Interior Points and Open Sets

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

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).

Interior Points

Definition. A point $z_0 \in S$ is an interior point of a subset $S \subseteq \mathbb{C}$ if there exists a positive radius $R > 0$ such that: $D_R(z_0) = \{z \in \mathbb{C} : |z - z_0| < R\} \subseteq S$

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 $z_0$ a little in any direction without ever leaving the set.

Definition: The interior of a set S, denoted $\text{int}(S)$ or $S^\circ$, is the set of all interior points of $S$: $\text{int}(S) = \{z \in S : \exists R > 0, B(z; R) \subseteq S\}$

Set $S$	Point $z_0$	Interior Point?	Reason
$B(0; 1)$ (open disk)	$0.5$	Yes	Any small disk around $0.5$ stays inside
$B(0; 1)$	$0.99$	Yes	Can still fit a tiny disk (radius $< 0.01$)
$\overline{B(0; 1)}$ (closed disk)	$1$	No	Any disk around $1$ extends outside
$\mathbb{R} \subset \mathbb{C}$ (real axis)	$0$	No	Any disk $B(0; \varepsilon)$ contains non-real points (e.g., $i\frac{\varepsilon}{2}$)
$\mathbb{C}$ (entire plane)	Any $z$	Yes	Any disk stays in $\mathbb{C}$
$\{0\}$ (singleton)	$0$	No	No disk around $0$ is contained in $\{0\}$
$\{1, i, −1, −i\}$ (discrete set)	1	No	$B(1; \varepsilon)$ contains points not in the set
$\{z : \mathcal{Re}(z) \gt 0 \}$ (discrete set)	1 + i	Yes	$B(1 + i; 0.5) \subseteq \{z : \mathcal{Re}(z) \gt 0 \}$

Key Properties:

1. Subset relation: $\text{int}(S) \subseteq S$.
2. Largest open subset: $\text{int}(S)$ is the largest open set contained in S.
3. Idempotence, $\text{int}(\text{int}(S)) = \text{int}(S)$.
4. Monotonicity. If $S \subseteq T$, then $\text{int}(S) \subseteq \text{int}(T)$.

Open Sets

Definition. A subset $S \subseteq \mathbb{C}$ is open if every point of S is an interior point. Formally, $\forall z_0 \in S, \exists R > 0 : B(z_0; R) \subseteq S$ -the ball $B(z_0; R)$ sits entirely inside S. In words, an open set contains all its interior points, $S = \text{int}(S)$, i.e., every point has a neighborhood entirely within the set.

Alternative Characterizations

• Interior equals set, S is open $\iff S = \text{int}(S)$.
• No boundary points, S is open $\iff S \cap \partial S = \emptyset$
• Complement closed, S is open $\iff \mathbb{C} \setminus S$ is closed.

Intuition: What Makes a Set Open?

We could think of open sets as having these four properties:

1. No edges included: The boundary is never part of an open set.
2. Wiggle room everywhere: Every point has space to move.
3. Like open intervals: Just as $(a, b)$ excludes endpoints, open sets exclude their boundaries.
4. Round bubbles: Open disks are the prototypical examples.

Facts, Examples and counterexamples

1. The empty set $\emptyset$ is vacuously open - there is no point to check!
2. $\mathbb{C}$ is an open set. For any $z \in \mathbb{C}$, take any $R > 0$ (say $R = 1$). Then, $B(z; R) = B(z; 1) \subseteq \mathbb{C}$. Every point has a neighborhood in $\mathbb{C}$ $\blacksquare$.
3. Open disks are open, $B(a; r) = \{z \in \mathbb{C} : |z - a| < r\}$ is an open set.
4. Closed disks are not open. $\overline{B(a; r)} = \{z \in \mathbb{C} : |z - a| \leq r\}$ is not an open set.
5. Every point of an open set is an interior point.
   This is just restating the definition: if S is open, then by definition every point of S has a small disk fully contained in S; that is, every point is interior.
6. $S = \{x + iy \in \mathbb{C} : x, y \in \mathbb{Q} \}$, S is the set of complex numbers whose real and imaginary parts are both rational numbers. No point of S is interior.
   Take any point $z_0 = a + ib \in S$ and any radius r > 0. The open disk $B(z_0; r)$ contains uncountably many points where the real or imaginary part (or both) are irrational, hence not in S. Thus, $B(z_0; r) \nsubseteq S \quad \forall r > 0$. So no point in S has a neighborhood fully contained in S; therefore, $\operatorname{int}(S) = \emptyset$, and S is not open.
   

   For any $r > 0$, the disk $B(z_0; r)$ contains the point: $w = \left(a + \frac{r}{2\sqrt{2}}\right) + i\left(b + \frac{r}{2\sqrt{2}}\right)$. Since $\frac{r}{2\sqrt{2}}$ is irrational (for $r \neq 0$), at least one of $\text{Re}(w)$ or $\text{Im}(w)$ is irrational ($w \notin S$, but $w \in B(z_0; r)$).

7. A mixed example: open disk + real axis. Consider $S = B(2 + 2i; 1) \cup \{x + 0i : x \in \mathbb{R}\}$. Points in the open disk B(2 + 2i; 1) are interior (there is a smaller disk entirely inside B(2 + 2i; 1), hence inside S).
   Points on the real axis (e.g., 0, 1, 2, etc.) are not interior: any disk around such points contains non-real points with $\operatorname{Im}(z) \neq 0$, and those points are not in S (except maybe if they happen to fall inside the separate disk around 2 + 2i). For points far from 2 +2i, their neighborhoods leave S immediately. Therefore, S is not open because some of its points are not interior.
8. Intersection of two open sets is open.
9. Arbitrary unions of open sets are open.
10. The exterior of a closed disk is an open set.
11. An annulus centered at a with inner radius $r_1$ and outer radius $r_2$ is the region (ring) between two concentric circles. $A(a; r_1, r_2) = \{z \in \mathbb{C} : r_1 < |z - a| < r_2 \}.$ The annulus $A(a; r_1, r_2)$ is open.
12. The Upper Half-Plane is Open. $\mathbb{H} = \{z \in \mathbb{C} : \text{Im}(z) > 0\}$ is an open set.

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

Proof (Figure 5).

Let $z \in B(a; r)$ be arbitrary. Then $|z - a| < r$.

Define $\delta = r - |z - a|$. Since $|z - a| < r$, we have $\delta > 0$.

Claim: $B(z; \delta) \subseteq B(a; r)$

Let $w \in B(z; \delta)$. Then $|w - z| < \delta$.

By the triangle inequality: $|w - a| = |(w - z) + (z - a)| \leq |w - z| + |z - a| < \delta + |z - a|$

$$= (r - |z - a|) + |z - a| = r$$

So $|w - a| < r$, meaning $w \in B(a; r)$.

Hence, $B(z; \delta) \subseteq B(a; r)$, proving $z$ is an interior point. Since $z$ was arbitrary, $B(a; r)$ is open $\blacksquare$.

Proposition: Closed disks are not open. $\overline{B(a; r)} = \{z \in \mathbb{C} : |z - a| \leq r\}$ is not an open set.

Proof:

Consider a boundary point $z$ with $|z - a| = r$. We aim to show that $z$ is not an interior point.

For any $\delta > 0$, consider the point: $w = z + \frac{\delta}{2} \cdot \frac{z - a}{|z - a|} = z + \frac{\delta}{2} \cdot \frac{z - a}{r}$

This point $w$ lies on the ray from $a$ through $z$, at distance $\frac{\delta}{2}$ beyond $z$.

Let’s compute the distance from $w$ to $a$: $|w - a| = \left|z - a + \frac{\delta}{2} \cdot \frac{z - a}{r}\right| = \left|\left(1 + \frac{\delta}{2r}\right)(z - a)\right|$

$= \left(1 + \frac{\delta}{2r}\right)|z - a| = \left(1 + \frac{\delta}{2r}\right) \cdot r = r + \frac{\delta}{2} > r$

So $w \notin \overline{B(a; r)}$, but $w \in B(z; \delta)$.

This means no disk around $z$ is contained entirely in $\overline{B(a; r)}$, so $z$ is not interior. Therefore $\overline{B(a; r)}$ is not open. $\blacksquare$

Proposition. The exterior of a closed disk is an open set. $\{z \in \mathbb{C} : |z - a| > r\}$ is an open set.

Proof:

Take any point $z_0 \in S = \{z \in \mathbb{C} : |z - a| > r\}$, so $|z_0 - a| > r$. Set $d = |z_0 - a| - r > 0.$

Consider the open disk $B(z_0; d)$. Take any $z \in B(z_0; d)$, so $|z - z_0| < d$. Then, by the reverse triangle inequality:

$|z - a| = |(z - z_0) + (z_0 - a)| \geq \big||z_0 - a| - |z - z_0|\big| > |z_0 - a| - d = r.$

So |z - a| > r, i.e., $z \in S$, hence: $B(z_0; d) \subseteq S.$ Thus, every point of S is interior, and S is open. $\blacksquare$.

Proposition. An annulus $A(a; r_1, r_2)$, 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: $A(a; r_1, r_2) = \{z : |z - a| > r_1\} \cap \{z : |z - a| < r_2\}$ (Figure 4).

1. $\{z \in \mathbb{C} : |z - a| > r_1\}$ is the exterior of a closed disc — The exterior of a closed disk is an open set.
2. $\{z \in \mathbb{C} : |z - a| < r_2\}$ is the interior of the larger disc — also open.
3. The finite intersection of open sets is open, so the annulus is open, too $\blacksquare$.

Proposition. The upper half-plane $\mathbb{H} = \{z \in \mathbb{C} : \text{Im}(z) > 0\}$ is an open set.

Proof:

Take any $z = x + iy \in \Pi$. Then, $Im(z) = y > 0$. Let $\delta = y/2 > 0$. Consider $B(z; \delta)$.

Take any $w = u + iv \in B(z; \delta)$, so: $|w - z| = |(u - x) + i(v - y)| < \delta = \frac{y}{2}$.

Then, in particular: $|v - y| \le |w - z| < \frac{y}{2} \quad \Rightarrow \quad -\frac{y}{2} < v - y < \frac{y}{2}$

$\Rightarrow \quad \frac{y}{2} < v < \frac{3y}{2} \quad \Rightarrow \quad v > 0.$

So $\Im(w) = v > 0$, hence $w \in \Pi$. Therefore, $B(z; \delta) \subseteq \Pi.$

Since every point of $\Pi$ is interior, $\Pi$ is open $\blacksquare$.

The reverse triangle inequality is a neat little companion to the classic triangle inequality (for any real numbers x and y, $|x + y| \le |x|+|y|$), but instead of telling us how big the sum can get, it tells us how small the difference can be.

Lemma (Reverse Triangle Inequality): For any complex numbers $x$ and $y$: $\boxed{\big||x| - |y|\big| \leq |x - y|}$. The difference of magnitudes is at most the magnitude of the difference.

Proof:

1. From the triangle inequality: $|x| = |(x - y) + y| \leq |x - y| + |y|$. Rearranging: $|x| - |y| \leq |x - y|$
2. Similarly: $|y| = |(y - x) + x| \leq |y - x| + |x| = |x - y| + |x|$. Rearranging: $|y| - |x| \leq |x - y|$
3. So both inequalities hold: $|x| - |y| \le |x - y|, \quad |y| - |x| \le |x - y|.$ Combine using absolute value: if $t \ge a$ and $t \ge -a$, then $t \ge |a|$. Here t = |x -y| and a = |x| - |y|, so: $|x - y| \ge \big||x| - |y|\big|$ $\blacksquare$.

Useful Corollary

1. $|x| - |y| \leq |x - y|$
2. $|y| - |x| \leq |x - y|$

Intuition

• An interior point has wiggle room in all directions.
• Disks are round “bubbles.” Every point inside has a little wiggle room: you can move a tiny bit in any direction and stay within the disk. The boundary circle is not included in the open disk —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 intersection.
• You can take any collection —finite or infinite— of open sets (disks, half-planes, strips, etc.) and their union is still open. Unions stitch together many bubbles into a bigger, possibly a tangled, irregular, or wavy shape —yet it remains open. That’s due to the fact that every point in the union belongs to at least one open set, so there’s always a tiny neighborhood around it contained in the union.
• Open sets never contain their boundary. Boundaries —circles, lines, curves— sit “on the edge” between a set and its complement. In an open set, you can approach the boundary from inside, but the boundary points themselves are always missing.
• Finite intersections are open. Overlap of bubbles still has wiggle room. However, infinite intersections are not necessarily open can shrink to a point or boundary.