Compactness, Cantor's Intersection Theorem, and Connectedness

Silence is a source of great strength, Lao Tzu

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).

Compact sets

Compactness and sequential compactness are not equivalent in a general topological space. First, we must distinguish both concepts:

• Topological Compactness. A set $S$ is compact if, whenever you cover it with a collection of open sets $\{U_\alpha\}$, you can pick a finite number of them that still cover $S$ (Intuitively: The set is “small enough or kinda of finite” that you don’t need infinite resources to cover it).
• Sequential Compactness. A set $S$ is sequentially compact if every infinite sequence $\{z_n\}$ in $S$ contains a subsequence $\{z_{n_k}\}$ that converges to a limit $L$, and that limit $L$ is inside $S$. (Intuitively: Sequences get trapped inside the set.)

Theorem: For a subset $S \subseteq \mathbb{C}$ equipped with the standard metric, the following are equivalent:

1. $S$ is Compact (Topologically).
2. $S$ is Sequentially Compact.
3. $S$ is Closed and Bounded (Heine-Borel).

Proof.

⇒ (Compact ⇒ Sequentially Compact):

1. Assume that $S$ is compact.
2. By the Heine-Borel theorem, since $S$ is compact, $S$ must be closed and bounded.
3. Take a Sequence: Let $\{z_n\}$ be any infinite sequence of points contained in $S$.
4. Since $S$ is bounded, the sequence $\{z_n\}$ is bounded. The Bolzano-Weierstrass Theorem states that every bounded sequence in $\mathbb{C}$ has a convergent subsequence $\{z_{n_k}\}$ that converges to some limit $L$.
5. Since $S$ is closed, it contains all its limit points. Therefore, the limit $L$ must belong to $S$.
6. Conclusion: We found a subsequence that converges to a point in $S$. Thus, $S$ is sequentially compact. $\quad \checkmark$

⇐ (Sequentially Compact ⇒ Compact):

Step A. S is bounded.

1. Suppose for the sake of contradiction that $S$ is unbounded.
2. Then, for every integer $n$, we can find a point $z_n \in S$ such that $|z_n| > n$.
3. Consider the sequence $\{z_1, z_2, z_3, \dots\}$. As $n \to \infty$, $|z_n| \to \infty$.
4. Does this sequence have a convergent subsequence? No. Any subsequence $\{z_{n_k}\}$ would also have norms approaching infinity ($|z_{n_k}| \to \infty$), so it cannot converge to a finite complex number.
5. This contradicts the assumption that $S$ is sequentially compact. Therefore, $S$ must be bounded.

Step B: Prove $S$ is Closed.

1. Let $L$ be a limit point of $S$. We need to show $L \in S$.
2. By the definition of a limit point, there exists a sequence $\{z_n\}$ in $S$ such that $z_n \to L$.
3. Since $S$ is sequentially compact, this sequence must have a subsequence that converges to a point in $S$.
4. Since the original sequence converges to $L$, every subsequence must also converge to $L$.
5. Therefore, the limit point $L$ must be in $S$.
6. Since $S$ contains all its limit points, $S$ is closed.

Conclusion: Since $S$ is both closed and bounded, by the Heine-Borel Theorem, $S$ is compact. $\quad \blacksquare$

Cantor’s Intersection Theorem

Cantor’s Intersection Theorem. Let (X, d) be a complete metric space (such as $\mathbb{R}^n$ or $\mathbb{C}$). Let $\{ K_j \}_{j \in \mathbb{N}}$ be a nested sequence of non-empty compact sets: $K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots$ (i.e., the sets are nested and decreasing). Then, the intersection is non-empty: $\bigcap_{j=1}^{\infty} K_j \neq \emptyset$.

Proof

Assume for the sake of contradiction that $\bigcap_{j=1}^{\infty} K_j = \emptyset$.

1. Construct an open cover of K₁. Taking complements: $\left( \bigcap_{j=1}^{\infty} K_j \right)^c = \bigcup_{j=1}^{\infty} K_j^c = \mathbb{C}$
   Since K₁ ⊆ ℂ, we have: $K_1 \subseteq \bigcup_{j=1}^{\infty} K_j^c$
2. Why each K_j^c is open? Each K_j is compact (hence closed in ℂ), so each K_j^c is open.
3. Apply compactness of K₁. The collection {K_j^c : j ∈ ℕ} is an open cover of K₁.
   By compactness, there exists a finite subcover:
   $K_1 \subseteq K_{j_1}^c \cup K_{j_2}^c \cup \cdots \cup K_{j_n}^c$ where we may assume j₁ < j₂ < ··· < j_n.
4. Exploit the nested structure. Since the K_j are nested (decreasing): $K_{j_n} \subseteq K_{j_{n-1}} \subseteq \cdots \subseteq K_{j_1}$.
   Taking complements (which reverses inclusions): $K_{j_1}^c \subseteq K_{j_2}^c \subseteq \cdots \subseteq K_{j_n}^c$
   Therefore: $\bigcup_{m=1}^{n} K_{j_m}^c = K_{j_n}^c$
5. Derive contradiction. We have K₁ ⊆ K_jₙ^c, which means K₁ ∩ K_jₙ = ∅.
   But K_jₙ ⊆ K₁ (since the sequence is nested and j_n ≥ 1), so: $K_{j_n} = K_{j_n} \cap K_1 = \emptyset$
   This contradicts the assumption that all K_j are non-empty. ∎

Uniqueness When Diameters Shrink

Corollary. If additionally $diam(K_j) \to 0$, then $\bigcap_{j=1}^{\infty} K_j$ contains exactly one point.

Proof:

We know the intersection is non-empty. Suppose it contains two distinct points p, q.

Then, $\forall j: p, q \in K_j$, so: $0 \le |p - q| \leq \text{diam}(K_j), \text{diam}(K_j) \to 0$

By the Squeeze theorem, this forces |p - q| = 0, hence p = q. $\blacksquare$

Counterexamples: Why Each Hypothesis Matters

1. Nested but Not Compact (Closed Only).
   Consider in ℝ: K_n = [n, ∞).
   Nested: K₁ ⊇ K₂ ⊇ ··· Closed: Yes. Bounded: No.
   Intersection: $\bigcap_{n=1}^{\infty} [n, \infty) = \emptyset$
   Closedness without boundedness allows escape to infinity.
2. Nested but Not Compact (Bounded Only).
   Consider in ℝ: K_n = (0, 1/n).
   Nested: K₁ ⊇ K₂ ⊇ ··· Bounded: Yes (all contained in [0,1]).
   Closed: No (missing boundary point 0). Intersection: $\bigcap_{n=1}^{\infty} (0, 1/n) = \emptyset$
   Boundedness without closedness allows the intersection point to “leak out.”
3. Example in $\mathbb{Q}$ which is not complete. Define $K_n = \{ q \in \mathbb{Q}: \sqrt{2} - \frac{1}{n} \le q \le \sqrt{2}+\frac{1}{n} \}, n \in \mathbb{N}$.
   Each $K_n$ is closed in $\mathbb{Q}$ because it is the intersection of the closed interval $[\sqrt{2} - \frac{1}{n}, \sqrt{2} + \frac{1}{n}] \subset \mathbb{R}$ with $\mathbb{Q}$, which is closed in the subspace topology.
   Each $K_n$ is non-empty: By density of $\mathbb{Q}$ in $\mathbb{R}$, there exist rational numbers arbitrarily close to $\sqrt{2}$
   The sets are nested: $K_{n+1} \subseteq K_n$ for all n.
   But the intersection is empty: $\bigcap_{n=1}^{\infty} K_n = \{ q \in \mathbb{Q}: q = \sqrt{2} \} = \emptyset$ since $\sqrt{2}$ is irrational.
   Completeness of the ambient space matters!

Corollary (Nested Closed Intervals). If [a_n, b_n] is a sequence of closed intervals in ℝ with $[a_1, b_1] \supseteq [a_2, b_2] \supseteq [a_3, b_3] \supseteq \cdots$. Then, $\bigcap_{n=1}^{\infty} [a_n, b_n] \neq \emptyset$. Moreover, if $(b_n - a_n) \to 0$, the intersection is a single point.

Important remark: This is equivalent to completeness of ℝ and is often taken as an axiom in constructing the real numbers.

Closed Subsets of Compact Sets

Proposition. Let K be a compact set in a topological space X (which could be ℝⁿ, ℂ, or any metric space). If C ⊆ K is closed, then C is compact.

Compactness is a “hereditary” property, but only for closed subsets.

Proof.

To show that C is compact, we must show that every open cover of C has a finite subcover.

Consider an arbitrary open cover of C. Let {U_α}_α∈A be an open cover of C, where α belongs to some index set A. This means that: C ⊆ ⋃_α∈A U_α

Construct an open cover of K: Since C is a subset of K, we can extend the open cover of C to an open cover of K. Because C is closed, its complement in K, C^c (the set of all points that are in K but not in C), is open. Now, consider the collection of open sets: $\mathbf{U}$ = {U_α}_α∈A ∪ {C^c}. This collection forms an open cover of K

Claim: K ⊆ (⋃_α∈A U_α) ∪ C^c.

1. If x ∈ C, then it must be in one of the U_α sets (because {U_α} covers C).
2. Otherwise, if a point x is in K but not in C (i.e. x∈K and x∉C), then it must be in C^c.

Therefore, since K is compact, the open cover {U_α}_α∈A ∪ {C^c} has a finite subcover. Let’s denote this finite subcover as {U_α1, U_α2, …, U_αn, C^c}. This finite subcover covers K: K ⊆ (U_α1 ∪ U_α2 ∪ … ∪ U_αn) ∪ C^c

Now, consider the sets U_α1, U_α2, …, U_αn. These are a subset of the original open cover {U_α} of C and a finite subcover of C: $\forall x \in C, x \notin C^c$, the finite subcover of K must also cover x, so $x \in U_{\alpha_j}$ for some $j \in \{1, \cdots, n\}$.

Connectedness in ℂ

Definition. A set S ⊆ ℂ is connected if it cannot be written as the union of two disjoint non-empty open subsets of ℂ which have a non-trivial intersection with S (both meet S).

Formal Definition. A set S ⊆ ℂ is connected if there do not exist two open sets $U, V \subseteq \mathbb{C}$ such that:

1. Covering: $S \subseteq U \cup V$
2. Disjointness: $U \cap V \cap S = \emptyset$ (They don’t share points within S)
3. Non-triviality: $U \cap S \neq \emptyset$ and $V \cap S \neq \emptyset$

S ⊆ ℂ is connected if and only if: $\nexists \, U, V \subseteq \mathbb{C} \text{ open such that } U \cap V = \emptyset, U \cap S \neq \emptyset, V \cap S \neq \emptyset, S \subseteq U \cup V$

Equivalently, A disconnected set consists of pieces separated by a “gap” that can be filled by two disjoint open neighborhoods U and V. In this case, (U, V) is called a separation of S (Fig b).

For a topological space X (and specifically for subsets of ℂ, $S \subseteq \mathbb{C}$), the following are equivalent:

Formulation	Statement
Separation	No separation by disjoint, non-empty open sets U, V ⊆ S exists s.t. S = U ∪ V
Clopen	The only subsets of S that are both open and closed in the subspace topology are ∅ and S itself
Continuous functions	Every continuous function f: S → {0, 1} is constant
For open sets	S cannot be written as S = U ∪ V where U, V are non-empty, disjoint, and open in S

The “Transitivity of Openness” Lemma. If $A$ is open in a subspace $G$, and $G$ is open in the whole space $\mathbb{C}$, then $A$ is open in $\mathbb{C}$.

Proof of Lemma:

1. By definition of the subspace topology, if $A \subseteq G$ is open in $G$, there exists some set $U$ open in $\mathbb{C}$ such that $A = G \cap U$.
2. By assumption, $G$ is open in $\mathbb{C}$.
3. The intersection of two open sets ($G$ and $U$) is always open in $\mathbb{C}$. Therefore, $A$ is open in $\mathbb{C}$.

Proposition. Let $G \subseteq G$ be an open set. Then, G is connected if and only if G cannot be written as $G = A \cup B$, where $A, B$ are non-empty, disjoint, and open in $\mathbb{C}$.

$(\Rightarrow)$ Assume $G$ is connected.

1. Suppose for the sake of contradiction that $G$ can be written as $G = A \cup B$, where $A$ and $B$ are non-empty, disjoint, and open in $\mathbb{C}$.
2. We must check if $A$ and $B$ are open in the subspace $G$.
3. By definition, $A$ is open in $G$ if $A = G \cap U$ for some open $U$. Since $A$ is already a subset of $G$ and $A$ is open in $\mathbb{C}$, we can just pick $U=A$. Thus, $A$ is open in $G$. (Similarly, $B$ is open in $G$).
4. This means we have decomposed $G$ into two non-empty, disjoint sets that are open in the subspace topology.
5. This contradicts the definition of Connectedness. Therefore, such a separation cannot exist. $\blacksquare$

Direction 2: $(\Leftarrow)$. Assume $G$ cannot be split by sets open in $\mathbb{C}$.

Proof

1. Suppose for the sake of contradiction $G$ is disconnected.
2. By the standard definition of disconnectedness, there exist sets $A, B \subseteq G$ such that: (i) $A, B$ are non-empty; (ii) $A \cap B = \emptyset$; (iii) $G = A \cup B$; (iv) $A$ and $B$ are open in the subspace $G$.
3. Now we apply our Lemma (Transitivity of Openness).Since $A$ is open in $G$, and $G$ is open in $\mathbb{C}$ (given by the proposition’s premise), then $A$ is open in $\mathbb{C}$. Similarly, $B$ is open in $\mathbb{C}$.
4. This implies that $G$ can be written as the union of two non-empty, disjoint sets that are open in $\mathbb{C}$.
5. This contradicts our initial assumption (that $G$ cannot be split by such sets). Therefore, $G$ must be connected. $\blacksquare$

Examples of Connected Sets

1. Any singleton {z₀} is connected (vacuously —there’s no way to split one point into two non-empty pieces).
2. Open Disks B_r(z₀) = {z ∈ ℂ : |z - z₀| < r} are connected. \
3. Closed Disks $\bar{B}(z_0; r)$ = {z ∈ ℂ : |z - z₀| ≤ r} are connected.
4. Line Segments [z₁, z₂] = {(1-t)z₁ + tz₂ : t ∈ [0,1]} are connected.
   The function f: [0, 1] → [z₁, z₂] given by f(t) = (1-t)z₁ + tz₂ is continuous and surjective. Since [0, 1] ⊂ ℝ is connected, and continuous images of connected sets are connected, f([0, 1]) = [z₁, z₂] is connected. ∎
5. The Entire Plane ℂ.
6. The annulus A = {z : r < |z| < R} for 0 ≤ r < R is connected.
   Claim. A is path‑connected (hence connected).
   Take any two points $z_1, z_2 \in A$. Write them in polar form: $z_1 = \rho_1 e^{i\theta_1}, z_1 = \rho_2 e^{i\theta_2}, r \lt \rho_j \lt R$, j = 1, 2.
   A. First move radially from $z_1$ to $\rho_2 e^{i\theta_1}$: $\gamma_1(t) = [(1-t)\rho_1 +t\rho_2]e^{i\theta_1}, t \in [0, 1]$, This stays in A because the radius is always between r and R.
   B. Move along the circle of radius $\rho_2$ from angle $\theta_1$ to $\theta_2$: $\gamma_2(t) = \rho_2 e^{i \bigr((1-t)\theta_1 + t\theta_2 \bigl)}$ which stays in A since the radius is constantly $\rho_2$.
7. Star-Shaped Sets. A set S is star-shaped with respect to z₀ if for every z ∈ S, the segment [z₀, z] ⊆ S. Every star-shaped set is connected.
   Any two points z₁, z₂ ∈ S can be connected via z₀: the path [z₁, z₀] ∪ [z₀, z₂] lies in S. ∎
8. Convex Sets are connected. A set S is convex if for any z₁, z₂ ∈ S, the segment [z₁, z₂] ⊆ S.

Examples of Disconnected Sets

1. Two Disjoint Disks. S = B₁(0) ∪ B₁(5) is disconnected. Separation: G₁ = B₂(0), G₂ = B₂(5).
2. The Real Integers ℤ ⊂ ℂ. Every point is isolated, so ℤ is totally disconnected (only connected subsets are singletons).
3. The Punctured Plane with a Removed Line, S = ℂ \ ℝ = {z : ℑ(z) ≠ 0} is disconnected. Separation: G₁ = {z : ℑ(z) > 0} (upper half-plane), G₂ = {z : ℑ(z) < 0} (lower half-plane).
4. The Set {z : |z| ≠ 1}. S = B(0; 1) ∪ {z : |z| > 1} (the complex plane minus the unit circle; the unit circle acts as a “barrier” that splits the complex plane into two disjoint open regions) is disconnected. Separation: G₁ = B(0; 1) (the open unit disk; open, $0 \in G_1$), G₂ = {z : |z| > 1} (the exterior of the closed unit disk; open, $2 \in G_2$).
5. $\mathbb{Q}$ is totally disconnected. Between any two rationals $p, q \in \mathbb{Q}$ lies an irrational ($r\in \mathbb{R} \setminus \mathbb{R}: p \lt r \lt q$), providing a “gap”.