Open Covers and Compactness in the Complex Plane

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

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).

Open Covers and Compactness in $\mathbb{C}$

Definition (Open cover). Let $S \subseteq \mathbb{C}$. An open cover of $S$ is a collection of open sets $\mathbf{U} = \{U_i\}_{i \in I}$ such that $S \subseteq \bigcup_{i \in I} U_i.$

A finite subcover is a finite subcollection $\{U_{i_1},\dots,U_{i_n}\} \subset \mathbf{U}$ such that $S \subseteq \bigcup_{k=1}^n U_{i_k}$ (finite subcollection that still covers S).

Definition. A set S in the complex plane ($S \subseteq \mathbb{C}$) is compact if every open cover of S admits a finite subcover.

Heine–Borel Theorem. A subset S of the complex plane ($S \subseteq \mathbb{C}$) is compact if and only if S is both closed and bounded in $\mathbb{C}$

1. Closed. $\mathbb{C} \setminus S$ is open $\iff S = \bar{S}$ (the set equals its closure) $\iff S$ contains all of its limit points $\iff S$ contains all of its boundary points (No “escape routes” at the boundary).
2. Bounded. $\exists R > 0$ such that $|z| \le M$ for all $z \in S$ $\iff S$ is contained in some disk $B(0; R)$ (No “escape to infinity”).

Counterexamples (Non-Compact Sets)

• Bounded but Not Closed (Leaks at the Edge). The Open Unit Disk $B(0; 1) = \{z \in \mathbb{C} : |z| < 1\}$ is not compact because:

1. It is not closed. it does not contain its boundary $\{z : |z| = 1\}$. This boundary consists entirely of limit points not contained in the set. For instance, the sequence {z_n} = {1 - 1/n} lies in B(0; 1), yet $z_n \to 1 \notin B(0; 1)$.
2. Consider the open cover $U = \{ B(0; r): 0 < r < 1 \}, \bigcup_{0 < r < 1} B(0; r) = B(0; 1)$ (every point in the open disk lies in some smaller concentric disk).

No finite subcover. Suppose $\{ B(0; r_1), \cdots, B(0; r_n) \}$ is a finite subcover. Let $r_{max} = max \{ r_1, \cdots, r_n \} \lt 1$. Then, $\bigcup_{k = 1}^n B(0; r_k) = B(0; r_{max}) ⊊ B(0; 1)$. Points with $r_{max} \lt |z| \lt 1$ are missing.

The finite subcover “leaks” toward its boundary. No finite collection of subdisks can reach all the way to the edge.

• Bounded but Not Closed (Leaks at the Edge). The Open Disk $B(0; r) = \{z \in \mathbb{C} : |z| < r\}$ is not compact because:

1. Closed: The boundary circle {z : |z| = r} consists entirely of limit points not contained in the set. For instance, the sequence $\{z_n \} = \{ r(1 - \frac{1}{n}) \}$ lies in B(0; r), yet $z_n \to r \notin B(0; r)$.
2. Explicit open cover with no finite subcover: Consider the nested family of open disks: $\mathcal{U} = \left\{ B_{r-\frac{1}{n}}(0) : n \in \mathbb{N}, n > \frac{1}{r} \right\}$.
   (i) $\mathcal{U}$ covers B_r(0): Take any arbitrary $z \in B(0; r)$. Then, |z| < r, so r - |z| > 0. By the Archimedean property, ∃N ∈ ℕ such that 1/N < r - |z|. Thus |z| < r - 1/N, meaning $z \in B(0; r-\frac{1}{n})$.
   (ii) However, no finite subcover exists: Suppose {B_r-1/n₁(0), B_r-1/n₂(0), …, B_r-1/nₖ(0)} is a finite subcollection with n₁ < n₂ < ··· < nₖ. Then, $\bigcup_{j=1}^{k} B_{r-\frac{1}{n_j}}(0) = B_{r-\frac{1}{n_k}}(0)$.
   The point z* = (r - 1/(2nₖ)) satisfies both |z*| < r and |z*| = r - 1/(2nₖ) > r - 1/nₖ, so z* ∈ B_r(0) but z* ∉ B_r-1/nₖ(0). ∎

   No matter how many finite disks you take, points near the boundary escape coverage (it leaks at the edge).

• Bounded but Not Closed (Leaks at the Edge). Open Interval (0, 1) in ℝ.
  Explicit counterexample cover: $\mathcal{U} = \left\{ \left(\frac{1}{n}, 1\right) : n \geq 2 \right\}$
  This covers (0,1) since for any x ∈ (0,1), we can always choose n such that n > 1/x, giving $x \in (\frac{1}{n}, 1)$.
  No finite subcover: If we take {(1/n₁, 1), …, (1/nₖ, 1)} with n₁ < ··· < nₖ, then ⋃ = (1/nₖ, 1), $\bigcup_{i = 1}^k (\frac{1}{n_i}, 1) = (\frac{1}{n_k}, 1) ⊊ (0, \frac{1}{n_k}]$

• Punctured Closed Disk: $D^* =\overline{\mathbb{B}(0; r)} \setminus \{0\} = \{z \in \mathbb{C} : 0 \lt |z| \le r \}$.
  The set is bounded (contained in the disk of radius r), but not closed (0 is a limit point not in the set, consider $z_n = \frac{1}{n} \to 0$).
  Open cover without finite subcover: $\mathcal{U} = \{ U_n \} = \left\{ \left\{ z : \frac{1}{n+1} < |z| \leq r \right\} : n \in \mathbb{N} \right\}$
  Each set $U_n$ is open in D*, because it equals $D^* \cap A_n$, where $An = \left\{ z : \frac{1}{n+1} < |z| \lt r+1 \right\}$ is open in $\mathbb{C}$.
  However, no finite subcollection of U convers D*. If we take $U_{n_1},\cdots, U_{n_k}$ and let $N = max\{ n_1, \cdots, n_k \}$, then $\bigcup_{i = 1}^k U_{n_i} = U_N = \left\{ z : \frac{1}{N+1} < |z| \le r \right\}$. Points satisfying $0 \lt |z| \le \frac{1}{N+1}$ (e.g., $z = \frac{1}{2(N+1)}$) are in D* but no in this union.

• Infinite Discrete Set. The Natural Numbers $S = \{n + i·0: n \in \mathbb{N} \} \subseteq \mathbb{C}$ is not compact because:

1. It is not bounded. Its extends infinitely along the real axis.
2. Consider the open cover of “individual bubbles” $\mathbf{U} = \{ \mathbb{B}(n; 1/2): n \in \mathbb{N} \}$. Coverage: each $n \in S$ lies in its own ball $\mathbb{B}(n; 1/2)$, so $\mathbf{U}$ covers S, but no finite subcover. Suppose $\{ \mathbb{B}(n_1; 1/2), \cdots, \mathbb{B}(n_k; 1/2) \} $ is a finite subcover. Let $N = max\{ n_1, \cdots, n_k \} + 1$. Then, $N \in S$ but $|N - n_j| \ge 1 \gt \frac{1}{2} \forall j \implies N \notin \cup_{j=1}^k \mathbb{B}(n_j; 1/x)$, contradiction.

   The set stretches forever. Any finite collection of bounded neighborhoods can only capture finitely many points, leaving the rest stranded.

• Open Rectangle: (a, b) × (c, d) in ℝ² (or ℂ) is not closed because its boundary points are limit points not contained within the set.
  All four edges are “leaking” —limit points on the boundary are excluded.
  Edge points. Take (a, y) where y ∈ (c, d). The sequence $\{ (a + \frac{1}{n}, y) \}_{n=1}^{\infty}$ lies in (a, b) × (c, d) and converges to $(a, y)$. Thus, (a, y) is a limit point of (a, b) × (c, d) but not in (a, b) × (c, d). Corner points. The sequence $\{ (a + \frac{1}{n}, c + \frac{1}{n}) \}_{n=1}^{\infty}$ lies in (a, b) × (c, d) and converges to $(a, c)$.

• Open Annulus: A = {z : 1 < |z| < 2}
  Not closed: Both circles |z| = 1 and |z| = 2 consist of limit points not in A.
  Bounded: Yes, |z| < 2 for all z ∈ A.

• Rational Points in [0,1]: $\mathbb{Q} \cap [0, 1]$.
  Despite being bounded (contained in [0, 1]), this set is not closed. Every irrational number $\alpha$ in [0,1] is a limit point of rationals (by density of ℚ), yet irrationals are not in the set.
  By density of $\mathbb{Q}$, for each $n \in \mathbb{N}$, choose $x_n \in \mathbb{Q} \cap [0, 1]$ such that $|x_n - \alpha| \lt \frac{1}{n}$. Then, $x_n \to \alpha$ by the squeeze theorem. Thus S is not sequentially compact, hence not compact.

Closed but Not Bounded (Escape to Infinity)

• The Entire Complex Plane ℂ. It is closed because ℂ has no points outside itself, so vacuously, it contains all its limit points. Alternatively, $\mathbb{C}^c = \emptyset$ is open.
  Why it’s unbounded: For any M > 0, the point z = M + 1 satisfies |z| > M.
  Open cover without finite subcover: $\mathcal{U} = \{ B(0; n) : n \in \mathbb{N} \}$ covers ℂ (every complex point z has |z| < n for some n), but any finite subcollection $\cup_{k} B(0; n_k) = B(0; max(n_k))$ misses points with |z| > max(nₖ).

• Right Half-Plane: $\mathbb{H}$ = {z ∈ ℂ : ℜ(z) ≥ 0}
  Why it’s closed: The complement {z : ℜ(z) < 0} is open (for any point with negative real part, a small disk stays in the left half-plane).
  Why it’s unbounded: The sequence {z_n = n} satisfies $z_n \in \mathbb{H}$ and $|z_n| \to \infty$.
  Open cover without finite subcover: $\mathcal{U} = \left\{ \{z \in H : |z| < n\} : n \in \mathbb{N} \right\}$

• The Real Line $\mathbb{R} \subset \mathbb{C}$.
  Why it’s closed: ℝ contains all its limit points in ℂ. Let z_n be a sequence in $\mathbb{R}$ that converges in $\mathbb{C}$ to some limit L. $\Im(z_n) = 0, z_n \to L$, the imaginary part is a continuous function, so taking limits gives $\Im (L)=\lim _{n\rightarrow \infty }\Im (z_n)= \lim _{n\rightarrow \infty } 0 = 0$. Thus L has zero imaginary part, meaning $L\in \mathbb{R}.$
  Why it’s unbounded: ℝ extends infinitely in both directions. $|x|\rightarrow \infty \quad \mathrm{as\ }x\rightarrow \pm \infty$.

• The Imaginary Axis: $i\mathbb{R}$ = {iy : y ∈ ℝ}.
  Why it’s closed: Same reasoning as ℝ—limits of purely imaginary sequences are purely imaginary. Let z_n be a sequence in $i\mathbb{R}$ that converges in $\mathbb{C}$ to some limit L. $\Re(z_n) = 0, z_n \to L$, the real part is a continuous function, so taking limits gives $\Re (L)=\lim _{n\rightarrow \infty }\Re (z_n)= \lim _{n\rightarrow \infty } 0 = 0$. Thus L has zero real part, meaning $L\in i\mathbb{R}.$
  Why it’s unbounded: The sequence $\{in \}_{n=1}^{\infty} \to \infty$.

• Closed Strip: S = {z ∈ ℂ : 0 ≤ ℜ(z) ≤ 1}
  Why it’s closed: S = {z : ℜ(z) ≥ 0} ∩ {z : ℜ(z) ≤ 1}, an intersection of two closed sets.
  Why it’s unbounded: Points like i·n ∈ S have unbounded modulus, $\{in \}_{n=1}^{\infty} \to \infty$

• Lattice Points: $\mathbb{Z} x \mathbb{Z}$ = {m + in : m, n ∈ ℤ}
  Why it’s closed: This is a discrete set —every point is isolated. For any lattice point z, the disk $\mathbb{B}(z; \frac{1}{2})$ contains no other lattice points. Every point in the set is isolated, a discrete set has no accumulation points (so it contains all of them), making it closed.
  Why it’s unbounded: |m + in| = $\sqrt{m^2 + n^2} \to \infty \text{ as } m, n \to \infty$.

• Parabola: P = {t + it² : t ∈ ℝ} in $\mathbb{C}$ (identified with $\mathbb{R}^2$) is both closed and unbounded.
  Why it’s closed: P is the zero set of the continuous function f(x + iy) = y - x² (f is continuous as a polynomial in x, y), and preimages of closed sets under continuous functions are closed, $P = f^{-1}\{ 0 \}$
  Why it’s unbounded: As $t \to \pm \infty, |z| = \sqrt{t^2 + t^4} \to \infty$

Neither Closed Nor Bounded

• Open Half-Plane: {z ∈ ℂ : ℜ(z) > 0}
  Not closed: The imaginary axis consists of limit points not in the set (e.g., $\{ \frac{1}{n} \}, \frac{1}{n} \to 0$).
  Not bounded: Same as closed half-plane.

• Punctured Plane: $\mathbb{C} \setminus \{ 0 \} = \mathbb{C}^*$
  Not closed: 0 is a limit point ($\{ \frac{1}{n} \}, \frac{1}{n} \to 0$) not in the set.
  Not bounded: Contains points of arbitrarily large modulus.

• Exterior of Unit Disk: {z : |z| > 1}
  Not closed: Points on |z| = 1 are limit points not in the set.
  Not bounded: Contains points of arbitrarily large modulus.

• ℚ (Rationals in ℂ, viewed as ℚ ⊂ ℝ ⊂ ℂ)
  Not closed: Every irrational number is a limit point of rationals (by density of ℚ), yet irrationals are not in the set.
  Let’s choose any irrational $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ (e.g., $\alpha = \sqrt{2}$ or $\alpha = \pi$). By density of $\mathbb{Q}$, for each $n \in \mathbb{N}$, choose $x_n \in \mathbb{Q}$ such that $|x_n - \alpha| \lt \frac{1}{n}$. Then, $x_n \to \alpha$ by the squeeze theorem. Thus, $\mathbb{Q}$ is not sequentially compact, hence not compact.
  Not bounded: Contains arbitrarily large rationals. For any M > 0, there exists $q \in \mathbb{Q}$ with |q| > M (take $q = ⌈M⌉+1 \in \mathbb{Z} \subset \mathbb{Q}$).

Compact Examples

• The unit circle $C = \{z \in \mathbb{C} : |z| = 1\}$ is compact because:
  Closed:
  (i) Proof Via complement $\mathbb{C}\setminus C$ is open. Take any $w \notin C$, $|w| \ne 1$. Let $\delta = ||w| - 1| \gt 0$ and consider the open disk B(w; δ/2). For any $z \in \mathbb{B}(w; \delta/2)$ the reverse triangle inequality gives: $||z| - |w|| \le |z-w| \lt \delta/2$. If $z \in S, |z| = 1$, then ||z| - |w|| = |1 - |w|| = δ < δ/2 $\bot$. Hence, $\mathbb{B}(w; \delta/2) \subseteq \mathbb{C}\setminus C$. Thus, $\mathbb{C}\setminus C$ is open.
  (ii) Proof Via continuous functions $C = \{z : |z| = 1\}$ is the preimage of the closed set $\{1\}$ under the modulus function f(z) = |z| (a continuou map $\mathbb{C} \to \mathbb{R}$). The singleton {1} is closed in $\mathbb{R}$. Since the preimage of a closed set under a continuous function is closed, C = $f^{-1}(\{ 1 \})$.
  It is obviously bounded, e.g., by a disk of radius 2 centered at the origin.
  By the Heine–Borel theorem in $\mathbb{C}$ (or $\mathbb{R}^2$), a subset is compact if and only if it is closed and bounded. Since C satisfies both conditions, it is compact.

• Closed Disk: The set $\overline{B_r(0)} = \{z ∈ ℂ : |z| ≤ r\}$ (a disk of radius r including the boundary) is compact.
  Boundedness: For any $z \in \overline{B_r(0)}, |z| \le r$. Thus, the set is contained in the disk of radius r+1 centered at the origin, so it is bounded.
  Closed. Proof via complement. $\mathbb{C} \setminus \overline{B_r(0)} = \{z: |z| \gt r\}$ is open. Take an arbitrary $z_0, |z_0| \gt r$, select $\varepsilon = |z_0| - r \gt 0$. Then, $\forall z \in B(z_0; \varepsilon)$, the reverse triangle inequality yields $|z| = |z_0 - (z_0 - z)| \ge |z_0| - |z_0 - z| = |z_0| - |z - z_0| \gt |z_0| -\varepsilon = r \implies |z| \ge r \implies B(z_0; \varepsilon) \subseteq \mathbb{C} \setminus \overline{B_r(0)}$. Hence, the complement is open.
  Alternatively, the modulus function $f: \mathbb{C} \to \mathbb{R}$ defined by f(z)=∣z∣ is continuous, and $\overline{B_r(0)} = f^{-1}([0, r])$. Since [0, r] is closed in $\mathbb{R}$, the preimage of a closed set under a continuous function is closed.
  By the Heine–Borel theorem in $\mathbb{C}$ (or $\mathbb{R}^2$), since $\overline{B_r(0)}$ is both closed and bounded, it is compact.

• Closed Rectangle: Consider the set $R = \{z = x + iy : a \le x \le b,\ c \le y \le d\}$, where a, b, c, and d are real numbers with $a \le b$ and $c \le d$. This is a rectangle in the complex plane, including its edges. It is compact because it is both:
  Bounded: for any $z = x + iy \in R$, we have $|x| \le max \{|a|, |b| \}$ and $|y| \le max \{|c|, |d| \}$. By the triangle inequality, ∣z∣ ≤ ∣x∣ + ∣y∣ ≤ max{∣a∣, ∣b∣} + max{∣c∣,∣d∣}. Thus, R is contained in a disk of finite radius and is bounded.
  Closedness: The functions Re(z) = x and Im(z) = y are continuous. The set R is the preimage of the closed rectangle $[a, b] x [c, d] \subset \mathbb{R}^2$ under the continuous map $z \to (Re(z), Im(z))$. Since [a, b] x [c, d] is closed in $\mathbb{R}^2$, its preimage is closed in $\mathbb{C}$
  Since R is closed and bounded, it is compact.

• Any finite set $S = \{z_1,\dots,z_n\} \subset \mathbb{C}$ is compact:
  Bounded: e.g., take $M = \max_i |z_i|$, then $|z| \le M$ for each $z \in S$.
  Closed: finite sets have no limit points (each point is isolated, there’s no way for a small enough disk or neighborhood around any of these points to contain another distinct point from the set), so they trivially contain all of their limit points.
  Alternatively, its complement is open, for any $w \notin S$, let $\delta = min\{|w-z_i|: z_i \in S\} \gt 0$. The disk $B(w, \delta/2)$ does not intersect S, or $B(w, \delta/2) \cap S = \emptyset, B(w, \delta/2) \subseteq \mathbb{C} \setminus S$, $\mathbb{C} \setminus S$ is open.
  Example: $S = \{1,i,-1,-i\}$ is compact.
  Since S is both closed and bounded in $\mathbb{C}$, it is compact.
  Directly from the definition of compactness: An open cover of S is a collection of open sets whose union contains S. For each $z_i \in S$, choose an open set $U_i$ from the cover that contains $z_i$. Then, {$U_1, \cdots, U_n$} is a finite subcover. Thus, S is compact.

• The set T = $\{z \in \mathbb{C} : \Re(z) \ge 0,\ |z| \le 1\} = \{z : \Re(z) \ge 0\} \cap \{z : |z| \le 1\}$ is the intersection of the closed right half-plane $\{z : \Re(z) \ge 0\}$ (this is closed because its complement $\{z : \Re(z) \lt 0\}$ is open) and the unit disk $\{z : |z| \le 1\}$ (closed and bounded).
  The intersection of closed sets is closed.
  $T$ is bounded (since $T \subset \overline{B(0; 1)}$).
  Hence, T is compact.

These examples illustrate the two pillars of Heine–Borel:no escape to infinity (boundedness) and no leaks at the edge (closedness). A set $S \subseteq \mathbb{C}$ is compact if and only if it satisfies both:

• Boundedness: A finite container ensures all points fit inside some ball B(0, R) of radius R > 0. Formally, S is bounded if there exists R > 0 such that ∣z∣ ≤ R for all z ∈ S.
• Closedness: A sealed container ensures no point “leak out” to the boundary (i.e., all limit points of S are contained in S). Formally, S is closed if for every convergent sequence $\{ z_n \} \subseteq S$, the limit $\lim_{n \to \infty} z_n$ lies in S.

A compact set is like a sealed finite container -nothing escapes to infinity, and nothing leaks through the walls.