JustToThePoint English Website Version
JustToThePoint en español
Colaborate with us

Bounded and compact sets

To err is human, to blame it on someone else is even more human, Jacob’s Law

Topology and Limits

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:

  1. It is not closed (it does not contain its boundary { z ∈ ℂ : ∣z∣ = 1}
  2. $\mathbf{L_g}$ = {B(0, r): r ∈ [0, 1) ⊆ ℝ}, the open covering $∪_{G∈\mathbf{L_g}} ⊇ B(0, 1)$ has no finite sub-cover that covers B(0, 1).

S = {n + i·0: n ∈ ℕ} ⊆ ℂ is not compact because

  1. It is not bounded (the points extend infinitely along the real axis).
  2. The opening covering $∪_{G∈\mathbf{L_g}}$ = U{G ∈ B(0; n): n ∈ ℕ} ⊇ S has no finite sub-cover that covers S.

Heine-Borel Theorem

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.

  1. S is closed ↭ The complement of S, Sc is an open set ↭ $S = \bar{S} = S ∪ ∂S$ ↭ S contains all of its limits points ↭ S contains its boundary (all of its boundary points).
  2. S is bounded: This means that S can be contained within a disk of finite radius. In other words, there exists some real number M such that |z| ≤ M for all z in S.

Examples

Counterexamples (Non-Compact Sets)

Cantor Theorem

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).

Regions and Domains

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.

Regions and Domains

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

Sequences of complex numbers

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 + 1n; 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.

Cauchy’s criterion

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

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.

Bitcoin donation

JustToThePoint Copyright © 2011 - 2025 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.