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

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, $f: D \to \mathbb{C}$.

• Domain D: A subset of $\mathbb{C}$ on which the function is defined. Every $z \in D$ is called a point of the domain.
• Function $f: D \to \mathbb{C}$: A rule that assigns to each $z \in D$ a unique complex number w = f(z).
• Range (or image) f(D): The set of all actual outputs {f(z): z ∈ D}.
• Decomposition into real and imaginary parts: If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts (real‑valued functions of two real variables).

Exterior, Limit, and Isolated Points

Definition. A point $z_0$ is an exterior point of $S \subseteq \mathbb{C}$ if there exists a positive radius $R > 0$ such that the disk $B(z_0; R)$ contains no points of $S$.

Equivalent formulations:

• $\exists R > 0 : B(z_0; R) \cap S = \emptyset$
• $\exists R > 0 : B(z_0; R) \subseteq \mathbb{C} \setminus S$ ($B(z_0; R)$ is entirely contained within the complement of S).

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

Definition: The exterior of $S$, denoted $\text{ext}(S)$, is the set of all exterior points: $\text{ext}(S) = \{z \in \mathbb{C} : \exists R > 0, B(z; R) \cap S = \emptyset\}$.

Key relationship: $\boxed{\text{ext}(S) = \text{int}(\mathbb{C} \setminus S)}$ In words, the exterior of S is the interior of its complement!

Fundamental Principle: Every point $z \in \mathbb{C}$ falls into exactly one of three categories relative to any set $S$:

1. Interior. Some disk around $z$ lies entirely in S, $z \in \text{int}(S)$.
2. Exterior. Some disk around $z$ lies entirely in $\mathbb{C} \setminus S, z \in \text{ext}(S)$.
3. Boundary. Every disk around $z$ meets both S and $\mathbb{C} \setminus S, z \in \partial S$

Partition of the plane: $\mathbb{C} = \text{int}(S) \sqcup \partial S \sqcup \text{ext}(S)$ (where $\sqcup$ denotes disjoint union).

Definition. A point $z_0$ (not necessarily in S) is a limit (accumulation) point of S if every punctured neighborhood of $z_0$ contains at least one point of $S: \forall R > 0: B'(z_0; R) \cap S \neq \emptyset$. Equivalently: every neighborhood of $z_0$ contains a point of $S$ different from $z_0$ itself.

Isolated Point: A point $z_0 \in S$ is an isolated point of $S$ if there exists $R > 0$ such that $B'(z_0; R) \cap S = \emptyset$. In words, $z_0$ is the only point of S in some neighborhood around it — it “stands alone, sad, and lonely like a deserted ship.”

Key Classification Within a Set

Every point $z_0 \in S$ is either:

• A limit point of $S$ (has other points of $S$ arbitrarily close), or
• An isolated point of $S$ (has a neighborhood with no other points of $S$)

These are mutually exclusive!

Fundamental Relationships

• Every interior point is a limit point. Proposition: If $z_0$ is an interior point of $S$, then $z_0$ is a limit point of $S$.

Proof.

Let $z_0$ be an interior point of $S$. An interior point of a set S is a point that has a neighborhood entirely contained in S, $\exists R > 0 : B(z_0; R) \subseteq S$.

For any $\varepsilon > 0$, let $r = \min\{R, \varepsilon\}$.

The disk $B(z_0; r) \subseteq S$ contains infinitely many points (since disks in $\mathbb{C}$ are uncountable).

In particular, $B'(z_0; \varepsilon) \cap S \supseteq B'(z_0; r) \neq \emptyset$.

So every punctured neighborhood of $z_0$ meets $S$, making $z_0$ a limit point. $\blacksquare$

• A limit point does not need to be interior (the converse of the previous fact is false). Counterexample: $S = [0, 1] \subset \mathbb{R} \subset \mathbb{C}$

The point $z_0 = 1$ is a limit point of S. Every neighborhood $B(1; \varepsilon)$ contains points like $1 - \frac{\varepsilon}{2} \in S$

However, $z_0 = 1$ is NOT an interior point of $S$:. Any disk $B(1; \varepsilon)$ contains points like $1 + \frac{\varepsilon}{2} \notin S$. Hence, no disk around $1$ is contained entirely in $S$

• Isolated points are never interior points. Proposition: If $z_0$ is an isolated point of $S$, then $z_0$ is not an interior point of $S$.

Proof:

Suppose $z_0$ is both:

• An isolated point of S. Then, there exists $ \epsilon \gt 0$ such that $\mathbb{B}(z_0; \varepsilon) \cup S = \{ z_0 \}$
• An interior point of S. Then, there exists $\delta \gt 0$ such that the open ball $B(z_0, \delta) = \{ z \in \mathbb{C}: |z - z_0| \lt \delta\}$ is entirely contained in S.

Let $r = min \{ \varepsilon, \delta \}$. From isolation: $\mathbb{B}(z_0; r) \cup S \subseteq \mathbb{B}(z_0; \varepsilon) \cup S = \{ z_0 \}$. From interior, $\mathbb{B}(z_0; r) \subseteq B(z_0, \delta) \subseteq S$. Combining both: $\mathbb{B}(z_0; r) \cup S = \{ z_0 \}$

But, this is absurd because $\mathbb{B}(z_0; r)$. contains infinitely many points (uncountably many, in fact). For example, $z_0 + \frac{r}{2} \in \mathbb{B}(z_0; r), z_0 + \frac{r}{2} \ne z_0$

• Interior Points Are Never Isolated. Proposition: If $z_0$ is an interior point of $S$, then $z_0$ is not an isolated point of $S$.

Proof:

If $z_0$ is interior, then by previous proposition, $z_0$ is a limit point. Limit points and isolated points are mutually exclusive by definition.

Therefore $z_0$ is not isolated. $\blacksquare$

• Boundary Points Are Either Isolated or Limit Points. Proposition: Every boundary point of $S$ is either an isolated point of $S$ or a limit point of $S$, e.g., every boundary point of $S = \{3\} \cup B(0; 1)$ is:
  Either an isolated boundary point: $z = 3$ (no other points of $S$ nearby).
  Or a limit boundary point: Points on circle $|z| = 1$ (infinitely many nearby points of $S$).

• A limit point need not be a boundary point, 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 of Open Sets

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

Let $z_0 = x_0 + iy_0 \in S$ be arbitrary. Then $x_0 > 2$. We need to show that there exists an open disk centered at z₀ that is entirely contained within S, $B(z_0, R) \subseteq S$.

Let $z \in B(z_0, R)$, then we have $|z - z_0| = \sqrt{(x - x_0)^2 + (y - y_0)^2} < R$. Squaring both sides (since both sides are positive): $(x-x_0)^2 + (y -y_0)^2 \lt R^2$. This implies: $(x - x_0)^2 < R^2$

Therefore, $|x - x_0| < R \implies -R < x - x_0 < R \implies x_0 - R < x < x_0 + R$

Let’s choose $R = x_0 - 2$: $x_0 - (x_0 - 2) < x < x_0 + (x_0 -2) \implies 2 < x < 2x_0 -2$. More importantly, we have shown that $\text{Re}(z) = x > 2$, meaning $z \in S$.

Since $z_0$ was arbitrary, S is open. $\blacksquare$

• Lower Half-Plane (Horizontal Strip Variant), S = {z = x + iy : y < 5}

Let $z_0 = x_0 + iy_0 \in S$. Then, $y_0 < 5$. We need to show that there exists an open disk centered at $z_0$ that is entirely contained within S, $B(z_0, R) \subseteq S$.

Let $z_0 \in B(z_0, R)$. Then, $\sqrt{(x - x_0)^2 + (y - y_0)^2} < R$. This implies, $|y - y_0| < R$ (same argument as we did previously), $-R < y - y_0 < R \implies y_0 - R < y < y_0 + R$

Let’s choose $R = 5 - y_0 > 0$. Then, $y_0 -(5 - y_0) \lt y \lt y_0 + (5 - y_0) \implies 2y_0 - 5 \lt y \lt 5$. More importantly, $\text{Im}(z) = y < 5$, meaning $z \in S$.

Since $z_0$ was arbitrary, S is open. $\blacksquare$

• Vertical Strip are open sets. $S = \{z : x_1 < \text{Re}(z) < x_2\}$

Let $z_0 = x_0 + iy_0 \in S$. Then $x_1 < x_0 < x_2$.

Choose radius: $R = \min(x_0 - x_1, x_2 - x_0) > 0$. This is the smaller distance from $z_0$ to either boundary.

We aim to show that $D_R(z_0) \subseteq S$

Let $z = x + iy \in D_R(z_0)$. Then, $|z - z_0| < R$.

Since $|x - x_0| \leq |z - z_0| < R$: $-R < x - x_0 < R$

$x > x_0 - R \geq[R = \min(x_0 - x_1, x_2 - x_0), R \leq x_0 - x_1] x_0 - (x_0 - x_1) = x_1$. $x < x_0 + R \leq[R = \min(x_0 - x_1, x_2 - x_0), R \leq x_2 - x_0] x_0 + (x_2 - x_0) = x_2$

So $x_1 < x < x_2$, meaning $z \in S$. $\blacksquare$

• Horizontal Strip are open sets. $S = \{z : y_1 < \text{Im}(z) < y_2\}$

Proof: Completely analogous, but with radius: $R = \min(y_0 - y_1, y_2 - y_0)$

• Rectangular Region are open sets. $S = \{z : x_1 < \text{Re}(z) < x_2 \text{ and } y_1 < \text{Im}(z) < y_2\}$

Proof:

$S$ is the intersection of two open strips: $S = \{x_1 < \text{Re}(z) < x_2\} \cap \{y_1 < \text{Im}(z) < y_2\}$

Since finite intersections of open sets are open, $S$ is open.

• Open annulus are open sets. 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.

Proof:

Write $A$ as an intersection: $A = \{|z - a| > r_1\} \cap \{|z - a| < r_2\}$ where $\{|z - a| < r_2\}$ is an open disk — open; and $\{|z - a| > r_1\}$ is the exterior of a closed disk, which is also open.

Finite intersection of open sets is open. $\blacksquare$

• Complement of a closed set. If $S$ is closed, then $\mathbb{C} \setminus S$ is open, e.g., the punctured plane $\mathbb{C} \setminus \{0\}$ is open.

Closed ⇒ Complement open: Assume S is closed. Let $z \in \mathbb{C} \setminus S$. Since $z \notin S$ and S contains all its limit points, z is not a limit point of S. Thus, there exists $\varepsilon \gt 0$ such that the open ball $\mathcal{B}(z, \varepsilon)$ contains no point of S, i.e., $\mathcal{B}(z, \varepsilon) \subseteq \mathbb{C} \setminus S$. Hence, every point of the complement is an interior point, so $\mathbb{C} \setminus S$ is open.

Complement open ⇒ Closed: Assume $\mathbb{C} \setminus S$ is open. Let z be a limit point of S. If $z \notin S$, then $z \in \mathbb{C} \setminus S$, which is open, so there exists $\varepsilon \gt 0$ such that $\mathcal{B}(z, \varepsilon) \subseteq \mathbb{C} \setminus S$. But this is a contradiction, $\mathcal{B}(z, \varepsilon)$ contains no points of S and yet z is supposedly a limit point. Therefore, $z \in S$, so S contains all its limit point, hence S is closed.

Note: $\{0\}$ is closed (finite sets are closed). Therefore, $\mathbb{C} \setminus \{0\}$ is open by the open-closed duality theorem. $\blacksquare$

Punctured Neighborhoods

Definition. A punctured (or deleted) neighborhood of a point a $\in \mathbb{C}$ (or more generally, in any topological space) is an open set surrounding a, but with the center point removed, $B'(a; r) = \{z \in \mathbb{C} : 0 < |z - a| < r\} = B(a; r) \setminus \{a\}$

This is the shell around a — all points close to a but not a itself. It formalizes the idea of arbitrarily close but distinct.

Properties

1. $B'(a; r)$ is open. Every point has room to wiggle.
2. It can be described as an annulus $B'(a; r) = A(a; 0, r)$ with inner radius 0.
3. $\overline{\mathbb{B}'(a; r)} = \overline{\mathbb{B}(a; r)}$. The closure “heals” the puncture; the limit point a returns.
4. The puncture is not a bug. In complex analysis, we often need to consider behavior near a point (local behaviour) without including the point itself: limits ($\lim_{z \to a} f(z)$ considers $z \neq a$, f(a) itself is irrelevant or may not even exist), limit points (a is a limit point of S if every $\overline{\mathbb{B}'(a; r)} \cap S \ne \emptyset$; it captures accumulation, points of S “pile up” near a without requiring $a \in S$), isolated singularities ($f$ analytic on $B'(a; r)$ but undefined or problematic at $a$; The function is “nice” nearby but singular at the center; classification -removable, pole, essential- depends on behavior in punctured neighborhoods), and residues (integrate around a, not thorough it).

Examples of Non-Interior Points

• The Rational Grid, $S = \{x + iy : x, y \in \mathbb{Q}\}$, $\text{int}(S) = \emptyset$ — no point of $S$ is interior.

Proof:

Let $z_0 = a + ib \in S$ where $a, b \in \mathbb{Q}$ and let $r > 0$ be an arbitrary radius.

1. Because the irrational numbers are dense in $\mathbb{R}$, there exists an irrational number δ such that 0 < δ < r.
2. Define the point w = (a+δ) + ib.
3. Check distance: $|w - z_0| = |(a + \delta) + ib -a -ib| = |\delta| = \delta \lt r \implies w \in \mathbb{B}(z_0; r)$.
4. The real part of w is a + δ. Since $a \in \mathbb{Q}$ and $\delta \notin \mathbb{Q}$, their sum is irrational, therefore $w \notin S$.
5. Since we found a point in the disk that is not in S, the disk $\mathbb{B}(z_0; r)$ is not (fully) contained in S.
6. Since this holds for any r > 0, $z_0$ is not an interior point. Since $z_0$ was arbitrary, $\text{int}(S) = \emptyset, \blacksquare$

• The Real Line in $\mathbb{C}$, $S = \mathbb{R} = \{x + 0i : x \in \mathbb{R}\}$, $\text{int}(S) = \emptyset$ (as a subset of $\mathbb{C}$).

Proof:

For any $x_0 \in \mathbb{R}$ and any $r > 0$, the point $w = x_0 + \frac{ir}{2}$ satisfies $|w - x_0| = \frac{r}{2} < r$.

So $w \in B(x_0; r)$, but $w \notin \mathbb{R}$.

No disk in $\mathbb{C}$ around any real number is contained in $\mathbb{R}$. $\blacksquare$

Relationships Between Point Types

If $z_0$ is…	Then $z_0$ is…	Then $z_0$ is NOT…
Interior	Always a limit point	Isolated, exterior
Exterior	Not in $S$	Interior, boundary (of $S$)
Isolated	In $S$, boundary	Interior, limit point of $S$
Limit point	—	Exterior
Boundary	Either isolated or limit	Interior, exterior