Basic Topology of the Complex Plane

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

Basic Topology of the Complex Plane

Recall that |z| is the distance between 0 and z in the complex plane, e.g., $|3+i| = \sqrt{10}, |z_1-z_2| = \sqrt{(x_1-x_2)² + (y_1-y_2)²} \text{ where } z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$. Hence, |z₁ -z| is the distance between the complex numbers z₁ and z₂ in the complex plane.

Definition. D_R(z₀) = {z ∈ ℂ | |z - z₀| < R} is an open disc centered at z₀ with radius R. An open disc centered at z₀ with radius R contains all points z in the complex plane such that the distance from z₀ is less than R.

Another notation is 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, e.g., B(i ; 1) = {z ∈ ℂ | |z - i| < 1} (Figure 2).

A deleted neighborhood is the area (neighborhood) surrounding a specific point, minus the point itself. This is denoted by B’(a; r) = {z ∈ ℂ | 0 < |z - a| < r} (Figure 3). A close ball of radius r around a is essentially a circle of radius r around a. It is denoted by $\overline{B(a, r)}$ = {z ∈ ℂ | |z - a| ≤ r}.

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 is the portion of the complex plane satisfying Im(z) > 0, Π = { z ∈ ℂ: Im(z) > 0} (it does not include the real line).

An annulus is the region between two concentric circles, formally {z ∈ ℂ: r₁ < |z -a| < r₂} (Figure 4).

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

Definition. A point z₀ is a boundary point of S, z₀∈ ∂S, if every open disk centered at z₀ (every epsilon-neighborhood) contains points both inside and outside (its complementary set) the set S. z₀ may or may not be an element of S↭ ∀ε > 0, B(z; ε) ∩ S ≠ ∅ & B(z; ε) ∩ (ℂ - S) ≠ ∅. These points are limit points of both the set and its complement, e.g., { z: |z| = 1} is a set of boundary points of B(0; 1) = {z: |z| < 1}; every point of ℂ is a boundary point of S = { x + iy: x, y ∈ ℚ}; the boundary set of S (the set of all its boundary points) = {z: 1 ≤ z < 2} is {z: |z| = 1 or |z| = 2}

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 their boundary points. The boundary is part of the set.

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 open.

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.

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 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 ⊥

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

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

Since these are the only possibilities for any disk D, every point z must be either interior, exterior, or a boundary point.

Theorem. Let S be a set of the complex numbers, S ⊆ ℂ, S 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. Since S 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). Since 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.

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: $\bar{S} = S ∪ ∂S$.
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 pont on S which is not a limit point is called an isolated point of S. B’(a; ε) ∩ S = ∅ for some ε > 0 where B’(a; ε) does not include a.
3. As the intersection of all closed sets containing S.

In particular, $\overline{D_R(z_0)} = ${z ∈ ℂ | |z - z₀| ≤ R }, $∂\overline{D_R(z_0)} = ${z ∈ ℂ | |z - z₀| = R }

• S = {¹⁄_k + i²⁄_k : k ∈ ℤ}, $\bar{S}$ = {¹⁄_k + i²⁄_k : k ∈ ℤ} ∪ {0}.

• Proposition. Let S ⊆ ℂ, $\bar{S}$ is a closed set in ℂ.

Claim: $\Complex - \bar{S}$ is an open set (Figure A).

z ∈ $\Complex - \bar{S}$, z ∉ S ($\bar{S} = S ∪ ∂S$), z is not a limit point of S $\leadsto \exist \epsilon > 0 s.t.~ B(z; \epsilon) ∩ S = ∅$

Claim: $B’(z; \epsilon) ∩ \bar{S} = ∅$

Suppose $B’(z; \epsilon) ∩ \bar{S} \ne ∅$ ⇒[$B(z; \epsilon) ∩ S = ∅$] there is a limit point of S in $B’(z; \epsilon)$, say w.

w ∈ $B’(z; \epsilon)$, ∃ε such that |w -z| < ε, pick δ: 0 < δ < ε - |w -z|. Since w is a limit point of S, then $B’(w, \delta) ∩ S \in \empty \leadsto B’(z; \epsilon) ∩ \bar{S} \ne ∅ \leadsto B’(z; \epsilon) ⊆ \Complex - \bar{S}, \Complex - \bar{S}$ is open in ℂ ⇒ $\bar{S}$ is a closed set.∎

Proposition. Let S ⊆ ℂ then the following statements are equivalent:
(i) S is closed in ℂ.
(ii) S contains all its limits points.
(iii) $\bar{S} = S$

Proof:

$(i) \leadsto (ii) \leadsto (iii) \leadsto (i)$

$(ii) \leadsto (iii): \bar{S} = S ∪ ∂S = S$

$(iii) \leadsto (i):$ Proposition. Let S ⊆ ℂ, $\bar{S}$ is a closed set in ℂ, then S = $\bar{S}$ is a closed set in ℂ.

$(i) \leadsto (ii)$. Suppose S is closed in ℂ.

Claim: S contains all its limits points ↭ Any arbitrary limit point of S cannot belong to ℂ - S.

Suppose for the sake of contradiction there is such a limit point z ∈ ℂ - S. S is closed ⇒[Theorem. Let S be a set of the complex numbers, S ⊆ ℂ, S S is closed if and only if its complementary set ℂ - S = {z ∈ ℂ | z ∉ S} is open.] ℂ - S is open.

Therefore, there exist ∃ε > 0: B(z; ε) ⊆ ℂ - S ⇒ B(z; ε) ∩ S = ∅ which contradicts the assumption that z is a limit point. ∎

Examples & exercises

• Any point with |z| = 1 is a limit point of the set {z∈ ℂ: |z| < 1}. 2 + 7i is an isolated point of the set {z∈ ℂ: |z| < 1} ∪ {2 + 7i}.

• Every complex number is a limit point of S = {x + iy ∈ ℂ: x, y ∈ ℝ}.

Let z = x + iy ∈ ℂ. Claim. B(z, ε) ∩ S ≠ ∅

Rational numbers are dense in real numbers. In simpler terms, between any two real numbers, you can always find a rational number. In particular, there are rational numbers x₀, y₀ such that $x < x_0 < x + \frac{\epsilon}{\sqrt{2}},\text{ and } y < y_0 < y + \frac{\epsilon}{\sqrt{2}}$ (Figure B). Hence, x₀ + iy₀ ∈ B(z, ε) ∩ S.

• 0 is the only limit point of S = {¹⁄_k + i²⁄_k: k ∈ ℤ} (Figure C)

Consider B(0, ε), ε > 0.

The Archimedean property states that For any two positive numbers, there exists a positive integer such that the larger number is greater than the smaller number multiplied by that integer: a < n⋅b or $\frac{a}{n} < b$

In our particular case, $\frac{1}{4n} < \frac{1}{2n} < \epsilon \leadsto \frac{1}{4n} + \frac{i}{2n} ∈ B(0, \epsilon) ∩ S$, that is, 0 is a limit point. Any other point z ∈ ℂ is not a limit point.

For z = x + iy ∈ C, not of the form x + 2ix for some x ∈ ℝ, then the perpendicular distance of the point from the line, say ε_z is positive and it satisfies B(z, ε_z) ∩ S = ∅, so z is not a limit of S.

Let z = x + i2x ∈ C, x > 1 or x < -1, x ∈ ℝ, then take ε_z = |x -1| or ε_z = |x +1| respectively, B(z, ε_z) ∩ S = ∅

If z = x + i2x ∈ C, -1 ≤ x ≤ 1 and z ∉ S, then there are an integer k ∈ ℤ such that $\frac{1}{k+1} < x < \frac{1}{k},$ pick ε_k = min{$|x-\frac{1}{k+1}|, |x-\frac{1}{k}|$}, then B(z, ε_z) ∩ S = ∅ (Figure D)

If z = x + 2ix ∈ C, -1 ≤ x ≤ 1 and z ∈ S, then there are an integer k ∈ ℤ such that $\frac{1}{k} = x$, pick ε_k = $\frac{1}{(k+1)²}$, B’(z, ε_z) ∩ S = ∅ where B’(z, ε_z) does not include the point z itself.

Definition. A set S in the complex plane is bounded if you can find an open set that completely contains it. Formally, if there exist a radius ∃R > 0 and a point z₀∈ ℂ such that the entire set S is contained within the open disc centered at z₀​ with radius R, S ⊆ D_R(z₀) = { z ∈ ℂ ∣ ∣z − z₀∣ < R }

Definition. Let a, b ∈ ℂ. A closed interval defined as: [a, b] = { a + t(b - a) | t ∈ [0, 1]} is the interval that includes all points between a and b, including the endpoints. A polygonal path or broken line segments between two points p and q is defined as the union of several line segments connecting a sequence of points z₀, z₁, … , z_n​: γ = [p = z₀, z₁] ∪ [z₁, z₂] ∪ … ∪ [z_n-1, z_n = q]. Here, each segment [z_i, z_i+1] is defined similarly to the closed interval, representing the straight line segment between the consecutive points z_i​ and z_i+1.

Definition. A set S in the complex plane is connected if for any two arbitrary elements of S, p, q ∈ S, there exists a polygonal path γ connecting p and q that lies entirely within S: γ = [p = z₀, z₁] ∪ [z₁, z₂] ∪ … ∪ [z_n-1, z_n = q] ⊆ S.

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.

Definition. An open connected set is a domain. This means a domain has two important properties:

1. Open: For every point z in the domain, there exists an open disk centered at z 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. This means a region can be:

• A domain (if it contains none of its boundary points).
• A closed set (if it contains all of its boundary points).
• Something in between (if it contains some, but not all, of its boundary points).

Examples:

1. D = {z | |z| < 1}. D is a domain (it’s open and connected) and also a region (it’s a domain plus 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| ≤ 1, z ≠ 1} (the closed disk excluding the point 1). R is not a domain (it’s not open). R is a region (it’s the domain D plus some of its boundary points, but not the point 1).

Definition. A set S in the complex plane is called star-shaped if there exists a point z₀ ∈ S (called a star center of S) such that for every other point z ∈ S, the line segment connecting z₀ and z is entirely contained within S. Mathematically, S is star-shaped if there exists a z₀ ∈ S such that for all z ∈ S and for all t ∈ [0, 1] (i.e., 0 ≤ t ≤ 1), the point (1 - t)z₀ + tz is also in S.