CALC-2.4

4-5 Hours

Subsequences, Completeness, and Cauchy Criterion

Advanced Topics

9 Sections

Foundational for Analysis

Learning Objectives

By the end of this course, you will be able to:

Define subsequences and understand their relationship to convergence
State and prove the Bolzano-Weierstrass Theorem
Understand cluster points and their characterization
Master the Cauchy Convergence Criterion
Understand equivalences among completeness properties
Apply Nested Interval Theorem and Heine-Borel Theorem

Section 1: Subsequences

Definition 4.1: Subsequence

Let $\{a_n\}$ be a sequence. If $\{n_k\} \subset \mathbb{N}$ with $n_1 < n_2 < n_3 < \cdots$ , then $\{a_{n_k}\}$ is a subsequence of $\{a_n\}$ .

Examples of Subsequences

Odd-indexed: $\{a_{2k-1}\} = a_1, a_3, a_5, \ldots$
Even-indexed: $\{a_{2k}\} = a_2, a_4, a_6, \ldots$
Prime-indexed: $\{a_{p_k}\} = a_2, a_3, a_5, a_7, a_{11}, \ldots$

Theorem 4.1: Subsequence Convergence

$\{a_n\}$ converges to $a$ $\Leftrightarrow$ Every subsequence of $\{a_n\}$ converges to $a$ .

Corollary 4.1: Odd-Even Test

$\{a_n\}$ converges $\Leftrightarrow$ Both $\{a_{2k-1}\}$ and $\{a_{2k}\}$ converge to the same limit.

Example 4.1: Proving Divergence via Subsequences

Application: Prove $\{(-1)^n\}$ diverges.

$\{a_{2k}\} = 1, 1, 1, \ldots \to 1$

$\{a_{2k-1}\} = -1, -1, -1, \ldots \to -1$

Since $1 \neq -1$ , the sequence diverges.

Section 2: Monotone Subsequences

Lemma 4.1: Every Sequence Has a Monotone Subsequence

Every sequence has a monotone subsequence.

Proof of Lemma 4.1:

Call index $n$ a "peak" if $a_n \geq a_m$ for all $m > n$ .

Case 1: Infinitely many peaks exist.

Let peaks be $n_1 < n_2 < n_3 < \cdots$ . Then $a_{n_1} \geq a_{n_2} \geq a_{n_3} \geq \cdots$ is decreasing.

Case 2: Finitely many peaks (possibly none).

Let $N$ be larger than all peak indices. Since $N$ is not a peak, $\exists n_1 > N$ with $a_{n_1} > a_N$ .

Since $n_1$ is not a peak, $\exists n_2 > n_1$ with $a_{n_2} > a_{n_1}$ . Continuing, we get strictly increasing subsequence.

∎

Section 3: Bolzano-Weierstrass Theorem

Theorem 4.2: Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Two Proof Methods

Method 1: Using Monotone Bounded Theorem

By Lemma 4.1, extract a monotone subsequence
This subsequence is bounded (inherited from original)
Bounded monotone sequence converges by Theorem 3.2

Method 2: Using Nested Intervals

Let $\{a_n\} \subset [A, B]$ . Bisect the interval
At least one half contains infinitely many terms
Continue bisecting to get nested intervals with length → 0
By Nested Interval Theorem, intersection is a single point
Pick one term from each interval to form convergent subsequence

Remark 4.1: Historical Note

Named after Bernard Bolzano (1781-1848) and Karl Weierstrass (1815-1897), this theorem is fundamental to real analysis and was crucial in the rigorization of calculus.

Section 4: Cluster Points

Definition 4.2: Cluster Point

Let $E \subset \mathbb{R}$ . A point $x_0 \in \mathbb{R}$ is a cluster point (accumulation point) of $E$ if:

\forall \delta > 0: U^0(x_0, \delta) \cap E \neq \emptyset

where $U^0(x_0, \delta) = (x_0 - \delta, x_0 + \delta) \setminus \{x_0\}$ is the punctured neighborhood.

Equivalent Characterizations

Every neighborhood of $x_0$ contains infinitely many points of $E$
There exists a sequence of distinct points $\{x_n\} \subset E$ with $x_n \to x_0$

Definition 4.3: Isolated Point

$x_0 \in E$ is isolated if $\exists \delta > 0$ : $U(x_0, \delta) \cap E = \{x_0\}$ .

Theorem 4.3: Cluster Point Theorem

Every bounded infinite subset of $\mathbb{R}$ has at least one cluster point.

Proof of Theorem 4.3:

Take a sequence of distinct points from the set. Apply Bolzano-Weierstrass to get a convergent subsequence. Its limit is a cluster point.

∎

Section 5: Cauchy Convergence Criterion

Definition 4.4: Cauchy Sequence

A sequence $\{a_n\}$ is a Cauchy sequence if:

\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n, m > N: |a_n - a_m| < \epsilon

Theorem 4.5: Cauchy Criterion

A sequence of real numbers converges if and only if it is a Cauchy sequence.

Proof of Theorem 4.5 (Necessity: Convergent ⟹ Cauchy):

Let $\lim a_n = L$ . Given $\epsilon > 0$ , $\exists N$ : $n > N \Rightarrow |a_n - L| < \frac{\epsilon}{2}$ .

For $n, m > N$ :

|a_n - a_m| \leq |a_n - L| + |L - a_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

∎

Proof of Theorem 4.5 (Sufficiency: Cauchy ⟹ Convergent):

Step 1: Cauchy sequences are bounded.

Take $\epsilon = 1$ . Get $|a_n - a_{N+1}| < 1$ for $n > N$ , so $|a_n| < |a_{N+1}| + 1$ .

Step 2: By Bolzano-Weierstrass, extract convergent subsequence $\{a_{n_k}\} \to L$ .

Step 3: Show original converges to $L$ .

Given $\epsilon > 0$ :

By Cauchy: $\exists N_1$ : $n, m > N_1 \Rightarrow |a_n - a_m| < \frac{\epsilon}{2}$
By subsequence: $\exists K$ : $k > K \Rightarrow |a_{n_k} - L| < \frac{\epsilon}{2}$

For $n > N_1$ , choose $k$ with $n_k > \max\{n, N_1\}$ :

|a_n - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| < \epsilon

∎

Example 4.2: Harmonic Series is NOT Cauchy

Problem: Show that $H_n = \sum_{k=1}^n \frac{1}{k}$ is not Cauchy.

Solution:

|H_{2n} - H_n| = \sum_{k=n+1}^{2n} \frac{1}{k} > n \cdot \frac{1}{2n} = \frac{1}{2}

No matter how large $N$ , taking $n > N$ and $m = 2n$ gives $|H_m - H_n| > \frac{1}{2}$ .

This violates the Cauchy criterion, so the harmonic series diverges.

Section 6: Nested Interval Theorem

Theorem 4.6: Cantor's Nested Interval Theorem

If $\{[a_n, b_n]\}$ is a sequence of closed intervals with:

$[a_n, b_n] \supseteq [a_{n+1}, b_{n+1}]$ for all $n$
$\lim_{n \to \infty} (b_n - a_n) = 0$

Then $\exists!\, \xi \in \mathbb{R}$ : $\{\xi\} = \bigcap_{n=1}^{\infty} [a_n, b_n]$

Remark 4.2: Open Intervals Fail

Counterexample: $\left(0, \frac{1}{n}\right)$

These are nested, and length → 0, but $\bigcap = \emptyset$ (every positive number is excluded by some interval).

Section 7: Equivalence of Completeness Properties

The Seven Equivalent Characterizations of ℝ Completeness

┌─────────────────────────────────────────────────────────────┐
│                COMPLETENESS EQUIVALENCES                     │
├─────────────────────────────────────────────────────────────┤
│  Supremum Axiom ⟺ Nested Interval Theorem                   │
│        ⇕                    ⇕                                │
│  Monotone Bounded ⟺ Bolzano-Weierstrass ⟺ Cauchy Criterion  │
│        ⇕                    ⇕                                │
│  Cluster Point Theorem ⟺ Heine-Borel Theorem                │
└─────────────────────────────────────────────────────────────┘

These theorems form the theoretical foundation of real analysis. Each can be derived from any other!

Supremum Axiom (Completeness)

Monotone Bounded Theorem

Bolzano-Weierstrass Theorem

Cauchy Criterion

Nested Interval Theorem

Cluster Point Theorem

Heine-Borel Theorem

Theorem 4.7: Supremum ⟹ Monotone Bounded

If every non-empty bounded above set has a supremum, then every monotone bounded sequence converges.

Proof of Theorem 4.7:

Let $\{x_n\}$ be increasing and bounded above. Let $S = \{x_n : n \in \mathbb{N}\}$ .

By the Supremum Axiom, $L = \sup S$ exists. We claim $x_n \to L$ .

Given $\epsilon > 0$ , since $L - \epsilon$ is not an upper bound, $\exists N$ : $x_N > L - \epsilon$ .

For $n > N$ : $L - \epsilon < x_N \leq x_n \leq L < L + \epsilon$ , so $|x_n - L| < \epsilon$ .

∎

Theorem 4.8: Monotone Bounded ⟹ Bolzano-Weierstrass

If every monotone bounded sequence converges, then every bounded sequence has a convergent subsequence.

Proof of Theorem 4.8:

Key Lemma: Every sequence has a monotone subsequence.

Call $x_n$ a peak if $x_n \geq x_m$ for all $m > n$ .

Case 1: Infinitely many peaks at $n_1 < n_2 < \cdots$ . Then $x_{n_k} \geq x_{n_{k+1}}$ , giving a decreasing subsequence.

Case 2: Finitely many peaks, last at $N$ . For $n_1 > N$ , $x_{n_1}$ is not a peak, so $\exists n_2 > n_1$ : $x_{n_2} > x_{n_1}$ . Continue to get an increasing subsequence.

The bounded monotone subsequence converges by assumption.

∎

Example 4.5: Finding Monotone Subsequences

Problem: Find a monotone subsequence of $\{x_n\} = \{1, 3, 2, 5, 4, 7, 6, 9, 8, ...\}$ .

Solution:

Peaks occur at: $n = 2$ (value 3), $n = 4$ (value 5), $n = 6$ (value 7), ...

These form the decreasing subsequence... wait, they're increasing! Let's recheck.

Actually, odd indices give $1, 2, 4, 6, 8, ...$ which is increasing.

Even indices give $3, 5, 7, 9, ...$ which is also increasing.

Remark 4.3: Why ℚ is Not Complete

Consider the sequence of rational approximations to $\sqrt{2}$ :

x_1 = 1, \quad x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)

This is a Cauchy sequence in $\mathbb{Q}$ , but $\sqrt{2} \notin \mathbb{Q}$ .

The sequence has no limit in $\mathbb{Q}$ — this is why we need $\mathbb{R}$ !

Using Completeness in Proofs

To show a sequence converges:

Show it's Cauchy (use Cauchy Criterion)
Show it's monotone and bounded (Monotone Bounded)
Show it's bounded and find a convergent subsequence (B-W)

To show a sequence diverges:

Find two subsequences with different limits
Show it's not Cauchy
Show it's unbounded

Example 4.6: Applying Cauchy Criterion

Problem: Show $x_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ (harmonic series) diverges.

Solution (using Cauchy Criterion):

Consider $x_{2n} - x_n$ :

x_{2n} - x_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \geq n \cdot \frac{1}{2n} = \frac{1}{2}

For $\epsilon = \frac{1}{4}$ , we can always find $m = n$ , $k = 2n$ with $k > m > N$ such that:

|x_k - x_m| = |x_{2n} - x_n| \geq \frac{1}{2} > \epsilon

So $\{x_n\}$ is NOT Cauchy, hence diverges.

Example 4.7: Convergent Subsequences

Problem: Find all cluster points of $\{x_n\} = \{\sin\frac{n\pi}{4}\}$ .

Solution:

The sequence cycles through: $\frac{\sqrt{2}}{2}, 1, \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, -1, -\frac{\sqrt{2}}{2}, 0, ...$

Cluster points are: $\{-1, -\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}, 1\}$

For example, $n = 2, 10, 18, ...$ gives $x_{n_k} = 1$ , a constant subsequence.

Theorem 4.9: Convergence via Subsequences

A bounded sequence $\{x_n\}$ converges to $L$ if and only if every convergent subsequence converges to $L$ .

Proof of Theorem 4.9:

(⟹) If $x_n \to L$ , every subsequence also converges to $L$ .

(⟸) Suppose every convergent subsequence has limit $L$ , but $x_n \not\to L$ .

Then $\exists \epsilon_0 > 0$ and infinitely many $n_k$ with $|x_{n_k} - L| \geq \epsilon_0$ .

The subsequence $\{x_{n_k}\}$ is bounded, so by Bolzano-Weierstrass, it has a convergent sub-subsequence $\{x_{n_{k_j}}\}$ .

This sub-subsequence converges to some $L' \neq L$ (since $|x_{n_{k_j}} - L| \geq \epsilon_0$ ). Contradiction!

∎

Example 4.8: Using Subsequence Characterization

Problem: Show $\{\frac{1 + (-1)^n}{n}\}$ converges.

Solution:

Even terms: $x_{2k} = \frac{2}{2k} = \frac{1}{k} \to 0$

Odd terms: $x_{2k-1} = \frac{0}{2k-1} = 0 \to 0$

Both subsequences converge to 0, so $x_n \to 0$ .

Summary: Completeness Toolkit

Theorem	Statement	Application
Bolzano-Weierstrass	Bounded ⟹ convergent subsequence	Existence proofs
Cauchy Criterion	Cauchy ⟺ Convergent (in ℝ)	Prove convergence without limit
Nested Intervals	Closed nested ⟹ common point	Bisection method
Cluster Point	Bounded has cluster point	Limit characterization

Section 8: Applications

Theorem 4.10: Contractive Mapping Principle

Let $f: [a,b] \to [a,b]$ be a contraction with ratio $0 < k < 1$ :

|f(x) - f(y)| \leq k|x - y| \text{ for all } x, y \in [a,b]

Then the sequence $x_{n+1} = f(x_n)$ converges to the unique fixed point of $f$ .

Example 4.9: Fixed Point Iteration

Problem: Show $x_{n+1} = \cos(x_n)$ converges for any $x_1 \in [0,1]$ .

Solution:

By Mean Value Theorem, $|\cos x - \cos y| = |\sin c||x - y| \leq |x - y|$ .

For $x, y \in [0,1]$ : $|\sin c| \leq \sin 1 \approx 0.84 < 1$ .

So $\cos$ is a contraction, and $x_n \to L$ where $\cos L = L$ .

Numerically: $L \approx 0.7391$ (Dottie number).

Theorem 4.11: Limsup and Liminf

For any bounded sequence $\{x_n\}$ :

\liminf_{n \to \infty} x_n = \lim_{n \to \infty} \inf_{k \geq n} x_k

\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup_{k \geq n} x_k

The sequence converges iff $\liminf = \limsup$ .

Example 4.10: Computing Limsup and Liminf

Problem: Find $\limsup$ and $\liminf$ of $x_n = (-1)^n + \frac{1}{n}$ .

Solution:

Even terms: $x_{2k} = 1 + \frac{1}{2k} \to 1$

Odd terms: $x_{2k-1} = -1 + \frac{1}{2k-1} \to -1$

Therefore:

$\limsup x_n = 1$
$\liminf x_n = -1$

Since $1 \neq -1$ , the sequence diverges.

Remark 4.4: Connection to Topology

Completeness of $\mathbb{R}$ is foundational to topology:

Compactness: A subset of $\mathbb{R}$ is compact iff closed and bounded (Heine-Borel)
Connectedness: $\mathbb{R}$ is connected (no "gaps")
Metric Spaces: $\mathbb{R}$ is a complete metric space

Example 4.11: Cantor Intersection

Problem: Prove $\bigcap_{n=1}^{\infty}[0, \frac{1}{n}] = \{0\}$ .

Solution:

Clearly $0 \in [0, \frac{1}{n}]$ for all $n$ , so $0 \in \bigcap$ .

If $x > 0$ , choose $N > \frac{1}{x}$ . Then $x > \frac{1}{N}$ , so $x \notin [0, \frac{1}{N}]$ .

Therefore $x \notin \bigcap$ , proving $\bigcap = \{0\}$ .

Looking Ahead: Chapter 3

In Chapter 3: Function Limits & Continuity, you will learn:

ε-δ definition of function limits
Continuous functions and their properties
Extreme Value Theorem & Intermediate Value Theorem
Uniform continuity

Example 4.12: Proving Boundedness

Problem: Prove that $\{\frac{n}{n+1}\}$ is bounded.

Solution:

For all $n \geq 1$ : $0 < \frac{n}{n+1} < 1$ .

Therefore the sequence is bounded with $m = 0$ , $M = 1$ .

Since it's also increasing, it converges to $\sup = 1$ .

Theorem 4.12: Dense Subsequences

If $\{x_n\}$ is bounded, then for any $a$ in the range of cluster points, there exists a subsequence converging to $a$ .

Remark 4.5: Constructing Cauchy Sequences

To show a sequence is Cauchy:

Estimate $|x_m - x_n|$ for $m > n$
Show this can be made $< \epsilon$ for large $m, n$
The estimate should not depend on specific values of $m, n$

Example 4.13: Cauchy Verification

Problem: Prove $x_n = \sum_{k=1}^{n}\frac{1}{k^2}$ is Cauchy.

Solution:

For $m > n$ :

|x_m - x_n| = \sum_{k=n+1}^{m}\frac{1}{k^2} < \sum_{k=n+1}^{m}\frac{1}{k(k-1)} = \frac{1}{n} - \frac{1}{m} < \frac{1}{n}

Given $\epsilon > 0$ , choose $N > \frac{1}{\epsilon}$ . For $m > n > N$ :

$|x_m - x_n| < \frac{1}{n} < \frac{1}{N} < \epsilon$ . ∎

Completeness Quick Reference

Subsequences

• Preserves order of indices
• Converges to same limit if original converges
• B-W: bounded ⟹ convergent subsequence

Cauchy Criterion

• Terms get close to each other
• In ℝ: Cauchy ⟺ Convergent
• Useful when limit is unknown

Example 4.14: Nested Interval Application

Problem: Find $\sqrt{2}$ using bisection.

Solution:

Start with $[1, 2]$ since $1^2 < 2 < 2^2$ .

Midpoint: $1.5$ . Since $1.5^2 = 2.25 > 2$ , take $[1, 1.5]$ .

Midpoint: $1.25$ . Since $1.25^2 = 1.5625 < 2$ , take $[1.25, 1.5]$ .

Continue... By Nested Interval Theorem, the intersection is $\{\sqrt{2}\}$ .

Theorem 4.13: Cluster Point Characterization

A point $c$ is a cluster point of $\{x_n\}$ iff for every $\epsilon > 0$ and every $N \in \mathbb{N}$ , there exists $n > N$ with $|x_n - c| < \epsilon$ .

Example 4.15: Finding All Cluster Points

Problem: Find all cluster points of $x_n = \sin\frac{n\pi}{4}$ .

Solution:

The sequence cycles: $\frac{\sqrt{2}}{2}, 1, \frac{\sqrt{2}}{2}, 0, -\frac{\sqrt{2}}{2}, -1, -\frac{\sqrt{2}}{2}, 0, \ldots$

Cluster points: $\{-1, -\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}, 1\}$

Remark 4.6: Completeness in Other Spaces

Completeness is not universal. In $\mathbb{Q}$ :

Cauchy sequences exist that don't converge in $\mathbb{Q}$
Example: $x_n = \sum_{k=0}^{n}\frac{1}{k!}$ is Cauchy but converges to $e \notin \mathbb{Q}$
This is why $\mathbb{R}$ was constructed: to "complete" $\mathbb{Q}$

Example 4.16: Subsequence Extraction

Problem: From $x_n = (-1)^n(1 + \frac{1}{n})$ , extract subsequences converging to 1 and -1.

Solution:

Even terms: $x_{2k} = 1 + \frac{1}{2k} \to 1$

Odd terms: $x_{2k-1} = -(1 + \frac{1}{2k-1}) \to -1$

Key Distinctions

Convergent vs Cauchy

In $\mathbb{R}$ : equivalent

In $\mathbb{Q}$ : Cauchy ⊋ Convergent

Limit vs Cluster Point

Limit: unique (if exists)

Cluster points: can be multiple

Example 4.17: Bolzano-Weierstrass Application

Problem: Prove $x_n = \sin(n)$ has a convergent subsequence.

Solution:

Since $-1 \leq \sin(n) \leq 1$ , the sequence is bounded.

By Bolzano-Weierstrass, there exists a convergent subsequence.

Note: The limit depends on which subsequence is chosen (the sequence itself diverges).

Theorem 4.14: Cauchy Sequence Properties

Every Cauchy sequence is bounded
A subsequence of a Cauchy sequence is Cauchy
If a Cauchy sequence has a convergent subsequence, the whole sequence converges

Remark 4.7: Completeness Applications

Completeness of $\mathbb{R}$ is used to prove:

Extreme Value Theorem
Intermediate Value Theorem
Existence of supremum/infimum
Convergence of monotone bounded sequences

Chapter 2.4 Mastery Checklist

Subsequence definition
Bolzano-Weierstrass Theorem
Cauchy criterion
Completeness of ℝ
Nested Interval Theorem

Example 4.18: Constructive Subsequence

Problem: Find a monotone subsequence of $x_n = \cos(n)$ .

Solution:

By the Monotone Subsequence Lemma, such a subsequence exists.

Constructively: Since $\{\cos(n)\}$ is bounded and has infinitely many values close to any point in $[-1,1]$ , we can extract a monotone subsequence approaching any cluster point.

Remark 4.8: Chapter 2 Summary

Key results from Chapter 2:

Definitions

ε-N definition, boundedness, monotonicity

Theorems

Squeeze, Monotone Bounded, B-W, Cauchy

Practice Quiz: Subsequences & Completeness

Questions

Correct

Accuracy

Which of the following is a subsequence of

\{n^2\} = \{1, 4, 9, 16, 25, ...\}

Easy

Not attempted

State the Bolzano-Weierstrass Theorem:

Easy

Not attempted

Prove that

\{(-1)^n + \frac{1}{n}\}

diverges using subsequences. The two limits are:

Medium

Not attempted

Show that

\{\sin(n)\}

has a convergent subsequence:

Medium

Not attempted

a_n = \sum_{k=1}^n \frac{1}{k^2}

a Cauchy sequence?

Medium

Not attempted

Use Cauchy criterion to prove

\sum_{n=1}^{\infty} \frac{1}{n^2}

converges. The key estimate is:

Hard

Not attempted

Find all cluster points of

\left\{\frac{(-1)^n n}{n+1}\right\}

Hard

Not attempted

Prove Nested Interval Theorem using Monotone Bounded. The key sequences are:

Hard

Not attempted

Frequently Asked Questions

What's the difference between a sequence and a subsequence?

A subsequence is formed by selecting terms from the original sequence while preserving their order. For example, from {1, 2, 3, 4, 5, ...}, the subsequence of even-indexed terms is {2, 4, 6, ...}. The indices must be strictly increasing: n₁ < n₂ < n₃ < ...

Why is the Bolzano-Weierstrass theorem so important?

It guarantees that every bounded sequence has a convergent subsequence. This is crucial for proving existence theorems in analysis, optimization (extreme value theorem), and fixed-point theorems. It's one of several equivalent formulations of completeness of ℝ.

How is a Cauchy sequence different from a convergent sequence?

In ℝ, they're equivalent! A Cauchy sequence is one where terms get arbitrarily close to EACH OTHER, while a convergent sequence has terms getting close to a specific LIMIT. The beauty is: in complete spaces like ℝ, being Cauchy is enough to guarantee a limit exists.

What is a cluster point vs a limit?

A limit is unique; a cluster point may not be. Every convergent sequence has one limit (= one cluster point). A divergent sequence like {(-1)^n} has two cluster points: -1 and 1 (via odd and even subsequences). Limit = unique cluster point.

Why do open intervals fail the Nested Interval Theorem?

Consider (0, 1/n). These are nested open intervals with length → 0, but their intersection is empty! Every point p > 0 is excluded by some interval. The closedness is essential to 'trap' a point in all intervals.

How do I use subsequences to prove divergence?

Find two subsequences with different limits. For {(-1)^n}: even terms → 1, odd terms → -1. Since 1 ≠ -1, the original sequence diverges. Or find one subsequence that diverges.

Chapter 2 Complete!

Congratulations! You have completed Chapter 2: Sequence Limits. You have mastered:

The ε-N definition of sequence limits
Fundamental limit properties and operations
Convergence theorems (Squeeze, Monotone Bounded, Stolz)
Completeness of ℝ (Bolzano-Weierstrass, Cauchy criterion)

Next: Chapter 3 will cover Function Limits and Continuity, building on these sequence limit foundations.