Complex Functions and Open Sets: Domains, Interior Points, and Topology in ℂ

Uncertainty is the only certainty there is, and knowing how to live with insecurity is the only security, John Allen Paulos

Introduction

This article will

1. Define what a complex-valued function is, its domain and range.
2. Introduce interior points and the formal notion of an open set in ℂ.
3. Work through examples and counterexamples to build intuition and real understanding.
4. Prove the fundamental facts that

• Disks B(a; r) are open,
• Intersections and arbitrary unions of open sets are open.
• Common “regions” (half‐planes, annuli, exteriors) are open.

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.

Interior Points

Definition. A point z₀∈ S is an interior point of a subset S ⊆ ℂ if ∃R > 0 (meaning, there exist a positive radius R) such that D_R(z₀) = {z ∈ ℂ : ∣z − z₀∣ < R} ⊆ S.

In words, there is an open disc around z₀​ that lies entirely within S (a neighborhood exists fully within the set) - you can wiggle your point z₀ a little in any direction without ever leaving the set.

Open Sets

Definition. A subset S ⊆ ℂ is open if every point of S is an interior point. Formally, ∀z₀ ∈ S, ∃R > 0 : (such that) D_R(z₀) ⊆ S -the ball D_R(z₀) sits entirely inside S. In other words, an open set contains all their interior points, i.e., every point has a neighborhood entirely within the set.

Alternative definition. A set A ⊆ ℂ is open if for any arbitrary element of the set ∀z ∈ A, there is a radius ∃r > 0 (possibly depending on z) such that an open disk or ball of radius r centered at it is completely contained in the set, B(z; r) ={w : ∣w − z∣ < r} ⊆ A.

Facts, Examples and counterexamples:

1. Every point of an open set is an interior point.
2. S = {z + iy ∈ ℂ: x, y ∈ ℚ}, S is the set of complex numbers whose real and imaginary parts are both rational numbers. No point of S is interior: any disc B(z₀, r) around any point $z_0 \in S, \text{ say } z_0 = a + ib$ contains points where the real or imaginary part (or both) are irrational — and therefore not in S. So every open disc around any point in S will contain points not in S, meaning no point in S is an interior point.
3. B(2+2i; 1) ∪ {x + 0i: x ∈ ℝ} is a set containing both, points in the open disk B(2+2i; 1) that are interior and real-axis points x + 0i are not interior (any disc around them picks up non-real points outside the vertical line).
4. The empty set $\emptyset$ is vacuously open - there is no point to check!
5. The complex set ℂ is open because around every z ∈ ℂ you can fit some disc D_R ⊂ ℂ.
6. B(a; r) is an open set.
7. Given two open sets A₁ and A₂ in the complex plane, their intersection A₁ ∩ A₂ is an open set (Figure 6).
8. Unions (arbitrary) of open sets remain open. Given {S_α}_α∈A an arbitrary collection of open sets in ℂ, then $\cup_{\alpha ∈ A} S_{\alpha}$ is also an open set.
9. The exterior of a closed disk is an open set. {z∈ ℂ: |z -a| > r} is an open set, too.
10. An annulus, the set of all complex numbers whose distance from a lies strictly between r₁ and r₂, is an open set.
11. $\Pi$ = {z ∈ ℂ: Im(z) > 0} is an open set.
12. The Closed Disk Is Not Open. $\overline{B(a; r)}$ = { z ∈ ℂ: |z -a| ≤ r} is not an open set. A natural candidate is a point on the boundary of the disk, $z \in \overline{B(a; r)}$ such that |z -a| = r, then there is no δ > 0 such that $B(z, δ) ⊆ \overline{B(a; r)}$

Concretely, consider w = z + δ/2⋅u, where u is the unit vector $u = \frac{z-a}{|z-a|} = \frac{z-a}{r}$, then

$|w - a| = |z + δ/2⋅\frac{z-a}{r} -a| = |z - a + δ/2⋅\frac{z-a}{r}| = |(1+ \frac{\delta}{2r})⋅(z-a)| = |(1+ \frac{\delta}{2r})|⋅|(z-a)| = (1+ \frac{\delta}{2r})·r = r + \frac{\delta}{2} >[\text{Since δ > 0}] r \leadsto w \notin \overline{B(a; r)}$

Proposition. B(a; r) is an open set.

Proof (Figure 5).

∀z ∈ B(a; r), |z - a| < r ↭ 0 < r - |z -a|. Let δ be such that 0 < δ < r - |z -a|, we aim to demonstrate that B(z, δ) ⊆ B(a; r).

∀w ∈ B(z, δ) ⇒ |w -z| < δ ⇒ |w -z + z -a| ≤ |w -z| + |z -a| < δ + |z -a| < r ⇒ z ∈ B(z, r) ∎

Proposition. A₁ ∩ A₂ is an open set.

Let A₁ and A₂​ be open sets in ℂ. By definition, for every z∈A₁, there exists a radius r₁​ > 0 such that B(z; r₁) ⊆ A:. Similarly, for every z∈A₂​, there exists a radius r₂​ > 0 such that B(z; r₂) ⊆ A₂.

Let z∈A₁∩A₂​. Since z∈A₁​, there exists r₁ > 0 such that B(z; r₁) ⊆ A₁​. Similarly, since z∈A₂​, there exists r₂ > 0 such that B(z; r₂) ⊆ A₂

Let r be the smaller of the two radii r₁​ and r₂, r = min{r₁​, r₂}. Then: B(z; r) ⊆ B(z;r₁) ⊆ A₁ and B(z;r) ⊆ B(z; r₂) ⊆ A₂. Since B(z; r) is contained in both A₁​ and A₂​, it is also contained in their intersection: B(z; r) ⊆ B(z; r₁) ∩ B(z; r₂) ∎

Proposition. Unions (arbitrary) of open sets remain open. Let {S_α}_α∈A be an arbitrary collection of open sets in ℂ, then $\cup_{\alpha ∈ A} S_{\alpha}$ is also an open set.

Proof.

Suppose $z \in \cup_{\alpha ∈ A} S_{\alpha} \leadsto z \in S_{\alpha}$ for some α ∈ A. Since S_α is an open set, there is an r > 0 such that B(z; r) ⊆ S_α ⊆ $\cup_{\alpha ∈ A} S_{\alpha}$ ∎

The reverse triangle inequality is a neat little companion to the classic triangle inequality — but instead of telling us how big the sum can get, it tells us how small the difference can be.

Lemma. For any complex numbers (or vectors) x and y, the reverse triangle inequality states: $\left|~|x| - |y|\~\right| \leq |x - y|$

Proof:

|x| + |y − x| ≥ |x + y − x| = |y|

|y| + |x − y| ≥ |y + x −y| = |x|

Move |x| to the right hand side in the first inequality and |y| to the right hand side in the second inequality, |y − x| ≥ |y| − |x| and |x − y| ≥ |x| − |y|.

From absolute value properties, we know that |y − x| = |x − y|, and if t ≥ a and t ≥ −a, then then t ≥ |a|.

Combining these two facts together, we get the reverse triangle inequality: |x − y| ≥ ||x| − |y||∎

Proposition. {z∈ ℂ: |z -a| > r} is an open set.

Proof.

Pick an arbitrary point z₀ ∈ {z∈ ℂ: |z -a| > r}, |z₀ -a| > r

Set d = |z₀ -a| - r, and consider the open disk D(z₀, d). Take any z in the open disk D(z₀, d), that is, |z -z₀| < d, claim: z ∈ {z∈ ℂ: |z -a| > r} ↭ |z -a| > r

|z -a| = ∣(z − z₀) + (z₀ −a)∣ ≥[Apply the reverse triangle inequality] ||z₀ −a| − |z − z₀|| > ||z₀ −a| − d| = |r| =[r > 0] r, |z -a| > r ⇒ D(z₀, d) ⊆ S , proving S is open.∎

Proposition. An annulus, the set of all complex numbers whose distance from a lies strictly between r₁ and r₂, is an open set.

Proof.

An annulus is an open set because it is the intersection of two open sets: {z ∈ ℂ: |z - a| > r₁} ∩ {z ∈ ℂ: |z - a| < r₂} where r₁ < r₂ = {z ∈ ℂ: r₁ < |z - a| < r₂} (Figure 4).

1. {z ∈ ℂ: |z - a| > r₁} is the exterior of a closed disc — The exterior of a closed disk is an open set.
2. {z ∈ ℂ: |z - a| < r₂} is the interior of the larger disc — also open.
3. The finite intersection of open sets is open, so the annulus is open too ∎

$\Pi$ = {z ∈ ℂ: ℑ(z) > 0} is an open set.

Proof.

Take any z = x + iy $\in \Pi$. By definition, ℑ(z) > 0. Set δ = ℑ(z) = y, which is strictly positive, we claim B(z, δ) $\subseteq \Pi$.

Let w be an arbitrary element w ∈ B(z, δ), |w - z| < δ. Write w = u + iv, so:

$|w - z| = |(u - x) + i(v - y)| = \sqrt{(u - x)^2 + (v - y)^2} < \delta = y$

In particular, the imaginary part of w, v satisfies: $|v - y| < y \Rightarrow -y < v - y < y \Rightarrow 0 < v < 2y \Rightarrow \operatorname{Im}(w) = v > 0 \Rightarrow w \in \Pi$

Therefore, every point $z \in \Pi$ has a neighborhood (specifically, an open ball) entirely contained in $B(z, δ) \subseteq \Pi$, which proves that $\Pi$ is open.

Intuition

• Disks are round “bubbles.” Every point inside has a little wiggle room: you can move a tiny bit and stay within the disk. No circle or boundary line is included —just the interior.
• When two disks overlap, their intersection is also open. The intersection of two overlapping bubbles contains the “lens” region (an almond shape region). Around each point in that lens you can still fit a small disk fully inside the overlap.
• You can take any collection —finite or infinite— of disks and form their union. Unions stitch together many bubbles into a bigger, possibly a tangled, irregular shape—yet it remains open. That’s due to the fact that every point in the union belongs to at least one disk, so there’s always a tiny neighborhood around it contained in the union.
• By definition, an open set never contains its boundary. Boundaries —circles, lines, arcs— are not part of an open set. Even though the circle “closes” the bubble, those exact points sit outside the open disk.