Regions, Connectivity, and Components in the Complex Plane

Behind this mask there is more than just flesh. Beneath this mask there is an idea… and ideas are bulletproof, Alan Moore

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).

Region (Domain)

Definition. A region (or domain) G is an open, non-empty, connected subset of ℂ. All three properties are essential:

• Non-empty: G ≠ ∅, avoids the trivial case.
• Open: for every $z \in G$, there exists $\epsilon > 0$ such that $B(z, \epsilon) \subset G$ (this ensures “room to move” so we can take derivatives in any direction without hitting a boundary).
• Connected: G cannot be split as G = G₁ ∪ G₂ where G₁, G₂ are disjoint, non-empty, open in ℂ.

Examples: Open disk, B(z₀; r); open (right) half-plane, {z : ℜ(z) > 0}; Open vertical strip, {z : a < ℜ(z) < b}; Punctured disk (disk with center removed), B(z₀; r) \ {z₀}; Punctured plane (complex plane minus origin), ℂ \ {0} = ℂ*; Slit plane (plane minus non-positive reals) ℂ \ (-∞, 0]; Annulus (ring between two circles), {z : r < |z| < R}; Upper half-plane (above the real axis), ℍ = {z : ℑ(z) > 0}.

Non-Examples: Closed disk $\bar{B}_r(z_0)$ (it is not open because its boundary is included); ∅; ℂ \ ℝ (it is not connected); B₁(0) ∪ B₁(5) (not connected); {0} (not open); [0, 1] (no open, it has no interior in ℂ).

The “Polygonal Path” Theorem

While the topological definition of connectedness (involving disjoint open sets) is rigorous, it is hard to visualize. For open sets, there is a much more intuitive equivalent: path connectivity.

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

Important Remark. A polygonal path is a continuous path formed by joining a finite number of line segments end-to-end. More specifically, a polygonal path from p to q is a path consisting of finitely many line segments: $\gamma = [z_0, z_1] \cup [z_1, z_2] \cup \cdots \cup [z_{n-1}, z_n]$ where z₀ = p and z_n = q and the segment [z₁, z₂] = {(1-t)z₁ + tz₂ : t ∈ [0, 1]}.

Definition. A set S is path-connected if any two points in S can be joined by a path lying entirely in S.

Definition. A set S is polygonally connected if any two points can be joined by a polygonal path in S.

Proof

Part 1. Connectedness ⇒ Polygonal Connectivity

We will show that the set of points reachable from a starting point $a$ is both open and closed relative to G. Since $G$ is connected, the only non-empty subset that is both open and closed is $G$ itself.

1. Suppose G is a region, fix a point $a \in G$.
2. Let $U = \{z \in G : \text{there exists a polygonal path from } a \text{ to } z \text{ in } G\}$.
3. Let $V = G \setminus U$. V is the set of points in $G$ not reachable from a.
4. U is Open. Let $z \in U$. Since $G$ is open, there exists a ball $B(z; r) \subset G$. For any point $w \in B(z; r)$, the straight line segment $[z, w]$ lies entirely in the ball (and thus in $G$). Since we can get from $a$ to $z$ (by definition of $U$), we can extend that path to $w$. Thus, $w \in U$. Therefore, $B(z; r) \subseteq U$, so $U$ is open.
5. V is Open. Let $z \in V$. Since $G$ is open, there exists $B(z; r) \subset G$. Claim: $B(z; r) \subseteq V$.
   Assume for contradiction there is a point $w \in B(z; r)$ such that $w \in U$ (meaning we can reach $w$ from $a$). We can draw a line segment from $w$ to $z$ inside the ball. Connecting the path a to w with $[w, z]$ gives a path $a \to z$. This would mean $z \in U$, which contradicts $z \in V$. Thus, no point in the ball is in $U$, so the whole ball is in $V$. $V$ is open.
6. $G = U \cup V$, where $U$ and $V$ are disjoint open sets. Since $G$ is connected, it cannot be split. Therefore, one of them must be empty. Since $a \in U$, $U \neq \emptyset$. Thus, $V = \emptyset$ and $U = G$. Every point is reachable. ∎

Part 2: Polygonal Connectivity ⇒ Connectedness

Strategy: We rely on the fact that a line segment is a connected set.

1. Assume for the sake of contradiction that $G$ is not connected. Then, $G = U \cup V$, where $U, V$ are disjoint, non-empty, open sets.
2. Pick $p \in U$ and $q \in V$.
3. By hypothesis, there exists a polygonal path $\gamma$ from $p$ to $q$ inside $G$. $\gamma = [z_0, z_1] \cup [z_1, z_2] \cup \cdots \cup [z_{n-1}, z_n]$ with z₀ = p ∈ U and z_n = q ∈ V.
4. Since $p \in U$ and $q \in V$, the path must “cross over” from $U$ to $V$ at some point.
5. Finding the crossing segment: Let j be the smallest index such that z_j ∈ V. (Such j exists since z_n ∈ V.) Then, (i) z_j-1 ∈ U (by minimality of j, since z₀,…,z_j-1 ∉ V, and they’re in G = U ∪ V); (ii) z_j ∈ V.
   Consider the segment ℓ = [z_j-1, z_j].
6. The segment ℓ is connected: The map f: [0,1] → ℓ defined by f(t) = (1-t)z_j-1 + t·z_j is continuous.
   The interval [0, 1] is connected in ℝ.
   The continuous image of a connected set is connected.
   Therefore, ℓ = f([0,1]) is connected in ℂ.
7. The segment ℓ satisfies: (i) ℓ ⊂ G = U ∪ V (since the polygonal path lies in G); (ii) z_j-1 ∈ ℓ ∩ U ≠ ∅; (iii) z_j ∈ ℓ ∩ V ≠ ∅.
8. Define: U’ = U ∩ ℓ (open in subspace topology of ℓ) and V’ = V ∩ ℓ (open in subspace topology of ℓ).
9. Then, U’ ∪ V’ = (U ∩ ℓ) ∪ (V ∩ ℓ) = (U ∪ V) ∩ ℓ = G ∩ ℓ = ℓ (since ℓ ⊂ G); U’ ∩ V’ = (U ∩ V) ∩ ℓ = ∅ ∩ ℓ = ∅; U’ ≠ ∅ (contains z_j-1) and V’ ≠ ∅ (contains z_j).
10. This means (U’, V’) is a separation of ℓ, contradicting that ℓ is connected! ⚡
11. Therefore no such polygonal path exists, and G is not polygonally connected. ∎

Characterization of Regions. Let G be a non-empty, open subset of ℂ. The following are equivalent:

1. G is connected (no separation exists).
2. G is path-connected (any two points joined by a continuous path).
3. G is polygonally connected (any two points joined by a polygonal path).

Sketch of Proof

If a set $G \subseteq \mathbb{C}$ is polygonally connected (meaning any two points can be connected by a polygonal chain), then $G$ is path-connected.

A polygonal path is a “broken-line” path, which is clearly continuous (no jumps or breaks). Since path-connectedness only requires the existence of some continuous path, polygonally connected spaces satisfy this trivially.

Proof.

1. Let $p, q \in G$. Since $G$ is polygonally connected, there exists a finite sequence of points $z_0, z_1, \dots, z_n \in G$ such that: $z_0 = p, z_n = q$. For each $1 \le i \le n$, the line segment $[z_{i-1}, z_i]$ is entirely contained in $G$.
2. Construct the Parameterization. We define a path $\gamma: [0, 1] \to G$ by dividing the time interval $[0, 1]$ into $n$ equal sub-intervals. On the $i$-th sub-interval, we traverse the segment from $z_{i-1}$ to $z_i$.
   For $t \in [\frac{i-1}{n}, \frac{i}{n}]$, we define $\gamma(t)$ using the linear interpolation formula: $\gamma(t) = \underbrace{\left(1 - n\left(t - \frac{i-1}{n}\right)\right)}_{\text{weight for } z_{i-1}} z_{i-1} + \underbrace{n\left(t - \frac{i-1}{n}\right)}_{\text{weight for } z_i} z_i$
   Note: The term $n(t - \frac{i-1}{n})$ essentially rescales the small interval $[\frac{i-1}{n}, \frac{i}{n}]$ to the standard unit interval $[0, 1]$ for interpolation.
3. Verify Continuity. A piecewise function is continuous if its components are continuous and they agree at the boundaries (the “knots” and glue between segments).
   Interior of segments: Inside any open interval $(\frac{i-1}{n}, \frac{i}{n})$, $\gamma(t)$ is a linear polynomial in t, which is obviously continuous.
   At the junctions: Let’s check the junction $t = \frac{i}{n}$ between segment $i$ and segment $i+1$.
   From the Left (End of segment $i$): Plug $t = \frac{i}{n}$ into the formula for segment $i$:
   $\gamma\left(\frac{i}{n}\right) = (1 - n(\frac{1}{n}))z_{i-1} + n(\frac{1}{n})z_i = (1-1)z_{i-1} + (1)z_i = \mathbf{z_i}$.
   From the Right (Start of segment $i+1$): Plug $t = \frac{i}{n}$ into the formula for segment $i+1$ (where we replace $i$ with $i+1$ in the bounds):
   $\gamma(t) = \left(1 - n\left(t - \frac{i}{n}\right)\right) + n\left(t - \frac{i}{n}\right) z_i$, $\gamma\left(\frac{i}{n}\right) = (1 - n(0))z_i + n(0)z_{i+1} = (1)z_i + 0 = \mathbf{z_i}$
   Since the path approaches the same value $z_i$ from both sides, there are no “jumps.” The function is continuous at every junction.
4. The function $\gamma$ is continuous on $[0,1]$, $\gamma(0) = p$, $\gamma(1) = q$, and the image lies entirely in $G$ (as it is a union of segments in $G$). Therefore, $G$ is path-connected.

Path-connected ⟹ Connected:

A path-connected space is “all in one piece” because you can “walk” continuously from any point to any other. If it were disconnected, you could separate the space into two disjoint open sets, but then your path would have to “jump” from one set to the other, violating continuity.

Proof sketch:

1. Assume for the sake of contradiction that is path-connected but disconnected. Then, there exists a separation G = U ∪ V, where U and V are non-empty, disjoint, and open in G.
2. Let p ∈ U to q ∈ V. Since G is path-connected, there exists a continuous path γ: [0,1] → G such that γ(0) = p and γ(1) = q.
3. Define A = γ⁻¹(U) and B = γ⁻¹(V).
4. Since γ is continuous and U, V are open in G, A and B are open in [0, 1] (by the subspace topology).
5. Besides, A ∩ B = ∅ (since U ∩ V = ∅), and A ∪ B = [0, 1] (since γ([0, 1]) ⊆ G = U ∪ V).
6. Thus, A and B form a separation of [0, 1]. But [0, 1] is connected (a standard result in topology), so this is a contradiction.
7. Therefore, G cannot be disconnected, and hence G is connected.

For open subsets of $\mathbb{C}$ (or $\mathbb{R}^n$), the equivalence between being connected and path-connected holds. Outside this setting, path-connectedness is strictly stronger.

The Canonical Counter-Example: Topologist’s Sine Curve

Definition: $S = \underbrace{\left\{ \left(x, \sin\frac{1}{x}\right) : x \in (0, 1] \right\}}_{\text{The oscillating part}} \cup \underbrace{\{(0, 0)\}}_{\text{The limit point}}$

As $x \to 0^+$, the curve $\sin(1/x)$ oscillates infinitely fast between $-1$ and $+1$. The point $(0,0)$ is glued to this chaos as a limit point, but no path can “navigate” through the oscillations to reach it continuously.

Properties:

• S is connected.

1. The oscillating part $G = \left\{ \left(x, \sin\frac{1}{x}\right) : x \in (0, 1] \right\}$ is connected (The map $ x\mapsto \left( x,\sin \frac{1}{x}\right)$ is continuous on (0, 1], so its image $G = \left\{ \left( x,\sin \frac{1}{x}\right) :x\in (0,1]\right\}$ is connected).
2. The closure of a connected set is connected, the closure $\overline{G}=G\cup \{ (0,0)\} =S$ add the limit point (0, 0).
3. Therefore, S is connected.

• S is NOT path-connected: there is no continuous path from (0, 0) to any point (x, sin(1/x))

Suppose that there is a continuous path $\gamma: [0, 1] \to S, \gamma(t) = (x(t), y(t))$ with $\gamma(0) = (0, 0)$ and $\gamma(1) = (1, sin(1))$

Since $\gamma(0) = (0, 0)$, we have x(0) = 0 and x(1) = 1. By the Intermediate Value Theorem, x(t) takes all values in [0, 1].

In particular, for any sequence $x_n\rightarrow 0^+$, there exist $t_n\rightarrow 0^+$ such that $x(t_n)=x_n$.

For all t > 0, the point $\gamma(t)$ lies on the graph, so $y(t)=\sin \left( \frac{1}{x(t)}\right)$.

But as $x(t)\rightarrow 0^+$, the quantity $\frac{1}{x(t)}$ blows up, and $\sin \left( \frac{1}{x(t)}\right)$ oscillates between -1 and 1 infinitely often. Thus, the sequence $y(t_n)$ has no limit.

Continuity of $\gamma$ requires $y(t)\rightarrow y(0)=0\quad \mathrm{as\ }t\rightarrow 0^+$. But we just saw that y(t) oscillates without converging.

This contradiction shows that no continuous path can connect (0,0) to any point of the oscillating curve.

Therefore, S is not path‑connected.

• Why doesn’t this contradict our theorem?.. or Why Openness Saves the Day?

The topologist’s sine curve is not open. Our theorem specifically requires G to be open. $S$ contains no open ball around $(0,0)$ Any neighborhood of $(0,0)$ in $S$ is a “thickened oscillation” that remains disconnected by the wild behavior.

Hierarchy of Connectedness Properties

$$\text{Convex} \subset \text{Star-shaped} \subset \text{Path-connected} \subset \text{Connected}$$

For open subsets of $\mathbb{R}^n$ or $\mathbb{C}$: $\text{Connected} = \text{Path-connected}$. However, for general topological spaces, the last inclusion is strict.

Definition and Properties of Connected Components

Definition. Let S be a subset of a topological space (e.g., S $\subseteq \mathbb{C}$ or $\mathbb{R}^n$). For a point $z \in S$, the connected component of z in S is $C_z=\bigcup \{ C\subseteq S: C \mathrm{\ is\ connected\ and\ } z\in C\}$. This is the largest connected subset of S containing z.

Why “largest” makes sense:

• If $C_1, C_2, \dots$ are connected subsets of S all containing z, then their union is also connected.
• Therefore, the union of all such sets is connected and contains every other connected set containing z.
• This union is unique, so the connected component is well-defined.

Properties:

• Connected components form a partition of the space.

Every point belongs to exactly one connected component. Thus, $S=\bigcup_{z \in S}C_z$, and the sets $C_z$ are pairwise disjoint.

If we index the distinct components by some index set I, we can write: $S=\bigsqcup _{\alpha \in I}C_{\alpha }.$

This is a genuine partition: (i) Covering: every point lies in some component. (ii) Disjointness: if $C_{\alpha }\cap C_{\beta }\neq \emptyset$, then they must be the same connected set, so $\alpha =\beta$.

• Connected Components of Open Sets in $\mathbb{C}$ are Open (Regions). If $G \subseteq \mathbb{C}$ is open, then each connected component $C \subseteq G$ is also open.
  Proof
  Let $z \in C$. Since G is open, there exists $\varepsilon \gt 0$ such that $B(z; \varepsilon) \subseteq G$.
  The disk $B(z; \varepsilon)$ is connected (as it is path-connected), and since C is the largest connected subset containing z, $B(z; \varepsilon) \subseteq C$.
  Thus, C is open. Since C is both open and connected, it is a region.

• Connected components are maximal connected sets. Each component $C_z$ is connected, and no strictly larger connected subset of S contains it.
• A set is connected if and only if it has exactly one connected component.

Proof

(⇒) If S is connected, then it has exactly one connected component

1. Assume S is connected. Every point $z \in S$ lies in some connected component $C_z$.
2. Each $C_z$ is a connected subset of S containing z.
3. However, since S itself is connected and contains z, maximality forces $C_z = S$. Thus all components coincide, so there is exactly one connected component.

(⇐) If S has exactly one connected component, then S is connected

1. Connected components partition S, $S=\bigsqcup _{\alpha \in I}C_{\alpha }.$
2. If there is exactly one component, say C, then S = C.
3. Since C is connected (by definition of a connected component) and S = C, it follows that S is connected.

Examples

• Connected Components of G = $\mathbb{C} \setminus \mathbb{R}$.

The set G = {z : ℑ(z) ≠ 0} (the complex plane minus the real axis) is not connected because it can be split into two disjoint open sets:
(i) The upper half-plane: $\mathbb{H} = \{z : ℑ(z) > 0 \}$
(ii) The lower half-plane: $\mathbb{H}^- = \{z : ℑ(z) < 0 \}$

1. Each Half-Plane is Connected. Both $ℍ$ and $ℍ^-$ are path-connected (hence connected). $\forall z_1, z_2 \in \mathbb{H}$, the straight line segment $[z_1, z_2| \subseteq \mathbb{H}$ is a continuous path within $\mathbb{H}$ (Similarly for $\mathbb{H}^-$).
2. No Larger Connected Subset Contains Either Half-Plane. Suppose $C \subseteq G$ is connected and contains $\mathbb{H}$. If C also contained a point $z_0 \in \mathbb{H}^-$, then C would contain a path from $z_0$ to some $z_1 \in \mathbb{H}$. However, any such path must cross $\mathbb{R}$, which is not in G, so C cannot be connected (a contradiction).
3. Both $\mathbb{H}$ and $\mathbb{H}^-$ are open in $\mathbb{C}$ (and hence in G), so they are regions.

• Discrete Set. Let $S = \{ \frac{1}{n}: n \in \mathbb{N} \} \subseteq \mathbb{C}$. Each singleton $\{ \frac{1}{n} \}$ is trivially connected (since it has only one point). No larger connected subset exists because any two distinct points $\frac{1}{n} \ne \frac{1}{m}$ cannot be connected by a path in S (since S is discrete). Thus, the connected components are the singletons $\{ \frac{1}{n} \}$.

• Union of Two Disjoint Disks. Let $S = B(0; 1) \cap B(2, 1) \subseteq \mathbb{C}$. B(0; 1) and B(2, 1) are both connected and open. No path exists in S connecting a point in B(0; 1) to a point in B(2; 1) (since both disks are disjoin). Thus, the connected components are B(0; 1) and B(2, 1).

• The real line $\mathbb{R} \subseteq \mathbb{C}$. $\mathbb{R}$ is connected (any two real numbers can be joined by a straight line segment in $\mathbb{R}$). Thus, its only connected component is $\mathbb{R}$ itself.

Simply Connected Regions

Definition. A region $G \subseteq \mathbb{C}$ is simply connected if it is open, connected, and every closed curve (loop) in G.

A simply connected region has no topological “holes” or “obstacles.” Loops can be smoothly shrunk to a point without leaving the region, analogous to a rubber band contracting freely.

Examples: Open disk B_r(z₀) (all loops can contract to the center z₀); ℂ (no holes; loops shrink to any point); Upper half-plane ℍ; ℂ \ (-∞, 0] (the slit lies on the boundary, so loops avoid it and contract freely); Convex Sets (e.g., polygons, star-shaped regions).

Counterexamples: Punctured disk B_r(0) (A loop encircling 0 cannot contract without crossing the hole); Annulus {z : r < |z| < R} (Hole in center); ℂ \ {0} (Similar to the punctured disk; hole at origin obstructs contraction); Torus (surface of a doughnut).

Importance: Simply connected regions are where:

• Every holomorphic function on a simply connected domain has an antiderivative.
• Cauchy’s theorem. If f is holomorphic on a simply connected domain Ω, then for any closed curve $\gamma$ in Ω: $\int_{\gamma}f(z) dz = 0.$