Limit Threshold Slides The Limit Threshold
Rigorous Proofs in Analysis: Lesson 1
Mathematical Rigor
The Problem of "Approach"
Imagine Achilles racing a tortoise. To reach the tortoise, he must first reach the halfway point.
Then, the halfway point of the remaining distance. This continues infinitely.
"If he must always cover half of the remaining distance, does he ever actually arrive?"
START
TARGET
1/2
3/4
7/8
We need a language for "closeness" that logic can verify.
The \(\epsilon-N\) Definition
A sequence \((a_n)\) converges to \(L\) if for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n > N\),
\[|a_n - L| < \epsilon\]
The Challenge
"\(\forall \epsilon > 0\)"
You give me any error margin, no matter how tiny.
The Response
"\(\exists N \in \mathbb{N}\)"
I find a starting point in the sequence.
The Guarantee
"\(\forall n > N\)"
After that point, every term stays within the margin.
Visualizing the "Tube"
\(L + \epsilon\) \(L - \epsilon\)
Threshold \(N\)
\(L\)
The number \(N\) is the moment the sequence "enters the tube" and never leaves.
The Game of N
If I give you the sequence \(a_n = \frac{1}{n}\) and a target limit \(L = 0\)...
My Epsilon 0.1
What is your \(N\)?
My Epsilon 0.001
How does \(N\) change?
Goal: Find a formula for \(N\) based on \(\epsilon\).
Limit Definition Workshop Limit Definition Workshop
Real Analysis | Lesson 1: The Limit Threshold
Name:
Date:
1. Decoding the Quantifiers
Write the formal definition of the limit of a sequence \( (a_n) \to L \) in symbolic logic using \(\forall\), \(\exists\), and implications. Then, translate each part into "plain English" in the space provided.
\(\forall \epsilon > 0\)
\(\exists N \in \mathbb{N}\)
\(n > N \implies |a_n - L| < \epsilon\)
2. The Epsilon Game
Consider the sequence \(a_n = \frac{n+1}{n}\). We suspect the limit is \(L = 1\). For each \(\epsilon\) provided by the "challenger," find the smallest integer \(N\) that satisfies the definition.
| Challenger's \(\epsilon\) | Inequality: \(|a_n - 1| < \epsilon\) | Smallest \(N\) |
| --- | --- | --- |
| 0.1 | | |
| 0.05 | | |
| 0.002 | | |
3. Finding the Formula
Using the same sequence \(a_n = \frac{n+1}{n}\), solve for \(n\) in terms of an arbitrary \(\epsilon\). Show your scratch work below to find a general formula for \(N(\epsilon)\).
4. Conceptual Check
"If a sequence enters an epsilon-tube at \(N=50\), must it also be inside the tube for \(N=100\)? Explain using the quantifiers."
Proof Blueprint Slides Proof Blueprint
The Anatomy of a Convergence Proof
Scratch Work → Formal Logic
The Writing Process
Phase 1: Scratch Work
Assume the goal: \(|a_n - L| < \epsilon\)
Work backward to isolate \(n\)
Express \(n\) as a function of \(\epsilon\)
*Note: This part is NEVER shown in the final proof.*
Phase 2: The Proof
State "Let \(\epsilon > 0\) be given."
Define \(N\) based on your scratch work.
Show that for \(n > N\), the inequality holds.
Q.E.D.
Case Study: Proving \( \lim_{n \to \infty} \frac{1}{n} = 0 \)
Scratch Work
\(|\frac{1}{n} - 0| < \epsilon\)
We found our threshold! Any \(N\) larger than \(1/\epsilon\) will work.
The Formal Proof
Let \(\epsilon > 0\).
\(n > \frac{1}{\epsilon} \implies \frac{1}{n} < \epsilon\)
Thus, \(\lim_{n \to \infty} \frac{1}{n} = 0\).
Fake Proof Detective
"Claim: \(\lim_{n \to \infty} 1 = 2\)."
Let \(\epsilon = 0.5\). We need \(|1 - 2| < 0.5\).
The Challenge:
A single \(\epsilon\) can destroy a limit claim, but a single \(\epsilon\) cannot prove it. Why?
Proof Construction Workshop Proof Construction Workshop
Real Analysis | Lesson 2: Constructing Convergence Proofs
Name:
Date:
The Workflow
For each problem, you must complete two distinct stages. The Scratch Work is where you find the threshold \(N\). The Formal Proof is where you present the logic clearly to a reader, starting with \(\epsilon\) and concluding with the convergence statement.
1
The Constant Sequence: \(a_n = C\)
Prove that for any constant \(C \in \mathbb{R}\), \(\lim_{n \to \infty} C = C\).
Phase 1: Scratch Work
Phase 2: Formal Proof
Let \(\epsilon > 0\)...
2
Algebraic Manipulation: \(a_n = \frac{3n+2}{n}\)
Prove that \(\lim_{n \to \infty} \frac{3n+2}{n} = 3\).
Phase 1: Scratch Work
Phase 2: Formal Proof
3
Slightly Harder: \(a_n = \frac{1}{n^2+1}\)
Prove that \(\lim_{n \to \infty} \frac{1}{n^2+1} = 0\). Hint: Use the fact that \(n^2+1 > n\) for all \(n \ge 1\).
Phase 1: Scratch Work
Phase 2: Formal Proof
"Analysis is the art of making the invisible epsilon visible."
Cauchy Completeness Slides The Cauchy Criterion
Convergence without a Target
Lesson 3: Cauchy & Completeness
The Problem with the Standard Definition
"To prove a sequence converges via \(\epsilon-N\), you must already know the limit ."
But what if you suspect a sequence converges, but you don't know the value \(L\)?
Can we define convergence based only on the internal behavior of the sequence?
Internal Check
Terms getting closer to each other .
External Check
Terms getting closer to a target \(L\) .
The Cauchy Definition
A sequence \((a_n)\) is Cauchy if for every \(\epsilon > 0\), there exists \(N \in \mathbb{N}\) such that for all \(n, m > N\),
\[|a_n - a_m| < \epsilon\]
"No matter which two terms you pick after the threshold \(N\), their distance is less than \(\epsilon\)."
Completeness
The Rationals (\(\mathbb{Q}\))
Are NOT complete. A Cauchy sequence of rational numbers can 'approach' a gap (like \(\sqrt{2}\)) that isn't in \(\mathbb{Q}\).
The Reals (\(\mathbb{R}\))
Are COMPLETE . Every Cauchy sequence in \(\mathbb{R}\) converges to a limit that is also in \(\mathbb{R}\).
\(\pi\)?
Completeness means the number line has no "missing pixels."
Rational Escape Activity Rational Escape Activity
Real Analysis | Lesson 3: Cauchy & Completeness
Name:
Date:
The Babylonian Algorithm
We define a sequence of rational numbers \((x_n)\) by the recursion: \(x_1 = 1\) and \(x_{n+1} = \frac{1}{2}(x_n + \frac{2}{x_n})\). This sequence is designed to approximate \(\sqrt{2}\).
A. Iterative Calculation
Calculate the first four terms of the sequence as simplified fractions. Verify that each term is a rational number.
\(x_1\)
1
\(x_2\)
\(x_3\)
\(x_4\)
B. Cauchy Behavior
Observe the difference between consecutive terms. As \(n\) increases, what happens to \(|x_{n+1} - x_n|\)?
Escaping \(\mathbb{Q}\)
"Assume this sequence converges to a limit \(L\). Taking the limit of both sides of the recursion formula gives \(L = \frac{1}{2}(L + \frac{2}{L})\). Solve for \(L\)."
Solve for \(L\):
The Paradox:
The sequence consists entirely of rational numbers . It is Cauchy (the terms bunch up). However, is the limit \(L\) a member of \(\mathbb{Q}\)? Explain what this implies about the "completeness" of the set of rational numbers.
Conceptual Synthesis
"If we only lived in the world of Rational numbers, would this sequence converge?"
Subsequence Extraction Slides Signal & Noise
Subsequences and Accumulation Points
Lesson 4: Patterns in Oscillation
Rigorous Extraction
A subsequence is formed by deleting some terms from the original sequence while preserving the order.
Original indices: \(n = 1, 2, 3, 4, 5, \dots\)
Sub-indices: \(n_1 < n_2 < n_3 < n_4 < \dots\)
Key Requirement: The indexing function \(k \mapsto n_k\) must be strictly increasing .
Visual Selection
\(a_1\)
\(a_2\)
\(a_3\)
\(a_4\)
\(a_5\)
\(a_6\)
\(a_2\)
\(a_5\)
...
Limit vs. Accumulation Point
Limit
The point where the entire sequence eventually settles.
Example: \(1/n \to 0\)
Accumulation Point
A value that is the limit of at least one subsequence .
Example: \((-1)^n\) has accumulation points \(\{-1, 1\}\)
"A sequence converges to \(L\) if and only if every subsequence converges to \(L\)."
Analyzing Oscillation
\(a_n = \sin\left(\frac{n\pi}{2}\right) + \frac{1}{n}\)
Is there a single limit? Or multiple accumulation points?
Subsequence \(n = 4k\)
\(a_{4k} \to 0\)
Subsequence \(n = 4k+1\)
\(a_{4k+1} \to 1\)
Subsequence \(n = 4k+3\)
\(a_{4k+3} \to -1\)
Accumulation Points: \{ -1, 0, 1 \}
Signal and Noise Workshop Signal and Noise Workshop
Real Analysis | Lesson 4: Subsequences & Accumulation Points
Name:
Date:
1. Defining the Sub-signal
Let \((a_n)\) be a sequence. Define a subsequence \((a_{n_k})\) such that \(n_k\) is the \(k\)-th prime number.
List the first 5 terms of \(n_k\):
2, 3, 5, 7, 11
Is \(k \mapsto n_k\) strictly increasing?
2. Finding Limits in the Noise
Consider the oscillating sequence: \[a_n = \cos\left(\frac{n\pi}{3}\right) + \frac{(-1)^n}{n}\] This sequence does not converge, but it contains several convergent "sub-signals."
Subsequence Index \(n_k\) Simplified \(a_{n_k}\) Sub-Limit \(L_k\) \(n_k = 6k\) \(n_k = 6k+1\) \(n_k = 6k+3\)
The Accumulation Set
List all distinct accumulation points found above:
3. Subsequence Logic
Theorem: If \( (a_n) \to L \), then every subsequence \( (a_{n_k}) \to L \).
Sketch a proof for this. Use the fact that since \( n_k \) is strictly increasing, \( n_k \ge k \).
Bolzano Weierstrass Slides The Crowded Box
The Bolzano-Weierstrass Theorem
Real Analysis: Final Lesson
The Bounded Infinity
Imagine an infinite number of points trapped inside a finite box.
"If the box never grows, but the points keep coming, they must eventually bunch up somewhere."
This "bunching up" is the existence of an accumulation point—and thus, a convergent subsequence.
Bolzano-Weierstrass Theorem
Every bounded sequence contains at least one convergent subsequence .
Input
Boundedness:
There exist \(m, M\) such that \(m \le a_n \le M\) for all \(n\).
Output
Convergence:
There exists a subsequence \((a_{n_k})\) that converges to some \(L \in [m, M]\).
Proof by Bisection
Start with interval \([a, b]\) containing all points.
Divide the interval in half.
At least one half must contain infinitely many points.
Repeat for that half, creating nested intervals.
The intersection is our limit point!
INTERVAL 1
Infinite Points Here
Zooming in on Infinity...
Theorem Architect Guide Theorem Architect Guide
Real Analysis | Lesson 5: The Bolzano-Weierstrass Theorem
Name:
Date:
The Fundamental Claim
"Every bounded sequence in \(\mathbb{R}\) has a convergent subsequence."
Proof by Nested Intervals
Let \((a_n)\) be a sequence such that \(m \le a_n \le M\) for all \(n \in \mathbb{N}\). We construct our convergent subsequence by following these logical steps:
1
Step 1: The First Bisection
Let \(I_1 = [m, M]\). Divide \(I_1\) into two equal sub-intervals: \([m, \frac{m+M}{2}]\) and \([\frac{m+M}{2}, M]\). Explain why at least one of these intervals must contain infinitely many terms of the sequence.
2
Step 2: Subsequence Selection
Let \(I_2\) be the sub-interval from Step 1 containing infinitely many terms. Pick a term \(a_{n_1}\) from the original sequence. In Step 2, pick \(a_{n_2}\) from \(I_2\) such that \(n_2 > n_1\). Why is it always possible to find such an \(n_2\)?
3
Step 3: The Convergence Argument
As we repeat this, the length of our interval \(I_k\) is \(\frac{M-m}{2^{k-1}}\). As \(k \to \infty\), what happens to the length? How does this force the subsequence to converge?
Critical Application
"Is it possible for a bounded sequence to have infinitely many accumulation points? Use Bolzano-Weierstrass to support your reasoning."
