Closure, Limit Points, and Connectedness in the Complex Plane

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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 II

• Proposition. For any S ⊆ ℂ, its closure $\bar{S}$ is a closed set of ℂ.

Recall that a set A is closed iff its complement $\Complex - \bar{A}$ is open.

Let z ∈ $\Complex - \bar{S}$ (an arbitrary element). By definition of closure $\bar{S} = S ∪ ∂S$ where ∂S is the boundary of S. Thus, z ∉ S and z ∉ ∂S. In particular, z is not a limit point of S (the closure of a set is the set of all its limit points).

A point z is a limit point of S if every neighborhood of z intersects S in at least one point other than z. A point w ∈ ∂S if every neighborhood of w intersects both S and ℂ − S.

Since z is not a limit point of S, there exists some epsilon, $\exist \epsilon > 0 ~s.t.~ B(z; \epsilon) ∩ S = ∅$ (🚀) where B(z;ϵ) is the open ball centered at z with radius ϵ.

$B(z; \epsilon) ∩ \bar{S} = B(z; \epsilon) ∩ (S ∪ ∂S) = (B(z; \epsilon) ∩ S) ∪ (B(z; \epsilon) ∩ ∂S) = ∅ ∪ (B(z; \epsilon)∩ ∂S) = B(z; \epsilon)∩ ∂S$

Claim: $B(z; \epsilon)∩ ∂S = ∅$

Suppose for contradiction that w ∈ B(z; ϵ) ∩ ∂S.

Since w ∈ ∂S, every neighborhood of w intersects S. Choose δ < ϵ − ∣w − z∣, so by construction B(w; δ) ⊆ B(z; ϵ). Then, B(w; δ)∩S ≠ ∅ (every neighborhood of w intersects S), contradicting B(z;ϵ)∩S = ∅ (🚀) ⊥

Thus, $B(z; \epsilon) ∩ \bar{S} = B(z; \epsilon)∩ ∂S = ∅ \leadsto B(z; \epsilon) ⊆ ℂ - \bar{S}$

$\forall z \in \Complex - \bar{S}, \exist \epsilon \gt 0 \text{ such that } B(z; \epsilon) ⊆ ℂ - \bar{S}\leadsto ℂ - \bar{S}$ is open. Therefore, $\bar{S}$ is closed. ∎

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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:

• $(i) \leadsto (ii)$

Suppose 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. For the sake of contradiction, suppose z were a limit point of S but z ∉ S, then by openness (ℂ∖S is open) there is an ε > 0 with B(z;ε) ⊂ ℂ∖S ⇒ B(z;ε) ∩ S = ∅. That contradicts z being a limit point ⊥

• $(ii) \leadsto (iii)$

By definition, $\bar{S}$ = S ∪ ∂S. But every point of ∂S is a limit point of S, so if by assumption, S already contains all its limit points then ∂S ⊆ S. Thus $\bar{S} = S ∪ ∂S = S$.

$(iii) \leadsto (i):$ Previous Proposition. For any S ⊆ ℂ, its closure $\bar{S}$ is a closed set of ℂ. If $\bar{S} = S$, the S itself is closed. ∎

Examples & exercises

• Any point with |z| = 1 is a limit point of the set {z∈ ℂ: |z| < 1}.
• The point 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}. In words, the complex numbers with rational coordinates are dense in ℂ!

S = {x + iy ∈ ℂ: x, y ∈ ℝ} is the set of all complex numbers with rational real and imaginary parts.

We want to show that every complex number is a limit point of S. That is, for any complex number $z = x + iy \in \mathbb{C}$ and any $\varepsilon > 0$, there exists a point in S that lies within $\varepsilon$-distance of z.

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

Since the rationals rational numbers $\mathbb{Q} $ are dense in the real numbers $\mathbb{R} $, 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).

Define $w = x_0 + i y_0 \in S$. Then: $|z - w| = \sqrt{(x - x_0)^2 + (y - y_0)^2} < \sqrt{\left(\frac{\varepsilon}{\sqrt{2}}\right)^2 + \left(\frac{\varepsilon}{\sqrt{2}}\right)^2} = \varepsilon$, which shows w = x₀ + iy₀ ∈ B(z, ε) ∩ S ∎

• 0 is the only limit point of S = {¹⁄_k + i²⁄_k: k ∈ ℤ ∖{0}} has exactly one limit point, namely 0. (Figure C)

0 is a limit point of S

Consider B(0, ε), ε > 0 an arbitrary small number. Since $\frac{\sqrt{1^2+2^2}}{k} = \frac{\sqrt{5}}{k} \leadsto 0 \text{ as } |k| \leadsto \infin$, there exists a large enough number N that $\forall |k| \gt N, |\frac{1}{k}+i\frac{2}{k}| = \frac{\sqrt{5}}{k} \lt ε$

Hence, for infinitely many k, the point $\frac{1}{k}+i\frac{2}{k}$ lies in B(0; e). Thus, every neighborhood of 0 meets S \ {0}, so 0 is indeed a limit point.

No other point z ∈ ℂ is a limit point.

Fix any z = x + iy ∈ C, not of the form x + 2ix for some x ∈ ℝ, then the perpendicular distance of the point from the line L = {w: ℑw = 2ℜw}, say ε_z is positive and it satisfies B(z, ε_z) ∩ S = ∅, so z is not a limit of S (Figure C).

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 is 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.

Observe that the next‐closest reciprocal to $\frac{1}{k} = x$ is either $\frac{1}{k+1} \text{ or } \frac{1}{k-1}$

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Definition. A subset S ⊂ ℂ 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 center 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 - the straight line joining a and b, including 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 ⊂ ℂ 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 (every points of S can be joined by a polygonal path lying entirely in S): γ = $\bigcap_{i=0}^{n-1} [z_i, z_{i+1}]$ = [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 subset of ℂ is a domain. This means a domain has two important properties:

1. Open: For every point z in the domain, z ∈ D there exists an open disk centered at z (ε-disk around it) 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 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/straight 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 lies entirely inside S.