Closure, Compactness, and Connectedness in the Complex Plane

I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.

Closure of a Set

Definition. The closure of a set S, denoted $\bar{S}$ or $\text{cl}(S)$, is the smallest closed set containing S. It can be defined in a few equivalent ways:

1. As the union of the set and its boundary: $\bar{S} = S \cup \partial S$. where $\partial S$ is the set of boundary points of S. Boundary points are those where every neighborhood intersects both S and its complement.
2. As the set of all limit points of S, $\bar{S} = S \cup S'$ where $S'$ is the set of limit points. A point z is a limit point (accumulation or cluster point) of S if every open disk centered at z contains at least one point of S different from z itself. Formally, ∀ R > 0, D_R​(z) ∩ (S∖{z}) ≠ ∅. Equivalently, {w ∈ ℂ: 0 < |w -z| < r} ∩ S ≠ ∅ for every r > 0. A point “a” on S which is not a limit point is called an isolated point of S. Formally, there exists ∃ ε > 0 such that B’(a; ε) ∩ S = ∅ where B’(a; ε) denotes a punctured disk which does not include a.

   A limit point z of S is one where every punctured disk around z meets S.

3. As the intersection of all closed sets containing S, $\bar{S} = \bigcap \{C : C \text{ is closed and } S \subseteq C\}$

Properties

Property	Statement
$S \subseteq \bar{S}$	$S$ is contained in its closure
$\bar{S}$ is closed	The closure is always closed
$S$ is closed $\iff S = \bar{S}$	Closed sets equal their closure
$\bar{\bar{S}} = \bar{S}$	Closure is idempotent
$\overline{S \cup T} = \bar{S} \cup \bar{T}$	Closure of union = union of closures
$\overline{S \cap T} \subseteq \bar{S} \cap \bar{T}$	Closure of intersection ⊆ intersection of closures

Examples $S \to$ Closure $\bar{S}$: $B(0; 1) = \{z : |z| < 1\} \to \{z : |z| \leq 1\}$; $(0, 1) \subset \mathbb{R} \to [0, 1]$; $\mathbb{Q} \subset \mathbb{R} \to \mathbb{R}$; $\{x + iy : x, y \in \mathbb{Q}\}\to \mathbb{C}$; $\{1/n : n \in \mathbb{Z}^+ \} \to \{1/n : n \in \mathbb{Z}^+\} \cup \{0\}$; $\mathbb{H} = \{z : \text{Im}(z) > 0\} \to \{z : \text{Im}(z) \geq 0\}$

Let $S = \left\{\frac{1}{k} + i\frac{2}{k} : k \in \mathbb{Z}, k \neq 0\right\}$. (i) Each point of $S$ is isolated (discrete set). (ii) As $|k| \to \infty$: $\frac{1}{k} + i\frac{2}{k} \to 0$, so $0$ is a limit point of $S$. (iii) $0 \notin S$. (iv) $\bar{S} = S \cup \{0\} = \left\{\frac{1}{k} + i\frac{2}{k} : k \in \mathbb{Z}, k \neq 0\right\} \cup \{0\}$

Bounded and Compact Sets

Definition: A set $S \subseteq \mathbb{C}$ is bounded if it is contained in some disk: $\exists M > 0: S \subseteq B(0; M)$. In other words, a set $S \subseteq \mathbb{C}$ is bounded if there exists a real number $M > 0$ such that S is entirely contained within the disk of radius M centered at the origin, $B(0; M)$, $\forall z \in S, |z| \le M$ Equivalently: $\sup\{|z| : z \in S\} < \infty$.

A set is bounded if it has a “finite size.” It does not stretch out to infinity in any direction. If you can draw a circle (no matter how big) around the origin that completely swallows the set, the set is bounded.

Examples: The unit disk $\overline{B(0; 1)}$, $|z| \le 1$ is bounded (it fits inside a radius of 2, for instance). $B(0; 1)$, $\{z : |z| = 1\}$, [0, 1], and any arbitrary finite set are bounded. Non-Example: The set of integers $\mathbb{Z}$ is unbounded because the numbers go on forever ($1, 2, 3...$). You cannot draw a finite circle that contains all integers. $\mathbb{R}$ and $\mathbb{C}$ are also unbounded.

Definition: A set $S \subseteq \mathbb{C}$ is compact if every sequence $\{z_n\}$ in S has a subsequence that converges to a limit $L$, and crucially, that limit $L$ must also be in $S$.

Compactness is about “trapping” sequences. If you have an infinite list of points inside a compact set, they eventually have to “cluster” or “accumulate” somewhere. Compactness guarantees that this clustering point is inside the set, not on the edge or outside.

Theorem (Heine-Borel): A subset S of $\mathbb{C}$ is compact if and only if it is closed (it contains all its boundary points -limit points-) and bounded (it fits inside some disk).

Why? If it's not bounded, a sequence can escape to infinity (e.g., $1, 2, 3, 4...$). It never settles down, so it has no convergent subsequence. If it's not closed, a sequence can escape to the edge, e.g., consider the open interval $(0, 1)$, the sequence $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}...$ converges to $0$. But $0$ is not in the set $(0, 1)$. The limit fell out of the set.

Set	Closed?	Bounded?	Compact?
$\overline{B(0; 1)}$	Yes	Yes	Yes
$B(0; 1)$	No	Yes	No
$\{z : \vert z \vert = 1\}$	Yes	Yes	Yes
$[0, 1] \subset \mathbb{R}$	Yes	Yes	Yes
$\mathbb{R}$	Yes	No	No
$\mathbb{C}$	Yes	No	No
Finite set	Yes	Yes	Yes

Properties of Compact Sets

• Closed subsets of compact sets are compact. If $K \subseteq \mathbb{C}$ is compact and $F \subseteq K$ is closed, then F is compact, e.g., $K = \overline{B(0; 1)}$ (closed unit disk, compact in $\mathbb{C}$), F = {z: |z| = 1} (the boundary circle, closed), then F is compact.
• Continuous images of compact sets are compact. If $f: K \to \mathbb{C}$ is continuous and K is compact, then f(K) is compact, e.g., f(x) = $e^{2\pi ix}$ maps [0, 1] onto the unit circle $S^1$, then $S^1$ is compact because it is the continuous image of a compact set.

Proof.

1. Take any sequence in f(K): $\{ y_n \} \subset f(K)$.
2. Each $y_n$ can be written as $y_n=f(x_n)$ for some $x_n \in K$.
3. Since K is compact, the sequence $\{ x_n \}$ has a convergent subsequence $\{ x_{n_k} \}$ with $x_{n_k}\rightarrow x \in K$.
4. Since f is continuous, $f(x_{n_k})\rightarrow f(x)$,so the subsequence $\{ y_{n_k} \} = \{ f(x_{n_k}) \}$ converges to $f(x)\in f(K).$
5. Thus every sequence in f(K) has a convergent subsequence with limit in f(K), so f(K) is compact.

• Extreme Value Theorem. If $f: K \to \mathbb{R}$ is continuous and K is compact, then f attains both its maximum and minimum value on K, that is, there exist points $x_{max}, x_{min} \in K$ such that: $f(x_{min}) \le f(x) \le f(x_{max}), \forall x \in K$.

Example

Consider $K = \overline{\mathbb{B}(0; 1)} = \{ z \in \mathbb{R}^n: ∥z∥ \le 1\}$, the closed unit ball, and f(z) = ∥z∥.

1. The function grows as we move outward, achieving ∥z∥ = 1 on the boundary sphere S^{n-1} (the set of points in n-dimensional space $\mathbb{R}^n$ at a fixed distance (radius 𝑟) from a central point).
2. The norm is minimized at the origin where ∥0∥ = 0.

Intuition

On a compact set, a continuous function cannot “oscillate wildly” without settling down enough to achieve its extreme values.

More deeply, since K is compact (in $\mathbb{R}^n$, this means closed and bounded), the function cannot run away to infinity. The values are trapped in some finite interval [m, M]. If a sequence of points $x_n$ produces values $f(x_n)$ approaching the supremum M, the points $x_n$, themselves must accumulate somewhere. Compactness guarantees that at least one accumulation point x actually lies inside K. Continuity then ensures that f(x) = M.

Proof

1. f(K) is bounded, so the supremum exists. Since K is compact and f is continuous, the image $f(K) \subseteq \mathbb{R}$ is compact (continuous image of compact is compact). In $\mathbb{R}$, compact sets are closed and bounded. Thus, f(K) is bounded above, and by the completeness of $\mathbb{R}$, the supremum $M = \sup f(K) = \sup \{f(x) : x \in K \}$ exists and is finite.
2. Approximate the supremum with a sequence. By the definition of supremum, for every integer n ≥ 1, there exists some point $x_n \in K$ such that: $M -\frac{1}{n} \lt f(x_n) \le M$. This constructs a sequence $\{ x_n \} \subseteq K$ with the property that $f(x_n) \to M$ as $n \to \infty$.
3. Extract a convergent subsequence. Because K is compact, every sequence in K has a subsequence that converges to a point in K. Thus, there exists a subsequence $\{ x_{n_k} \}$ such that $x_{n_k}\rightarrow x\in K$.
4. Apply continuity. Since f is continuos at x, it preserves limits. Therefore, $f(x_{n_k})\rightarrow f(x)$.
5. Identify the limit. But the left-hand side tends to M (since it is a subsequence of a sequence converging to M). By the uniqueness of limits in $\mathbb{R}$, we must have f(x) = M. Thus, f attains its maximum at x.
6. The same argument works for the minimum, define g(x) = -f(x). Then g is continuous, so it attains a maximum at some point $x_{min}$. This means $-f(x_{min}) \ge -f(x), \forall k \in K$, implying $f(x_{min}) \le f(x), \forall k \in K$. Thus, f attains its minimum.

Why Both Hypotheses Are Necessary

The theorem fails catastrophically if either hypothesis is dropped:

1. Not Compact. f(x) = x on K = (0, 1) (open interval). $\sup_{x \in K} f(x) = 1$, but $f(x) \le 1, \forall x \in K$. No maximum is ever attained.
2. Not Bounded. $f(x) = x^2$ on $K = \mathbb{R}$, f is unbounded and no maximum exists.
3. Not Continuous. $f(x) = \begin{cases} x, &x \lt 1 \\\\ 0, &x = 1 \end{cases}$ on K = [0, 1], $\sup_{x \in K} f(x) = 1$ (approached as $x \to 1^-$), but f(1) = 1. No maximum attained.

• Finite subcover. Every open cover of K has a finite subcover. Intuition: You can always shrink an infinite collection of open sets covering K to a finite subset that still covers K.

Let $K = [0, 1] \subseteq \mathbb{R}$. Consider the open cover $\mathcal{U}=\{ (n-1, n+1) \cap [0, 1]: n \in \mathbb{Z}\}$. A finite subcover is $\{ (-1,1)\cap [0,1],\ (0,2)\cap [0,1]\}$.

$\mathcal{U}$ is a open cover. (i) every set in $\mathcal{U}$ is open: (n-1, n+1) is an open interval in $\mathbb{R}$; the intersection of an open set with [0, 1] is open in [0, 1]. (ii) And the family $\mathcal{U}$ cover K = [0, 1].

Important Note:

1. The standard topology on $\mathbb{R}$ is defined using open intervals. A set is open if, around every point in it, you can fit a small open interval entirely within the set.
2. When we take a subset $K = [0,1]$ of $\mathbb{R}$, we don’t automatically inherit the definition of “open” from $\mathbb{R}$. Instead, we give $K$ the subspace (or relative) topology.
3. A set $V \subseteq K$ is open in K iff $V = U \cap L$ for some open set U in $\mathbb{R}$. Openness is relative.
4. Take an open interval in $\mathbb{R}$, for example $U = (a, b)$ or $(n-1, n+1)$. This is open in $\mathbb{R}$ by definition. Now, form the intersection $V = U \cap [0, 1]$, V is open in [0, 1] (by the very definition of the subspace topology stated above).

Connected and Path-Connected Sets

Definition: A set $S \subseteq \mathbb{C}$ (or any topological space) is connected if it cannot be written as the union of two disjoint, nonempty, relatively open subsets. In other words, S is connected if there do not exist two nonempty sets U, V such that:

1. $S = U \cup V$ (they cover S).
2. $U \cap V = \emptyset$ (they are disjoint).
3. Both U and U are relatively open in S (i.e., $U = S \cup U'$ and $V = S \cup V'$ for some open sets U’, V’ in the ambient space).

Intuitively: S is in one single piece, you can't separate it into two parts without cutting.

Definition: A set $S$ is path-connected if for any two points $z_1, z_2 \in S$, there exists a continuous curve $\gamma: [0, 1] \to S$ (a “path”) with $\gamma(0) = z_1$ and $\gamma(1) = z_2$.

Intuition 🔮. You can draw a continuous curve within S connecting any two points without lifting your pen.

Examples: $B(0; 1), \overline{B(0; 1)}, \mathbb{C}, \mathbb{H}$ (upper half-plane), $A(0; 1, 2)$ (annulus) and $\mathbb{C} \setminus \{0\}$ (we can go around 0).

Anti-examples: $B(0; 1) \cup B(3; 1)$ (Two disjoint open disks. Take U = left disk, V = right disk. They are separated by distance 2), $\{z : ∥z∥ \neq 1\}$ (this is the punctured plane at the unit circle: the interior ∥z∥ < 1 and exterior ∥z∥ > 1. These are two disjoint open sets separated by the “fence” ∥z∥=1), $\mathbb{R} \setminus \{0\}$ (this equals (−∞, 0) ∪ (0, ∞). You cannot jump from negative to positive without crossing 0).

In $\mathbb{R}$, connected sets are exactly intervals. Removing a point from an interval breaks it into two intervals.

$\text{Path-connected} \implies \text{Connected}$. If you can draw a continuous line (path) from any point to any other point in a set $S$, the set effectively consists of one piece. It cannot be split into two disjoint, non-empty open sets. This direction always holds.

Why the converse fails (Crucial Example)?

The Topologist’s Sine Curve is the canonical example of a set that is connected but not path-connected: $S = \underbrace{\{(x, \sin(1/x)) : 0 < x \le 1\}}_{\text{Curve } C} \cup \underbrace{\{(0, y): -1 \le y \le 1\}}_{\text{Vertical Segment } V}$

Why is $S$ Connected?

1. $C$ is Connected: The curve $C$ is the image of the interval $(0, 1]$ under the continuous function $f(x) = (x, \sin(1/x))$. Since the interval $(0, 1]$ is connected, its image $C$ must be connected.
2. The Closure Property states: If a set A is connected, then its closure $\bar{A}$ is also connected.
3. As $x \to 0$, the term $\sin(1/x)$ oscillates faster and faster, hitting every value between $-1$ and $1$ infinitely many times. Therefore, the “limit points” of the curve $C$ as it approaches the y-axis are all the points on the vertical segment from $(0, -1)$ to $(0, 1)$. This means the vertical segment V is part of the closure of C, $S = \bar{C}$. Since $C$ is connected, $S$ is connected.

Why is $S$ NOT Path-Connected?

We try to assume a path exists and prove it leads to a contradiction (Reductio ad absurdum).

Imagine trying to draw a continuous path $\gamma(t)$ starting at the origin $(0, 0)$ on the vertical line $V$ and ending at a point on the curve $C$, say $(a, \sin(1/a)), a \in (0, 1]$. Let $\gamma: [0, 1] \to S$ be this path, where: $\gamma(0) = (0, 0), \gamma(t) = (x(t), y(t)), \gamma(1) = (a, \sin(1/a))$ (e.g., (1/π, 0)).

1. The projection $\gamma_{x}(t)$ (the x-coordinate of γ(t)) must be continuous and map [0, 1] to [0, a]. By the Intermediate Value Theorem, $\gamma_{x}(t)$ takes all values in (0, a].
2. Since the path moves from the set $V$ (the wall) to the set $C$ (the curve), there must be a specific time $t_0$ that represents the last instant the path touches the wall before entering the curve: (i) For $t \le t_0$: The path is on the wall $V$ (so $x(t) = 0$); (ii) For $t > t_0$: The path is on the curve $C$, so $x(t) > 0$ and $y(t) = \sin(1/x(t))$.
3. For the path to be continuous at $t_0$, the limit of the path as we approach $t_0$ from the right (coming from the curve side) must equal the position at $t_0$ (on the wall). As $t \to t_0^+$, the path slides closer to the wall, so $x(t) \to 0$. For any $t > t_0$, the point is on the curve $C$, $\gamma_{y}(t) = sin(\frac{1}{\gamma_x(t)})$. As $t$ approaches $t_0$, $\gamma_x(t)$ gets smaller and smaller (approaching 0), $sin(1/\gamma_x(t))$ goes to infinity, and $\gamma_{y}(t) = sin(\frac{1}{\gamma_x(t)})$ oscillates violently between -1 and 1. This violates continuity because $\gamma_{y}(t)$ is oscillating infinitely fast as $t \to t_0$, it does not approach any single number. It doesn’t approach whatever y-value the path had at $t_0$.

Conclusion: So, even if you wander around the wall $V$ for awhile, the very first step you take off the wall into the curve $C$ requires a continuity that the geometry makes impossible.

Therefore, S is not path-connected, no continuous path can bridge the gap between the “wall” (V) and the “curve” (C).

Theorem. If a set $U \subseteq \mathbb{C}$ is open and connected, it is automatically path-connected.

Proof (Open + Connected ⟹ Path-Connected)

1. Suppose $\Omega$ is open and connected. Fix a base point $z_0 \in \Omega$.
2. Define A = set of points in $\Omega$ that can be joined to $z_0$ by a path within $\Omega$.
3. Define B = set of points in $\Omega$ that can be joined to $z_0$ by any path.
4. A is open. If $z \in A$, since $\Omega$ is open, there is a ball $\mathbb{B}(z; r) \subseteq \Omega$. Any point w in this ball can be joined to z by a straight line segment, then to $z_0$ via the path from z. Thus, $\mathbb{B}(z; r) \subseteq A$.
5. B is open, too. If $z \in B$, since $\Omega$ is open, there is a ball $\mathbb{B}(z; r) \subseteq \Omega$. If any point in this ball could connect to $z_0$, then z could connect to that point and then to $z_0$, contradiction. Hence, $\mathbb{B}(z; r) \subseteq B$.
6. Therefore, $\Omega = A \cup B$ is a disjoint union of two open sets. By assumption $\Omega$ is connected, and A is not empty ($z_0 \in A$), B must be empty, that implies that every point connects to $z_0$, making $\Omega$ path-connected.

Domains and Regions

Definition: A domain (or region) in the complex plane $\mathbb{C}$ is defined as a nonempty, open, connected subset of $\mathbb{C}$:

• Open: Every point $z_0$ in the set has a neighborhood (e.g., an open disk $\mathbb{B}(z_0; \varepsilon)$) entirely contained within the set.
• Connected: It cannot be partitioned into two disjoint nonempty open subsets. For open subsets of $\mathbb{}C$, this is equivalent to path-connectedness: any two points in the domain can be joined by a continuous polygonal path (or curve) lying entirely within the domain.

Domains are the natural setting for studying complex analysis because differentiation is a local property meaning that to determine the derivative of a function at a point, we need to consider values of the function in the vicinity of that point.

For integration, particularly contour integration, it is important that the domain is connected. This means that any two points in the domain can be joined by a continuous path that lies entirely within the domain. This property is vital for defining integrals over paths in the complex plane, as it allows for the evaluation of integrals without leaving the set.

Examples of Domains

Common examples of domains in $\mathbb{C}$ include:

• The entire plane $\mathbb{C}$; Open unit disk B(0; 1) = $\{ z \in \mathbb{C}: |z| < 1 \}$; Upper half-plane $\mathbb{H} = \{ z \in \mathbb{C}: \mathcal{Im}(z) \gt 0 \}$; Puncture plane $\mathbb{C} \setminus \{0\}$ (the complex plane with the origin removed).
• Open annulus $A(0; 1, 2) = \{ z \in \mathbb{C}: 1 < |z| < 2 \}$; Slit plane $\mathbb{C} \setminus (-\infty, 0]$ (the plane with the negative real axis removed).

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 $\mathcal{D}$.