Limits of sequences - supremum and infimum, topology of R, Heine–Borel theorem, Bolzano–Weierstrass theorem, compactness equivalence to closed and bounded

I. Limits of Sequences

A sequence is an ordered list of numbers, typically denoted as $\{a_n\}_{n=1}^\infty$, where $a_n$ is the $n$-th term of the sequence. For example, $\{1, 1/2, 1/3, 1/4, \dots\}$ is a sequence where $a_n = 1/n$. The concept of a limit of a sequence is fundamental in real analysis. It describes the behavior of the sequence as $n$ becomes very large.

Definition of a Limit: A sequence $\{a_n\}$ is said to converge to a real number $L$, denoted as $\lim_{n \to \infty} a_n = L$, if for every positive number $\epsilon$ (no matter how small), there exists a natural number $N$ such that for all $n > N$, the inequality $|a_n - L| < \epsilon$ holds. In simpler terms, as $n$ gets larger and larger, the terms $a_n$ get arbitrarily close to $L$. If such an $L$ exists, the sequence is called convergent; otherwise, it is called divergent.

Example: Consider the sequence $\{1/n\}$. We claim that $\lim_{n \to \infty} 1/n = 0$. To prove this, let $\epsilon > 0$ be given. We need to find an $N$ such that for all $n > N$, $|1/n - 0| < \epsilon$. This inequality simplifies to $1/n < \epsilon$, or $n > 1/\epsilon$. So, if we choose $N$ to be any integer greater than or equal to $1/\epsilon$ (e.g., $N = \lceil 1/\epsilon \rceil$), then for all $n > N$, we have $n > 1/\epsilon$, which implies $1/n < \epsilon$. Thus, the limit is indeed 0.

Properties of Convergent Sequences: If $\{a_n\}$ converges to $L$ and $\{b_n\}$ converges to $M$, then:

• The sequence $\{a_n + b_n\}$ converges to $L + M$.
• The sequence $\{a_n - b_n\}$ converges to $L - M$.
• The sequence $\{c \cdot a_n\}$ converges to $c \cdot L$ for any constant $c$.
• The sequence $\{a_n \cdot b_n\}$ converges to $L \cdot M$.
• If $M \neq 0$, the sequence $\{a_n / b_n\}$ converges to $L / M$.
• If $a_n \ge 0$ for all $n$, then $L \ge 0$.
• If $a_n \le b_n$ for all $n$, then $L \le M$.

Monotone Convergence Theorem: A monotone sequence (either non-decreasing or non-increasing) that is bounded converges. A sequence $\{a_n\}$ is non-decreasing if $a_n \le a_{n+1}$ for all $n$, and non-increasing if $a_n \ge a_{n+1}$ for all $n$. A sequence is bounded if there exist real numbers $m$ and $M$ such that $m \le a_n \le M$ for all $n$. This theorem is crucial as it guarantees the existence of a limit under certain conditions.

II. Supremum and Infimum

The concepts of supremum (least upper bound) and infimum (greatest lower bound) are essential for understanding the completeness of the real number system and are closely related to the Monotone Convergence Theorem.

Upper Bound: A real number $U$ is an upper bound for a set $S$ of real numbers if for every element $x \in S$, we have $x \le U$. A set can have infinitely many upper bounds.

Least Upper Bound (Supremum): The supremum of a set $S$, denoted as $\sup S$, is the smallest of all the upper bounds of $S$. If a set has a supremum, it must be unique. A set has a supremum if and only if it is non-empty and bounded above.

Lower Bound: A real number $L$ is a lower bound for a set $S$ of real numbers if for every element $x \in S$, we have $x \ge L$. A set can have infinitely many lower bounds.

Greatest Lower Bound (Infimum): The infimum of a set $S$, denoted as $\inf S$, is the largest of all the lower bounds of $S$. If a set has an infimum, it must be unique. A set has an infimum if and only if it is non-empty and bounded below.

Completeness Axiom of Real Numbers: Every non-empty set of real numbers that is bounded above has a supremum in the set of real numbers. Similarly, every non-empty set of real numbers that is bounded below has an infimum in the set of real numbers. This axiom distinguishes the real numbers from the rational numbers, for example. The set of rational numbers is not complete because, for instance, the set of rational squares less than 2 does not have a supremum within the rationals (its supremum is $\sqrt{2}$).

Relationship to Monotone Convergence Theorem: For a non-decreasing sequence $\{a_n\}$ that is bounded above, its limit $L$ is equal to its supremum, i.e., $L = \sup \{a_n\}$. Similarly, for a non-increasing sequence $\{a_n\}$ that is bounded below, its limit $L$ is equal to its infimum, i.e., $L = \inf \{a_n\}$.

Example: Consider the set $S = \{x \in \mathbb{R} \mid x^2 < 2\}$. This set is bounded above by 2, 3, or any number greater than $\sqrt{2}$. The least upper bound (supremum) is $\sqrt{2}$. It is also bounded below by -2, -3, or any number less than $-\sqrt{2}$. The greatest lower bound (infimum) is $-\sqrt{2}$.

III. Topology of $\mathbb{R}$

Topology deals with the properties of space that are preserved under continuous deformations. In the context of real numbers ($\mathbb{R}$), topology studies the properties of intervals and the notion of "closeness" without relying on distance directly, although distance is used to define topological concepts.

Open Set: A set $O \subseteq \mathbb{R}$ is called an open set if for every point $x \in O$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq O$. An open interval $(a, b)$ is defined as $\{x \in \mathbb{R} \mid a < x < b\}$. The key idea is that every point in an open set has "room around it" that is still within the set.

Example: The interval $(0, 1)$ is an open set. For any $x \in (0, 1)$, we can find a small $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subset (0, 1)$. For instance, if $x = 0.5$, we can choose $\epsilon = 0.1$, and $(0.4, 0.6) \subset (0, 1)$. The set $\mathbb{R}$ itself and the empty set $\emptyset$ are also considered open sets.

Closed Set: A set $F \subseteq \mathbb{R}$ is called a closed set if its complement, $\mathbb{R} \setminus F$, is an open set. In other words, a closed set contains all its "limit points."

Example: The interval $[0, 1]$ is a closed set because its complement, $(-\infty, 0) \cup (1, \infty)$, is an open set. The set $\{0\}$ is closed because its complement $(-\infty, 0) \cup (0, \infty)$ is open. The set $\mathbb{R}$ and the empty set $\emptyset$ are both open and closed.

Properties of Open and Closed Sets:

• The union of any collection of open sets is open.
• The intersection of any finite collection of open sets is open.
• The intersection of any collection of closed sets is closed.
• The union of any finite collection of closed sets is closed.

Neighborhood: A neighborhood of a point $x$ is any open set containing $x$. Sometimes, a neighborhood is defined more broadly as any set containing an open set that contains $x$. In the context of $\mathbb{R}$, an $\epsilon$-neighborhood of $x$ is the open interval $(x-\epsilon, x+\epsilon)$ for some $\epsilon > 0$. This formalizes the idea of "points close to $x$."

Interior Point: A point $x$ is an interior point of a set $A$ if there exists an open set $U$ such that $x \in U \subseteq A$. The set of all interior points of $A$ is called the interior of $A$, denoted by $\text{int}(A)$ or $A^\circ$. The interior of $A$ is always an open set.

Boundary Point: A point $x$ is a boundary point of a set $A$ if every neighborhood of $x$ contains at least one point in $A$ and at least one point not in $A$. The set of all boundary points of $A$ is called the boundary of $A$, denoted by $\partial A$. The boundary of a set is always a closed set.

Connected Set: A set $S \subseteq \mathbb{R}$ is connected if it cannot be written as the union of two disjoint non-empty open sets. In $\mathbb{R}$, the connected sets are precisely the intervals (open, closed, half-open, infinite).

IV. Heine–Borel Theorem

The Heine–Borel theorem is a cornerstone of real analysis, providing a powerful criterion for identifying compact sets in $\mathbb{R}^n$ (and specifically in $\mathbb{R}$).

Definition of Open Cover: An open cover of a set $S$ is a collection of open sets $\{U_\alpha\}_{\alpha \in I}$ such that $S \subseteq \bigcup_{\alpha \in I} U_\alpha$. A subcollection of these open sets is a finite subcover if it is finite and still covers $S$. That is, there exist finitely many indices $\alpha_1, \dots, \alpha_k$ such that $S \subseteq U_{\alpha_1} \cup \dots \cup U_{\alpha_k}$.

Heine–Borel Theorem Statement: A subset $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

Significance: This theorem is incredibly useful because it gives us a simple way to check for compactness. Instead of dealing with potentially infinite collections of open sets and trying to find a finite subcover, we only need to check if the set is closed and bounded.

Proof Idea (for $\mathbb{R}$):

1. If $K$ is closed and bounded, it is compact: Let $K$ be a closed and bounded subset of $\mathbb{R}$. Since $K$ is bounded, there exists an interval $[-M, M]$ such that $K \subseteq [-M, M]$. Let $\{U_\alpha\}$ be any open cover of $K$. Then $\{U_\alpha\} \cup \{ (-\infty, -M), (M, \infty) \}$ is an open cover of $[-M, M]$. It can be shown that any bounded closed interval is compact. This implies that there exists a finite subcollection of these sets that covers $[-M, M]$. By removing $(-\infty, -M)$ and $(M, \infty)$ if they are not needed, we obtain a finite subcover of $K$.
2. If $K$ is compact, it is closed and bounded: If $K$ is compact, it is bounded. Suppose $K$ is not bounded. Then for any $n \in \mathbb{N}$, there exists $x_n \in K$ such that $|x_n| > n$. The set $\{x_n\}$ is an infinite set in $K$. One can construct an open cover of $K$ for which no finite subcover exists if $K$ is not closed or not bounded. For instance, if $K$ is compact, it must be closed. If $K$ is not closed, there is a sequence $\{x_n\}$ in $K$ converging to a point $L \notin K$. This sequence has a convergent subsequence, and its limit must be in $K$ if $K$ is closed.

Example: The interval $[0, 1]$ is closed and bounded, so by the Heine–Borel theorem, it is compact. The interval $(0, 1)$ is bounded but not closed, so it is not compact. The set of integers $\mathbb{Z}$ is closed (as a union of singletons, each closed) but not bounded, so it is not compact. The set $[0, \infty)$ is closed but not bounded, so it is not compact.

V. Bolzano–Weierstrass Theorem

The Bolzano–Weierstrass theorem is another fundamental result that guarantees the existence of convergent subsequences.

Bolzano–Weierstrass Theorem Statement: Every bounded infinite subset of $\mathbb{R}$ has at least one limit point. (A limit point of a set $S$ is a point $x$ such that every neighborhood of $x$ contains a point of $S$ other than $x$ itself. In $\mathbb{R}$, limit points are the same as the boundary points of the closure of the set.)

Alternative Statement (for sequences): Every bounded sequence of real numbers has a convergent subsequence.

Significance: This theorem is crucial for proving the convergence of sequences and the existence of solutions to certain problems. It states that if you have an infinite number of points trapped within a finite interval, at least one of those points must be "crowded" by other points.

Proof Idea (for sequences): Let $\{a_n\}$ be a bounded sequence. If it takes on only finitely many values, then at least one value must be repeated infinitely often, forming a constant subsequence which converges. If it takes on infinitely many values, we can use a "bisection" method. Consider the interval containing the sequence. If the left half contains infinitely many terms, focus on that half; otherwise, focus on the right half. Repeat this process. This generates a sequence of nested intervals, each containing infinitely many terms of the original sequence. The intersection of these intervals is a single point, which can be shown to be a limit point of the sequence, and thus a convergent subsequence exists.

Example: Consider the sequence $\{(-1)^n\}$. This sequence is $ -1, 1, -1, 1, -1, 1, \dots $. It is bounded between -1 and 1. It does not converge. However, it has two convergent subsequences: the subsequence of terms at odd indices, $\{a_{2k-1}\} = \{-1, -1, -1, \dots\}$, which converges to -1, and the subsequence of terms at even indices, $\{a_{2k}\} = \{1, 1, 1, \dots\}$, which converges to 1. The Bolzano-Weierstrass theorem guarantees that at least one such subsequence exists.

Connection to Compactness: A set $K \subseteq \mathbb{R}$ is compact if and only if every infinite subset of $K$ has a limit point in $K$. This is another equivalent definition of compactness.

VI. Compactness Equivalence to Closed and Bounded

We have seen two major theorems: the Heine–Borel theorem and the Bolzano–Weierstrass theorem. Their relationship solidifies our understanding of compactness in $\mathbb{R}$.

Recap of Equivalences for $K \subseteq \mathbb{R}$ (or $\mathbb{R}^n$):

• Definition via Open Covers: $K$ is compact if every open cover of $K$ has a finite subcover.
• Heine–Borel Theorem: $K$ is compact if and only if $K$ is closed and bounded.
• Bolzano–Weierstrass Theorem (for sets): $K$ is compact if and only if every infinite subset of $K$ has a limit point in $K$.
• Bolzano–Weierstrass Theorem (for sequences): $K$ is compact if and only if every sequence in $K$ has a subsequence that converges to a point in $K$.

Why is this important? These equivalences mean that proving one property implies the others. For exam purposes, the most practical condition to check is whether a set is closed and bounded, thanks to the Heine–Borel theorem. If a set is closed and bounded, we immediately know it's compact, and therefore, it must satisfy the other properties (having finite subcovers, having limit points for infinite subsets, having convergent subsequences for any sequence within it).

Example Scenario: Suppose we are asked to prove that a certain set $A$ has a maximum value. If we can show that $A$ is compact (e.g., by showing it's closed and bounded), then we know that any sequence in $A$ has a subsequence converging to a point in $A$. Let this point be $x$. If $A$ is closed, then $x \in A$. If $A$ is bounded, the sequence is bounded. If $A$ contains its limit points (i.e., is closed), then the limit of the subsequence is in $A$. The existence of a convergent subsequence is often the first step in proving the existence of maximum or minimum values.

Summary Table:

In essence, the combination of being closed and bounded provides the necessary conditions for a set in $\mathbb{R}$ to possess the powerful properties associated with compactness. This makes the closed and bounded criterion the most frequently used test for compactness in introductory real analysis.