To err is human, to blame it on someone else is even more human, Jacob’s Law
Definition. A set S ⊆ ℂ is said to be bounded if there is an M > 0 such that |z| < M for every z ∈ s. This means all points in S lie within a disk of radius M centered at the origin.
Definition. A set S ⊆ ℂ is compact if it is both closed and bounded. This is the Heine-Borel theorem applied to ℂ.
Definition. Let S be a set in the complex plane, S ⊆ ℂ. An open covering of a set S is a collection of open sets $\mathbf{L_g}$ such that $∪_{G∈\mathbf{L_g}} ⊇ S$. A finite subcover is a finite subset of $\mathbf{L_g}$ that still covers S.
Definition. A set S in the complex plane (S ⊆ ℂ) is compact if every open covering of S has a finite subcover ↭[Heine-Borel Theorem] S is both closed and bounded in ℂ
B(0, 1) is not compact because:
S = {n + i·0: n ∈ ℕ} ⊆ ℂ is not compact because
Heine-Borel Theorem (in ℂ). A subset S of the complex plane S ⊆ ℂ is compact, that is, every open cover of S has a finite subcover if and only if the set S is both closed and bounded in ℂ.
The unit circle in ℂ, defined as {z ∈ ℂ : |z| = 1}, is compact because ℝ - ℂ is open, and it is bounded by the set {z ∈ ℂ, |z| = 1}. Thus, it is obviously bounded (e.g., by a disk of radius 2 centered at the origin).
Closed Disk: The set {z ∈ ℂ : |z| ≤ r} (a disk of radius r including the boundary) is compact. It’s closed because it contains its boundary (the circle |z| = r), and it’s bounded by definition.
Closed Rectangle: Consider the set {z = x + iy : a ≤ x ≤ b, c ≤ y ≤ d}, where a, b, c, and d are real numbers. This is a rectangle in the complex plane, including its edges. It is compact because it is both closed (it contains its boundary) and bounded (it fits within a sufficiently large disk).
Finite Set of Points: Any finite set of points in ℂ is compact. It is trivially bounded, and it is also closed because it contains all its limit points since finite sets have no limit points (there’s no way for a neighborhood around any of these points to contain another distinct point from the set indefinitely.).
The set {z ∈ ℂ : Re(z) ≥ 0 and |z| ≤ 1} This set is compact because it is the intersection of two closed sets (a half-plane and a closed disk) which is closed, and it is bounded by the disk of radius 1.
The set S = {1,i,−1,−i} ⊂ ℂ is compact because it is both bounded (Every point in S satisfies ∣z∣ ≤ 1) and closed (it is a finite set, it has no limit points).
Open Disk: The set {z ∈ ℂ : |z| < r} (a disk without the boundary) is not compact. It is bounded but not closed (it does not contain its boundary).
The Entire Complex Plane (ℂ): ℂ is closed (it contains all its limit points) but it is not bounded. Therefore, it is not compact.
The set {z ∈ ℂ : Re(z) ≥ 0} This set is closed but not bounded, therefore it is not compact.
Cantor Theorem. Let {Kj}j∈ ℕ be a family of non-empty, compact (closed and bounded) subsets of ℂ such that: K1 ⊃ K2 ⊃··· ⊃ Kn ⊃ ⋯ (i.e., the sets are nested and decreasing). Then, the intersection: $\cap_{j=1}^{\infty} K_j$ is not empty.
Proof.
Let’s assume for the sake of contradiction that $\cap_{i=1}^{\infty} K_i = \empty$, then its complement $\cup _{i=1}^{\infty} K_i^c = \Complex$. In particular, $K_1 ⊆ \cup _{i=1}^{\infty} K_i^c$.
Since K1 is compact, there is a finite subcover ⇒ ∃j1, j2, ···, jn and (this is somehow more restricted, but it still applies) j1 < j2 < ··· < jn such that $K_1 ⊂ \cup _{m=1}^{n}$ Kjₘc ⊂ Kjₙ+1c because by assumption Kjₙ+1 ⊂ Kjₙ ⊂ ··· ⊂ Kj₁
This leds to K1 ∩ Kjₙ+1 = Ø, this contradicts the given assumption of a nested subsets of ℂ.
Proposition. Let K be a compact set in a topological space (which could be ℝⁿ, ℂ, or a more general space). Let C be a closed subset of K, i.e., C ⊆ K. We want to show that C is compact.
Proof.
To show that C is compact, we need to show that every open cover of C has a finite subcover.
Consider an arbitrary open cover of C: Let {Uα} be an open cover of C, where α belongs to some index set A. This means that: C ⊆ ⋃α∈A Uα
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, Cc (the set of all points that are in K but not in C), is open. Now, consider the collection of open sets: {Uα} ∪ {Cc}. This collection forms an open cover of K
If x ∈ C, then it must be in one of the Uα sets (because {Uα} covers C).
Otherwise, if a point x is in K but not in C (i.e. x∈K and x∉C), then it must be in Cc.
Therefore, K ⊆ (⋃α∈A Uα) ∪ Cc.
Since K is compact, the open cover {Uα} ∪ {Cc} has a finite subcover. Let’s denote this finite subcover as: {Uα1, Uα2, …, Uαn, Cc}. This finite subcover covers K: K ⊆ (Uα1 ∪ Uα2 ∪ … ∪ Uαn) ∪ Cc
Now, consider the sets Uα1, Uα2, …, Uαn. These are a subset of the original open cover {Uα} of C because C ⊆ K and a finite subcover of C ∎
Definition. A set S ⊆ ℂ is said to be connected if it is not contained in the union of two disjoint non-empty open subsets of ℂ which have a non-trivial intersection with S. Mathematically, ∄G1, G2 such that G1∪G2 ⊃ S, G1 ≠ ∅, G2 ≠ ∅, G1∩G2 = ∅, G1 and G2 are open, G1∩S ≠ ∅, G2∩S ≠ ∅ (S is disconnected Fig b).
Definition. An open connected non-empty subset of ℂ is called a region or a domain.
Proposition. Let G a non-empty open subset of ℂ. Then, G is a region if and only if any two points of G can be connected by a polygonal path.
Proof (⇒)
Suppose G is a region, fix a point a ∈ G. Let G1 = {z ∈ G: there is a polygonal path from a to z in G}, G2 = G \ G1.
Claim: G1 and G2 are open sets, then either G1 = 0 or G2 = 0
Suppose z ∈ G1, then since G1 ⊂ G and G is open, there is an r>0 such that B(z; r) ⊂ G. Since a and z are connected by a polygonal path, we can extend this path to any point w∈ B(z; r) by joining an additional joining z and w (Figure C), hence w ∈ G1 ⇒ B(z; r) ⊂ G1 ⇒ G1 is open.
Suppose z ∈ G2. This means z is in G, but there is no polygonal path from a to z within G. Since G is open, there exists an r > 0 such that the open ball B(z; r) = {w ∈ ℂ : |w - z| < r} is entirely contained in G (i.e., B(z; r) ⊂ G).
Now, we will show that B(z; r) is also contained in G2. Suppose, for the sake of contradiction, that there exists a point w ∈ B(z; r) such that w ∈ G1. This would mean there is a polygonal path from a to w within G.
Since w is in B(z; r), we can connect z and w by a straight line segment (which is a simple polygonal path) that lies entirely within B(z; r), and therefore within G. If there’s a polygonal path from a to w and a polygonal path (the line segment) from w to z, then by concatenating these paths, we can construct a polygonal path from a to z within G ⊥
Therefore, G1 and G2 are open, then by assumption, G is a region ⇒ either G1 = ∅ or G2 = ∅, but we know that G1 ≠ ∅ (a ∈ G), then G = G1 ∎
The complex plane ℂ is equipped with the standard metric topology, where the distance between two complex numbers z1 and z2 is given by the modulus: d(z1, z2) = ∣z1 − z2∣. This metric induces a topology on ℂ, where open sets are unions of open disks: Bϵ(z0) = { z ∈ ℂ : ∣z − z0∣ < ϵ}.
Definition. A sequence of complex numbers, denoted by (an) or {an}n=1∞ is an ordered list of complex numbers such that for each natural number n, there is a corresponding n-th complex number an in the list, e.g., an := 12 + 1⁄n; a0 := 1, a1 = 1, an := an-1 + 2an-2 + i(an-1 -7an-2).
Definition. A sequence (an) is said to have a limit L∈ℂ if for every epsilon (∀ε > 0), there exist a natural number (∃ n0∈ ℕ) such that for each n ≥ n0, |an -L| < ε. A convergent sequence approaches a specific complex number L as n grows larger.
The limit L (if it exists) is unique for a convergent sequence.
Consider the sequence $a_:=\frac{1}{n}+i(1−\frac{1}{n})i$. As n→∞, an→0+i(1−0)=i. Thus, the limit L = i·L = i, and the sequence is convergent.
Interpretation: {an} converges to L in ℂ, means that the terms an> cluster around L as n grows larger.
A sequence of complex numbers can be visualized as a set of points in the complex plane. A sequence {an} converges to a limit L ∈ ℂ if, for every open neighborhood Bϵ(L) of L, there exists a natural number n0 ∈ ℕ such that for all n ≥ n0, the terms an lie within Bϵ(L). In other words: ∀ ϵ > 0, ∃ n0 ∈ ℕ such that n ≥ n0 ⇒ an ∈ Bϵ(L). This is equivalent to the standard definition $\lim_{n \to \infty}a_n = L \iff \forall \epsilon > 0, \exist n_0 \in ℕ \text{ such that } |a_n - L| < \epsilon, \forall n ≥ n_0$.
Definition. A sequence is said to be convergent if it has a limit. A sequence is said to diverge or be no convergent if for every M > 0, there exists a natural number n0∈ ℕ such that for all n ≥ n0, |an| > M. A divergent sequence grows without bound (in magnitude) as n increases.
Consider the sequence an: = n + in2. As n→∞, ∣an∣ = $\sqrt{n²+n⁴}$→∞. Thus, the sequence diverges.
A sequence (an) is divergent if it does not converge to any finite limit L ∈ ℂ, that is, the sequence “escapes” every bounded region of the complex plane. Formally: ∀M > 0, ∃ n0 ∈ ℕ such that ∀n ≥ n0 ⇒ ∣an∣ > M. This is equivalent to saying that the sequence (an) “goes to infinity” in the extended complex plane ℂ ∪ {∞}.
In complex analysis, a set is said to be compact if it is both closed and bounded. The Bolzano-Weierstrass theorem states that every bounded sequence in ℂ has a convergent subsequence. This is a key result in complex topology and analysis.
It provides a way to determine if a sequence converges without knowing its limit. This is particularly useful in situations where finding the limit directly is difficult or impossible.
Cauchy’s criterion in ℂ (or ℝ) A sequence (an) of complex numbers is convergent in ℂ if and only for every given epsilon ε > 0 (no matter how small) there exist a natural number n0∈ ℕ such that for all n, m, n ≥ n0, we have |am - an| < ε
ε represents an arbitrarily small positive number. It defines the “closeness” or “tolerance” we want between terms of the sequence. n0 is a point in the sequence beyond which all terms are “close enough” to each other. |am - an| < ε states that the absolute difference between any two terms am and an beyond n0 is smaller than our chosen tolerance ε. In other words, a sequence is Cauchy (and therefore convergent) if its terms eventually become arbitrarily close to each other.
Consider the sequence an = 1/n in ℝ. For any ε > 0, we can find an n0 such that for all m, n ≥ n0, |1/m - 1/n| < ε. For example, if ε = 0.01, we can choose n0 = 100. Then for any m, n ≥ 100, |1/m - 1/n| ≤ |1/m| + |1/n| ≤ 1/100 + 1/100 = 0.02 < 0.01. This sequence is Cauchy and converges to 0.
Next, analyze the complex sequence zn = (1 + i)/n. This sequence is Cauchy and converges to 0 in the complex plane.
Consider the sequence an = (-1)n in ℝ. The terms oscillate between -1 and 1. No matter how large we choose n0, we can always find m and n ≥ n0 such that |am - an| = |-1 - 1| = 2 (if m is odd and n is even). Since we can’t make the difference arbitrarily small, this sequence is not Cauchy and does not converge.
Bolzano-Weierstrass theorem. Every bounded sequence in ℝⁿ (or ℂ) has a convergent subsequence. Equivalently, every infinite subset of a compact set in ℝⁿ (or ℂ) has at least one limit point (accumulation point).
Recall that a sequence (xn) in ℝⁿ is bounded if there exists a number M > 0 such that ||xn|| ≤ M for all n, where ||.|| denotes the Euclidean norm. In ℂ, a sequence (zn) is bounded if there exists a number M > 0 such that |zn| ≤ M for all n. A subsequence of (xn) is a sequence formed by selecting some of the terms of (xn) in their original order. A point x is a limit (accumulation) point of a set S if every neighborhood of x contains infinitely many points of S.
The idea is as follows. If you have an infinite number of points crammed into a finite space (a bounded set), then there must be at least one point where the points accumulate or cluster.
Consider the bounded sequence an = (-1)n. This sequence itself does not converge. However, it has convergent subsequences:
Consider the set S = {(1/n, 1/m) : n, m ∈ ℕ}. This is an infinite subset of the compact set [0, 1] × [0, 1] in ℝ². The point (0, 0) is a limit point of S.
The second statement of the Bolzano-Weierstrass theorem directly connects it to compactness. In ℝⁿ (and ℂ), a set is compact if and only if it is closed and bounded (Heine-Borel theorem). Therefore, any infinite subset of a closed and bounded set has a limit point.