, f is unique.

   Conformal equivalence: A bijective analytic map with analytic inverse. Such maps preserve angles and orientation (conformal mappings).

Example: Half-plane to disk, $\mathcal{H} = \{ Im(z) \gt 0 \}$ is simply connected, map $z \to \frac{z-i}{z+i}$ sends $\mathcal{H}$ to $\mathcal{D}$. Strip to disk: $\mathcal{S} = \{ 0 \lt Im(z) \lt \pi \}$ is simply connected, map $z \to \frac{e^z -1}{e^z + 1}$ sends $\mathcal{S}$ to $\mathcal{D}$.