Open Sets in C. Unions, Intersections, Continuity, and Representation

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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).

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.

• Proposition. Open disk are open. $D_r(a) = \{z : |z - a| < r\}$.

Claim: $D_r(a)$ is open. Let $z_0 \in D_r(a)$ be arbitrary. Then, $|z_0 - a| < r$.

By the triangle inequality: $|z - a| = |(z - z_0) + (z_0 - a)| \leq |z - z_0| + |z_0 - a|$

Choose radius: $R = r - |z_0 - a| > 0$, we aim to show that $D_R(z_0) \subseteq D_r(a)$.

$\forall z \in D_R(z_0), |z - z_0| \lt R, |z - a| \leq |z - z_0| + |z_0 - a| < R + |z_0 - a| = (r - |z_0 - a|) + |z_0 - a| = r$

Therefore, $\forall z \in D_R(z_0), |z - a| < r$, meaning $z \in D_r(a)$ $\blacksquare$

Proposition. Unions (arbitrary) of open sets remain open. Let $\{S_\alpha\}_{\alpha \in A}$ be any collection (finite, countable, or uncountable) of open subsets of $\mathbb{C}$. Then, $S = \bigcup_{\alpha \in A} S_\alpha$ is open

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 $B(z; r) \subseteq S_{\alpha} \subseteq \cup_{\alpha ∈ A} S_{\alpha}$ Thus, z is interior in S, and since z was arbitrary, S is open $\blacksquare$.

• Infinite union of disks are open, e.g., $\cup_{n \in ℕ} D_{1/n}(a)$ is open. This is a “tail” of ever-shrinking open disks. It is open because arbitrary unions of open sets are open.

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

$S = \bigcup_{n=1}^{\infty} D_{1/n}(a) = D_1(a)$ since $D_{1/n}(a) \subseteq D_{1/m}(a)$ when $n > m$.

Proposition. Intersection of two open sets is open. Let $A_1, A_2$ be open subsets of $\mathbb{C}$. Then, $A_1 \cap A_2$ is open.

Proof:

Take any point $z \in A_1 \cap A_2$. Since:

• $z \in A_1$ and $A_1$ is open, $\exists r_1 > 0$ such that $B(z; r_1) \subseteq A_1$.
• $z \in A_2$ and $A_2$ is open, $\exists r_2 > 0$ such that $B(z; r_2) \subseteq A_2$.

Let $r = \min\{r_1, r_2\} > 0$. Then, $B(z; r) \subseteq B(z; r_1) \subseteq A_1, \quad B(z; r) \subseteq B(z; r_2) \subseteq A_2$. Thus, $B(z; r) \subseteq A_1 \cap A_2$. So every point of $A_1 \cap A_2$ is interior, hence $A_1 \cap A_2$ is open $\blacksquare$.

Corollary. If $A_1, A_2, \ldots, A_n$ are open sets, then $\bigcap_{i=1}^{n} A_i$ is open.

Proof.

Base case (n = 1). The intersection is just the single open set itself, which is open by hypothesis.

Base case (n = 2): It was previously demonstrated.

Inductive step: Assume that the intersection of any k open sets is open for some k ≥ 2. Consider k + 1 open sets $A_1, A_2, \ldots, A_{k+1}$. Their intersection can be written as: $\bigcap_{i=1}^{k+1} A_i = (\bigcap_{i=1}^{k} A_i) \cap A_{k+1}$.

By the Inductive Hypotheses, $\bigcap_{i=1}^{k} A_i$ is open. The set on the right is then the intersection of two open sets, which by the n = 2 case is open. Hence, by induction, the intersection of any finite collection of open sets is open. $\blacksquare$

Infinite Intersections May Not Be Open! Let $A_n = B(0; 1 + \frac{1}{n})$ for $n \in \mathbb{Z}^+$. Each $A_n$ is open (open disk), but $\bigcap_{n=1}^{\infty} A_n = \overline{B(0; 1)} = \{z : |z| \leq 1\}$. This is the closed unit disk, which is not open!

A similar example is $A_n = B(0; \frac{1}{n})$ for $n \in \mathbb{Z}^+$. Each $A_n$ is open (open disk), but $\bigcap_{n=1}^{\infty} A_n = \{ 0 \}$ which is not open (since no open disk around 0 is contained in {0}).

Theorem: A function $f: \mathbb{C} \to \mathbb{C}$ is continuous if and only if the preimage of every open set is open.

Proof.

“Only If” (⟹)

Assumption: 𝑓 is continuous. Let 𝑉 be an open set in the codomain. Claim: $f^{-1}(V)$ is open.

1. Let $z_0$ be an arbitrary point in the preimage $f^{-1}(V)$.
2. By definition of the preimage, $f(z_0) \in V$.
3. Since 𝑉 is open, there exists an open disk $\mathbb{B}_{\varepsilon}(f(z_0)) \subseteq V$ for some $\varepsilon$.
4. Because f is continuous, for this particular $\varepsilon$, there exists a $\delta$ such that if $|z - z_0| \lt \delta$, then $|f(z) - f(z_0)| \lt \varepsilon$.
5. This means $f(\mathbb{B}_{\delta}(z_0)) \subseteq \mathbb{B}_{\varepsilon}(f(z_0)) \subseteq V$.
6. Therefore, $f(\mathbb{B}_{\delta}(z_0)) \subseteq V$, so $z_0$ is an interior point of $f^{-1}(V)$.
7. Since for every point $z_0 \in f^{-1}(V)$, there is a disk around $z_0$ completely contained within $f^{-1}(V)$, the set $f^{-1}(V)$ is indeed open.

“If” (⟸)

Assumption: The preimage of every open set is open. Claim: f is continuous.

1. Let $z_0 \in \mathbb{C}$ and $\varepsilon \gt 0$. We want to show that f is continuous at $z_0$.
2. Consider the open disk V = $\mathbb{B}_{\varepsilon}(f(z_0))$ in the codomain.
3. By assumption, $f^{-1}(V)$ is an open set in the domain.
4. Since $f(z_0) \in V$, it follows that $z_0 \in f^{-1}(V)$.
5. Because $f^{-1}(V)$ is an open set, there exists a $\delta \gt 0$ such that $\mathbb{B}_{\delta}(z_0) \subseteq f^{-1}(V)$.
6. This means if $|z - z_0| \lt \delta$, then $f(z) \in V = \mathbb{B}_{\varepsilon}(f(z_0))$, which implies $|f(z) - f(z_0)| \lt \varepsilon$.
7. Therefore, f is continuous at $z_0$. Since $z_0$ was arbitrary, then f is continuous on $\mathbb{C}$.

Example. Consider f(z) = $\overline{z}$ (complex conjugation). This function is continuous, then for any open $V \subseteq \mathbb{C}$, $f^{-1}(V) = \{ z: \overline{z} \in V \}$ is the reflection of V over the real axis. Since reflections preserve openness, $f^{-1}(V)$ is open.

Theorem: Open Sets are Unions of Open Disks. A set $S \subseteq \mathbb{C}$ is open if and only if it is a union of open disks. Every open set is made by gluing together potentially infinitely many round bubble.

Proof:

($\Leftarrow$) Any union of open disks is open (since arbitrary unions of open sets are open).

($\Rightarrow$) Suppose $S$ is open. For each $z \in S$, there exists $r_z > 0$ with $B(z; r_z) \subseteq S$.

Then, $S = \bigcup_{z \in S} \{z\} \subseteq \bigcup_{z \in S} B(z; r_z) \subseteq S$

So $S = \bigcup_{z \in S} B(z; r_z)$ is a union of open disks. $\blacksquare$

Properties of Open/Closed Sets

Property	Statement
Arbitrary union	Any union of open sets is open
Finite intersection	Finite intersections of open sets are open
Infinite intersection	Infinite intersections of open sets need not be open!, e.g., $\bigcap_{n=1}^{\infty} B(0; 1/n) = \{0\}$ is not open.
$\emptyset$ and $\mathbb{C}$	Both are open and closed
Arbitrary intersection	Any intersection of closed sets is closed
Finite union	Finite unions of closed sets are closed
Infinite union	Infinite unions of closed sets need not be closed!, e.g., $\bigcup_{n=1}^{\infty} [1/n, 1] = (0, 1]$ is not closed