Basic Topology of the Complex Plane

The problem is not the problem. The problem is your attitude about the problem, Captain Jack Sparrow

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.

Basic Topology of the Complex Plane

Recall that |z| is the distance between 0 and z in the complex plane, e.g., $|3+i| = \sqrt{10}$. Recall that for two complex numbers $z_1 = x_1 + y_1, z_2 = x_2 + y_2$, their distance in the complex plane is

$|z_1-z_2| = \sqrt{(x_1-x_2)² + (y_1-y_2)²}$.

Open Discs and Balls

Definition. D_R(z₀) = {z ∈ ℂ | |z - z₀| < R} is an open disc (ball) centered at z₀ of radius R. An open disc centered at z₀ of radius R contains every point in the complex plane who distance from z₀ is strictly less than R.

Alternate notation: B(a; r) = {z ∈ ℂ | |z - a| < r}; an open disc or ball (Figure 1), the set of all points enclosed by the circle C of radius r centered at a. Its boundary circle is {z ∈ ℂ | |z - a| = r}; e.g., B(i ; 1) = {z ∈ ℂ | |z - i| < 1}.

Closed Discs

Definition. A closed disk of radius R centered at a point a is typically denoted by $\overline{B(a; R)}$. This includes all points whose distance from “a” is less than or equal to R. A close ball of radius r around a is essentially a circle of radius r around a, $\overline{B(a, r)}$ = {z ∈ ℂ | |z - a| ≤ r}.

The boundary of this disk — also called the circumference — is the set of points exactly at distance R from “a”: ∂B(a; R) = {z: ∣z−a∣ = R} = $\overline{B(a; R)}$ \ B(a; R). This expression means the boundary is the difference between the closed disk and the open disk (which excludes the boundary).

Deleted Neighborhood

A deleted neighborhood is the area (neighborhood) surrounding a specific point, minus the center point itself. This is denoted by B’(a; r) = {z ∈ ℂ | 0 < |z - a| < r} (Figure 3).

The set difference is $\overline{B(a, r)} - B(a, r)$ = {z ∈ ℂ | |z - a| = r} is the circumference around “a” of radius r.

The Upper Half–Plane

Definition. The upper half-plane is the portion of the complex plane satisfying Im(z) > 0, Π = { z ∈ ℂ: Im(z) > 0}. It consists of all points strictly above the real axis so it does not include the real line.

Annuli

Definition. An annulus centered at a with inner radius r₁ and outer radius r₂ is the region or ring between two concentric circles, formally A(a; r₁; r₂) = {z ∈ ℂ: r₁ < |z -a| < r₂} (Figure 4).

Isolated Points

Definition. A point z₀ ∈ S ⊆ ℂ is an isolated point of S if ∃R > 0 (meaning, there exist a positive radius) such that D_R(z₀) ∩ S = { z₀ }, that is, there exist an open disc centered at z₀ and the only point from S within this disk is z₀ itself. Isolated points stand alone within the set, with a neighborhood that does not include other points of the set.

z₀ sits in S with a small “bubble” around it containing no other points of S.

Boundary Points

Definition. A point z₀ is a boundary point of S, written z₀∈ ∂S, if every open disk centered at z₀ contains points both inside and outside (its complementary set ℂ∖S) the set S - every epsilon-neighborhood centered at z₀ contains at least one point in S and at least one point not in S.

z₀ may or may not be an element of S. Formally, $\forall \varepsilon > 0,\quad B(z_0; \varepsilon) \cap S \neq \emptyset \quad \text{and} \quad B(z_0; \varepsilon) \cap (\mathbb{C} \setminus S) \neq \emptyset$. These points are limit points of both the set and its complement.

Some examples are:

• The unit circle { z: |z| = 1} is the boundary of the open unit disk B(0; 1) = {z: |z| < 1}.
• Every point of ℂ is a boundary point of S = { x + iy: x, y ∈ ℚ}. In words, the set of complex numbers with rational real and imaginary parts has every point in $\mathbb{C}$ as a boundary point.
• The boundary set of S (the set of all its boundary points) = {z: 1 ≤ z < 2} is {z: |z| = 1 or |z| = 2}.

Open and Closed Sets

A set S ⊂ ℂ is open if every point of S has some open disc entirely contained in S.

Alternative definition. A set S is closed if its boundary is completely contained within S. Mathematically, this can be expressed as: ∂S ⊆ S. In other words, a closed set contains all its boundary points. The boundary is part of the set.

This means the set is “complete” in a topological sense —it doesn’t leave out any edge points.

Alternative definition. A set S is closed if its complement ℂ - S is an open set, e.g., $\overline{B(a; r)}$ is closed because $\Complex - \overline{B(a; r)}$ = {z ∈ ℂ: |z -a| > r} is clearly open.

These concepts — open/closed discs, neighborhoods, boundaries, isolated points, and basic regions like half-planes and annuli — provide the foundational “language” for limits, continuity, and convergence in complex analysis.

Theorem. Let S be a set of the complex numbers, S ⊆ ℂ, every point in ℂ is either an interior point, an exterior point, or a boundary point of S, and these categories are mutually exclusive.

This theorem is a cornerstone of topology in the complex plane. It formalizes the idea that every point in $\mathbb{C}$ has a well-defined relationship to any subset $S \subseteq \mathbb{C}$: it’s either inside, outside, or on the edge — and never more than one at a time.

Proof:

We need to show that these three possibilities are mutually exclusive and exhaustive (meaning every point z in ℂ must fall into exactly one of these categories).

Mutually exclusive:

1. A point cannot be both an interior point and an exterior point. If z were both, there would be disks D_1z ⊆ S and D_2z ⊆ ℂ - S. The intersection of these disks would be empty, but both contain z, which is obviously a contradiction.

2. A point cannot be both an interior point and a boundary point. If z were an interior point, there would be a disk D_z ⊆ S. This disk contains no points of ℂ - S, contradicting the definition of a boundary point.

3. A point cannot be both an exterior point and a boundary point. Suppose, for the sake of contradiction, that a point z is both an exterior point and a boundary point of a set S.

Because z is an exterior point, there exists an open disk D_ez (where the subscript ’e’ stands for exterior) centered at z such that D_ez ⊆ ℂ - S. Because z is a boundary point, every open disk centered at z must contain at least one point of S - which contradicts the assumption that the disk lies entirely in $\mathbb{C} \setminus S$ ⊥

Exhaustiveness: Let z be an arbitrary point in ℂ. Consider an arbitrary open disk D_z centered at z. There are three possibilities:

• D_z is entirely contained in S. Then z is an interior point of S.
• D_z is entirely contained in ℂ - S. Then z is an exterior point of S.
• D_z contains points of both S and ℂ - S. Then z is a boundary point of S.

Since these possibilities are mutually exclusive, we have already proved Exhaustiveness.

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Theorem. Let S be a set of the complex numbers, S ⊆ ℂ, S is closed if and only if its complementary set ℂ - S = {z ∈ ℂ | z ∉ S} is open.

Proof:

S is closed ⇒ ℂ - S is open:

Let z be an arbitrary point in ℂ - S. This means z ∉ S. By assumption, S is closed, meaning it contains all its boundary points, and z is not in S, z cannot be a boundary point of S. Therefore, z ∉ ∂S.

If z is not a boundary point of S, it must be an interior point of ℂ - S. This means there exists an open disk D_z centered at z such that D_z is entirely contained within ℂ - S (i.e., D_z ⊆ ℂ - S).

Since z was an arbitrary point in ℂ - S, we have shown that every point in ℂ - S is an interior point. By definition, this means that ℂ - S is open.

ℂ - S is open ⇒ S is closed, meaning ∂S ⊆ S.

Let z be a boundary point of S (i.e., z ∈ ∂S). We want to show that z must also be in S.

Suppose, for the sake of contradiction, that z is not in S (i.e., z ∈ ℂ - S). By assumption, ℂ - S is open, there must exist an open disk D_z centered at z such that D_z is entirely contained within ℂ - S (i.e., D_z ⊆ ℂ - S).

If D_z is entirely within ℂ - S, then D_z contains no points of S. But this contradicts the definition of a boundary point. A boundary point of S must have the property that every neighborhood around it contains points both in S and in ℂ - S.

Therefore, our assumption that z ∉ S must be false. Thus, z must be in S. Since z was an arbitrary boundary point of S, we have shown that all boundary points of S are contained in S. This means ∂S ⊆ S, and therefore, S is closed.

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Definition. The closure of a set S in the complex plane (S ⊆ ℂ) denoted by $\bar{S}$ (or sometimes 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: $\bar{S} = S ∪ ∂S$. where ∂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. 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, $\bar{S}$ = ⋂ { C: C is closed and S ⊂ C}.

Examples

• Let D_R​(z₀) = {z : |z - z₀| < R}. Its closure is the closed disk $\overline{D_R(z_0)} = ${z ∈ ℂ | |z - z₀| ≤ R }. The boundary of that closed disk is the circle: $∂\overline{D_R(z_0)} = ${z ∈ ℂ : |z - z₀| = R }

• Illustrative Sequence Example Let S = {¹⁄_k + i²⁄_k : k ∈ ℤ, k ≠ 0}. Each point of S is isolated except that as ∣k∣→∞, $\frac{1}{k} + i\frac{2}{k} \rightarrow 0$. Hence $\bar{S}$ = {¹⁄_k + i²⁄_k : k ∈ ℤ} ∪ {0} since 0 is the unique limit point not already in S.