Domains, Interior & Exterior Points, and Open Sets in ℂ

Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost, W.S. Anglin.

Introduction

This article covers:

1. Complex functions: what their domains and ranges are, and how we decompose (f(z)=u(x,y)+iv(x,y)).
2. Interior points —points around which we can draw a small disk completely inside the set.
3. Exterior points —points around which we can draw a small disk that avoids the set entirely.
4. Open sets —a set where every point within it is also an interior point.
5. A gallery of examples and counterexamples: half‐planes, strips, disks, annuli, rational‐part sets, and more.
6. Key propositions and proofs: disks are open, intersections and unions of open sets stay open, and the reverse triangle inequality in ℂ.

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.

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.

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.

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.

Exterior Points

Definition. A point z₀ is an exterior point of S ⊆ ℂ if ∃R > 0 (meaning, there exist a positive radius) such that D_R(z₀) contains no points of S ↭ ∃ε > 0 s.t. B(z; ε) ∩ S = ∅. In other words, D_R(z₀) is entirely contained within the complement of S (ℂ - S), D_R(z₀) ⊆ ℂ - S.

Put simply, you can draw an open disk around z₀ that does not meet S at all (avoids S entirely) —i.e., it lies entirely in the complement of S.

Note: Interior points lie inside S, exterior points lie outside S, and the remaining points —those neither interior nor exterior —are precisely the boundary points of S.

Facts

1. Every interior point is a limit point, but every limit point need not be an interior point.

   An interior point of a set S is a point that has a neighborhood entirely contained in S. Since every neighborhood of an interior point contains infinitely many points of S, it is also a limit point of S. However, a limit point of S is a point where every neighborhood contains at least one point of S other than itself, (it doesn’t guarantee you can fit a whole disk inside S). Example [0,1] ⊂ ℝ. The point 1 is a limit point (every ε-neighborhood contains points of [0, 1] other than 1) but 1 is not interior, since any small disk around 1 extends past 1 into (1, 1+δ) ⊄ [0, 1].

2. Isolated points are never interior points and interior points are never isolated points.

   An isolated point of a set S is a point x∈S such that there exists a neighborhood of x containing no other points of S. By definition, an isolated point cannot be an interior point because it lacks a neighborhood entirely contained in S. Conversely, an interior point has a neighborhood entirely contained in S, meaning it cannot be isolated (it has infinitely many nearby points in S).

3. Every boundary point of a set S (e.g., S = {3} ∪ B(0, 1)) is an isolated point (e.g., {3} itself has no nearby siblings in S) or a limit point (every point of the circle is both a limit point and a boundary point {z: |z| = 1} - every disk around such a point meets infinitely many other points of S).

4. A limit point need not lie on the boundary point.

   A point x (not necessarily in S) is a limit point of S if every neighborhood of x contains at least one point of S different from x itself. A point x is a boundary point of S if every neighborhood of x contains at least one point of S and at least one point not in S.

   However, a limit point does not have to lie on the boundary of S, e.g., every point in S = (0, 1) is a limit point because any open interval around a point in (0, 1) contains other points from S. However, none of them are boundary points because you can find neighborhoods around them that don’t intersect the complement $\mathbb{R} \setminus (0, 1).$

   The actual boundary points of (0,1) are 0 and 1 — even though they are not in S, every neighborhood around them contains points both in and out of S.

Illustrative Examples

• Half-plane, S = {z = x + iy | x > 2} is an open set.

We need to show that for every point z₀ in S, there exists an open disk centered at z₀ that is entirely contained within S.

Let z₀ be an arbitrary point in S. This means that z₀ can be written as z₀ = x₀ + iy₀, where x₀ > 2.

Choose a radius R. We need to find a radius R > 0 such that the open disk centered at z₀ with radius R, denoted by D_R(z₀) = {z ∈ ℂ | |z - z₀| < R} ⊆ S

Let’s choose R = x₀ - 2. Since x₀ > 2, R is strictly greater than 0, satisfying the condition for a radius.

We aim to prove that D_R(z₀) ⊆ S. Let z be an arbitrary point in D_R(z₀). This means |z - z₀| < R. We want to show that z is also in S.

Let z = x + iy. Then: |z - z₀| = |(x + iy) - (x₀ + iy₀)| = |(x - x₀) + i(y - y₀)| = $\sqrt{(x - x₀)² + (y - y₀)²}$

Since |z - z₀| < R, we have: $\sqrt{(x - x₀)² + (y - y₀)²} < R$

Squaring both sides (since both sides are positive): (x - x₀)² + (y - y₀)² < R²

Because squares are non-negative, we know that: (x - x₀)² < R², and taking the square root of both sides: |x - x₀| < R ↔ -R < x - x₀ < R ↔[Adding x₀ to both sides:] x₀ - R < x < x₀ + R ↔[Now, substitute our choice of R = x₀ - 2:] x₀ - (x₀ - 2) < x < x₀ + (x₀ - 2) ↔ 2 < x < 2x₀ - 2. More importantly, we have shown that x > 2, so z ∈ S

Therefore, ∀z ∈ D_R(z₀) ⇒ z ∈ S, hence D_R(z₀) ⊆ S. Since z₀ was an arbitrary point in S, we have shown that for every point in S, there exists an open disk centered at that point that is entirely contained within S. This is the definition of an open set. Therefore, S is an open set ∎

• Horizontal strip, S = {z = x + iy : y < 5}

Given an arbitrary z₀ with ℑ(z₀ = y₀) < 5, set R = 5 − y₀ > 0. Then, ∀z ∈ D_R(z₀), |z - z₀| < R forces $\sqrt{(x - x₀)² + (y - y₀)²} < R \leadsto (x - x₀)² + (y - y₀)² < R² \leadsto (y - y₀)² < R² \leadsto |y - y₀| < R ↔ -R < y - y₀ < R ↭ y₀ - (5 - y_0) < y < y_0 + (5 - y_0) ↔ 2y_0 - 5 < y < 5, ℑ(z) < 5, $ so z ∈ S ∎

• Open disk. D_r(a) = {z : ∣z−a∣ < r}. Each arbitrary point z₀ of D_r(a), ∣z₀−a∣ < r, set R = r - ∣z₀ − a∣ > 0.

Then, ∀z ∈ D_R(z₀), |z - z₀| < R $\Rightarrow |z - a| \leq |z - z_0| + |z_0 - a| < R + |z_0 - a| = r$ ∎ Thus, $z \in D_r(a)$, which shows that an open disk contains an open neighborhood around each of its points—proving it’s open!

• Strips are open sets {z : x₁ < ℜ(z) < x₂}, {z : y₁ < ℑ(z) < y₂}. At any z₀ = x₀ + iy₀, $d = \min(x_2 - x_0, x_0 - x_1) > 0$. Choosing R₁ < d, we can center a small open disk $D_R(z_0)$ that stays entirely inside the vertical strip. The same logic applies for horizontal strips using imaginary parts, R₂ < d’. Set R the smaller distance to either boundary, R = {R₁, R₂}

The open disk $D_R(z_0) = \{ z \in \mathbb{C} : |z - z_0| < R \}$ is fully contained in the rectangular region $S_V \cap S_H$. This proves that every point in the strip lies inside an open disk fully contained in the strip —meeting the formal definition of an open set in $\mathbb{C}$.

• Open annulus. An open annulus centered at a $\in \mathbb{C}$ is defined as: A = $\{ z \in \mathbb{C} : r_1 < |z - a| < r_2 \}$. This is the set of points strictly between two concentric circles —like a ring without its edges.

{z : r₁ < ∣z − a∣ < r₂} = {∣z − a∣ > r₁} ∩ {∣z − a∣ < r₂} is open because is the intersection of two open sets, namely:
{∣z − a∣ > r₁} is an open disk—open in $\mathbb{C}$
{∣z − a∣ < r₂} is the exterior of a closed disk, which is also open

• Complement of a closed set. If S ⊂ ℂ is closed, then ℂ \ S is open. For instance, ℂ \ {0} = { z ≠ 0} is open (a punctured plane).

This follows directly from the definitions of open and closed sets in topology. A set is closed if it contains all its limit points (i.e. its complement is open). {0} is closed because it is a finite set, and finite sets are closed in ℂ. Therefore, its complement —the punctured plane— is open

• Infinite union. $\cup_{n \in ℕ} D_{1/n}(a)$ is open, even though it’s not a single disk —it’s a “tail” of ever-shrinking ones. This is because infinite unions of open disks are open.

Specifically: $D_{1}(a), D_{1/2}(a), D_{1/3}(a), ...$. Each disk is open, and each one is contained in the next.

Definition. A punctured neighborhood of a point a $\in \mathbb{C}$ (or more generally, in any topological space) is an open set surrounding a, but with the point a itself removed. It captures everything near a except a itself.

Formally, a punctured neighborhood of a is a set of the form: { z $\in \mathbb{C} : 0 < |z - a| < r $} = $D_r(a) \setminus a$.

• Punctured plane ℂ \ {0} is open because {0} is closed.
• Rational grid {x + iy : x, y ∈ ℚ} has no interior points (every disk picks up irrationals).