This deck covers the three theorems that make "the iterates settle down" rigorous: the Monotone Convergence Theorem, the Bolzano-Weierstrass theorem, and the Cauchy criterion. It targets the misconceptions that monotonicity alone gives convergence, that every bounded sequence converges, that Cauchy implies convergent in any space, and that a subsequential limit is THE limit.
Subject: Foundations of Higher Mathematics · 106 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. State and prove the Monotone Convergence Theorem and use it to prove a recursive sequence converges.
2. State and prove Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence.
3. Define a Cauchy sequence and prove that in the reals, Cauchy and convergent mean the same thing.
4. Explain why the same Cauchy sequence fails to converge in the rationals, and state completeness three equivalent ways.
Warm-up
Discussion prompt
Before we open Monotone Convergence, Bolzano-Weierstrass & Cauchy: without looking back, what was the main idea of Sequences & Convergence, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
The epsilon-N definition of convergence read as a nested-quantifier game: given a tolerance, produce a threshold. Covers uniqueness of limits, convergent implies bounded, the algebra of limits, divergence to infinity, and the squeeze theorem.
Concept
A sequence is increasing if every term is at least as big as the one before it, and decreasing if every term is at most the one before it.
\[ \text{increasing: } a_{n+1} \ge a_n \text{ for all } n \qquad \text{decreasing: } a_{n+1} \le a_n \text{ for all } n \]
monotone — A sequence that is either increasing (for all n, the next term is at least the current) or decreasing (for all n, the next term is at most the current). One-directional forever.
Counterexample
Discussion prompt
A sequence is increasing if every term is at least as big as the one before it, and decreasing if every term is at most the one before it.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
If the inequality is always strict, the sequence is strictly increasing or decreasing; if equality is allowed, it is weakly (non-strictly) so.
\[ \text{strictly increasing: } a_{n+1} > a_n \qquad\quad \text{non-decreasing: } a_{n+1} \ge a_n \]
The single word monotone covers all four cases. Every theorem in this deck needs only weak monotonicity, so that is what we assume.
Analogy
Discussion prompt
Explain Strict versus weak, and the umbrella word by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
If the inequality is always strict, the sequence is strictly increasing or decreasing; if equality is allowed, it is weakly (non-strictly) so.
Picture it
Figure (svg): A rising staircase of five steps whose heights approach a dashed horizontal ceiling line labelled as the supremum.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture the terms as heights of steps read left to right. A monotone sequence is a staircase that only ever goes up (or only ever goes down). It never doubles back.
Intuition
Picture the terms as heights of steps read left to right. A monotone sequence is a staircase that only ever goes up (or only ever goes down). It never doubles back.
Figure (svg): A rising staircase of five steps whose heights approach a dashed horizontal ceiling line labelled as the supremum.
If a rising staircase also has a ceiling it can never cross, the steps have nowhere to go but to crowd up against some height. That height is the whole story of this deck.
Explain it
Discussion prompt
Explain A one-way staircase to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Picture the terms as heights of steps read left to right. A monotone sequence is a staircase that only ever goes up (or only ever goes down). It never doubles back.
Concept
A sequence is bounded above if some fixed number sits at or above every term, and bounded below if some fixed number sits at or below every term. Bounded means both.
\[ \text{bounded above: } \exists\, M \;\; \forall n \;\; a_n \le M \]
Boundedness is a statement about the whole set of terms at once. It is the second ingredient the Monotone Convergence Theorem needs.
Concept
Here is the first headline theorem. Monotone plus bounded is enough for convergence, and it even tells you the limit.
\[ (a_n)\text{ increasing and bounded above} \;\Rightarrow\; a_n \to \sup_n a_n \]
\[ (a_n)\text{ decreasing and bounded below} \;\Rightarrow\; a_n \to \inf_n a_n \]
Notice you do not need to guess the limit in advance. Completeness hands you the supremum, and monotonicity forces the terms onto it.
Intuition
The supremum is the least ceiling. Because it is a ceiling, no term ever passes it. Because it is the LEAST ceiling, dropping down by any tiny amount stops being a ceiling.
So just below the supremum there must already be a term. Once the increasing sequence reaches that term, it stays in the thin band between there and the ceiling forever.
\[ S - \varepsilon \;<\; a_N \;\le\; a_n \;\le\; S \;<\; S + \varepsilon \qquad (n \ge N) \]
Ranking
Put in order
Put the moves of Worked example: proving the Monotone Convergence Theorem into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The set of terms is nonempty and bounded above, so by the least-upper-bound (completeness) axiom it has a supremum.
Worked example
Claim: an increasing sequence that is bounded above converges to its supremum. Let us prove it straight from the completeness axiom.
Produce the candidate limit
Why: The set of terms is nonempty and bounded above, so by the least-upper-bound (completeness) axiom it has a supremum. Call it S.
\[ S = \sup\{\, a_n : n \in \mathbb{N} \,\} \]
Fix a tolerance and use 'least'
Why: Given any positive tolerance, S minus that tolerance is smaller than S, hence NOT an upper bound. So some term pokes above it.
\[ \forall \varepsilon > 0 \;\; \exists N \;\; a_N > S - \varepsilon \]
Propagate with monotonicity
Why: For every index past N, increasing terms are at least a_N, and S is an upper bound above them all. So they are trapped in a band of width epsilon.
\[ n \ge N \;\Rightarrow\; S - \varepsilon < a_N \le a_n \le S \]
Verify the definition of the limit is met
Why: The band gives |a_n - S| < epsilon for all n at least N, which is exactly convergence to S. Sanity check: nothing exceeds S, so the limit could not be larger; and it cannot be smaller since terms climb past S minus epsilon for every epsilon.
\[ n \ge N \;\Rightarrow\; |a_n - S| < \varepsilon \quad\therefore\quad a_n \to S \]
Picture it
Animation
Shows: Each line of the worked example "proving the Monotone Convergence Theorem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The band gives |a_n - S| < epsilon for all n at least N, which is exactly convergence to S. Sanity check: nothing exceeds S, so the limit could not be larger; and it cannot be smaller since terms climb past S minus epsilon for every epsilon.
Concept
Boundedness is not optional. An increasing sequence with no ceiling does not settle; it marches off to infinity.
\[ (a_n)\text{ increasing and unbounded above} \;\Rightarrow\; a_n \to +\infty \]
So for a monotone sequence there are exactly two fates: converge to a finite limit (if bounded) or diverge to infinity (if not). No oscillation is possible.
Step zero
Discussion prompt
Worked example: geometric partial sums by monotone convergence — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Show the sequence is increasing
Answer:
Worked example
Consider the running totals of a halving sum. We show they converge without summing the infinite series directly.
\[ a_n = 1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots + \tfrac{1}{2^n} \]
Show the sequence is increasing
Why: Each new term adds a strictly positive amount, so every total exceeds the one before it.
\[ a_{n+1} - a_n = \tfrac{1}{2^{n+1}} > 0 \]
Show it is bounded above by 2
Why: The closed form of the finite geometric sum is below 2 for every n, so 2 is a ceiling.
\[ a_n = \frac{1 - (1/2)^{n+1}}{1 - 1/2} = 2 - \tfrac{1}{2^n} < 2 \]
Conclude convergence
Why: Increasing and bounded above, so the Monotone Convergence Theorem applies; the limit is the supremum.
\[ a_n \to \sup_n a_n = 2 \]
Verify the limit against the closed form
Why: The explicit formula 2 minus one over two-to-the-n has one-over-two-to-the-n going to 0, confirming the limit is exactly 2.
\[ \lim_{n\to\infty}\left(2 - \tfrac{1}{2^n}\right) = 2 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "geometric partial sums by monotone convergence", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The explicit formula 2 minus one over two-to-the-n has one-over-two-to-the-n going to 0, confirming the limit is exactly 2.
Trap
The tempting shortcut: 'the sequence is increasing, so it must converge.'
\[ a_n = n : \quad 1, 2, 3, 4, 5, \ldots \]
This is strictly increasing, yet it converges to nothing. It runs away to infinity. Monotonicity by itself buys you no limit.
Monotone convergence needs BOTH monotone and bounded. Add a ceiling and the conclusion returns.
\[ a_n = 1 - \tfrac{1}{n} : \quad 0, \tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \ldots \to 1 \]
Increasing and bounded above by 1, so it converges, and the limit is the supremum 1. Always check the bound before invoking the theorem.
Concept
Many sequences are given not by a formula in n but by a starting value and a rule that builds the next term from the current one.
\[ a_1 = c, \qquad a_{n+1} = f(a_n) \]
For these, the Monotone Convergence Theorem is the tool of choice: prove the sequence is bounded and monotone, and it must converge even before you know the value.
Intuition
If the terms converge to some limit, feeding that limit through the rule must return the limit itself. The limit sits still under the update rule.
\[ a_n \to L \;\Rightarrow\; L = f(L) \]
So a converging recursion climbs toward a fixed point of its rule, like a ball rolling to the bottom of a valley. Solving the fixed-point equation reveals the wall the terms press against.
Estimation
Predict first
A classic. Start at 1 and repeatedly add 2 and take the square root. We prove it converges and find the limit.
Commit before you compute: what does Worked example: a recursive square-root sequence converges come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the limit satisfies the fixed point
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Discard the negative root because all terms are positive, leaving L equal to 2.
Worked example
A classic. Start at 1 and repeatedly add 2 and take the square root. We prove it converges and find the limit.
\[ a_1 = 1, \qquad a_{n+1} = \sqrt{2 + a_n} \]
Bound it above by 2 (induction)
Why: If a term is below 2 then the next is the square root of something below 4, hence below 2. The base term is 1, below 2, so the bound holds for all n.
\[ a_n < 2 \;\Rightarrow\; a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2 \]
Show it is increasing
Why: The gap a_{n+1} exceeds a_n exactly when 2 plus a_n exceeds a_n squared, i.e. when the quadratic is negative. On the range from 0 to 2 both factors give a negative product.
\[ a_{n+1} > a_n \iff a_n^2 - a_n - 2 < 0 \iff (a_n-2)(a_n+1) < 0 \]
Invoke monotone convergence
Why: Increasing and bounded above by 2, so by the theorem the sequence converges to some finite limit L, which we can now solve for.
\[ 0 < a_n < 2 \;\Rightarrow\; (a_n-2)(a_n+1) < 0 \;\Rightarrow\; a_{n+1} > a_n \]
Solve the fixed-point equation
Why: Both sides converge to L; passing to the limit in the recursion gives a quadratic. The negative root is impossible since every term is positive.
\[ L = \sqrt{2 + L} \;\Rightarrow\; L^2 - L - 2 = 0 \;\Rightarrow\; L = 2 \;\text{ or }\; L = -1 \]
Verify the limit satisfies the fixed point
Why: Discard the negative root because all terms are positive, leaving L equal to 2. Check: the square root of 2 plus 2 is the square root of 4, which is 2. The fixed point holds exactly.
\[ \sqrt{2 + 2} = \sqrt{4} = 2 \;\checkmark \qquad L = 2 \]
Picture it
Animation
Shows: Each line of the worked example "a recursive square-root sequence converges", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Discard the negative root because all terms are positive, leaving L equal to 2. Check: the square root of 2 plus 2 is the square root of 4, which is 2. The fixed point holds exactly.
Missing information
Discussion prompt
The Newton iteration for the square root of 2 gives a decreasing sequence bounded below. Same tool, opposite direction.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The average of a positive number and 2 over that number is at least the square root of their product by AM-GM, and that product is exactly 2.
Worked example
The Newton iteration for the square root of 2 gives a decreasing sequence bounded below. Same tool, opposite direction.
\[ a_1 = 2, \qquad a_{n+1} = \tfrac{1}{2}\!\left(a_n + \tfrac{2}{a_n}\right) \]
Bound it below by root two
Why: The average of a positive number and 2 over that number is at least the square root of their product by AM-GM, and that product is exactly 2.
\[ a_{n+1} = \tfrac{1}{2}\!\left(a_n + \tfrac{2}{a_n}\right) \ge \sqrt{a_n \cdot \tfrac{2}{a_n}} = \sqrt{2} \]
Show it is decreasing
Why: Once a term is at least the square root of 2, its square is at least 2, so the update subtracts a nonnegative amount. The sequence never climbs.
\[ a_{n+1} - a_n = \frac{2 - a_n^2}{2\,a_n} \le 0 \quad (\text{since } a_n^2 \ge 2) \]
Invoke monotone convergence and solve
Why: Decreasing and bounded below by the square root of 2, so it converges; the limit is a fixed point of the update.
\[ L = \tfrac{1}{2}\!\left(L + \tfrac{2}{L}\right) \;\Rightarrow\; L = \tfrac{2}{L} \;\Rightarrow\; L^2 = 2 \]
Verify the fixed point
Why: The positive root is the square root of 2. Check: averaging that value with 2 divided by it returns the same value, so the limit is exactly the square root of 2.
\[ \tfrac{1}{2}\!\left(\sqrt{2} + \tfrac{2}{\sqrt{2}}\right) = \tfrac{1}{2}\!\left(\sqrt{2} + \sqrt{2}\right) = \sqrt{2} \;\checkmark \]
Concept
A subsequence is what you get by keeping infinitely many terms of a sequence, in their original order, and throwing the rest away.
\[ n_1 < n_2 < n_3 < \cdots \qquad\Longrightarrow\qquad (a_{n_k})_{k \ge 1} \]
subsequence — A sequence formed by a strictly increasing choice of indices. Because the indices strictly increase, the k-th chosen index is always at least k, which is the fact every subsequence proof leans on.
Intuition
Think of the full sequence as a numbered list. A subsequence highlights infinitely many rows, never re-ordering them. You may skip as much as you like but you can never go back.
Figure (svg): A row of eight dots with the first, third, fourth, and seventh circled to indicate a chosen subsequence.
Circled dots are the kept terms. Read left to right, they form a brand new sequence living inside the old one.
Concept
A number is a subsequential limit (a cluster value) of a sequence if some subsequence converges to it. One sequence can have many.
A key fact: the whole sequence converges to a limit exactly when it is bounded and has that single value as its ONLY subsequential limit. Two different cluster values block convergence.
\[ a_n \to L \;\Rightarrow\; \text{every subsequence } a_{n_k} \to L \]
Step zero
Discussion prompt
Worked example: subsequential limits of oscillating sequences — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split the first by parity of the index
Answer:
Worked example
We compute the cluster values of two oscillating sequences to see why oscillation blocks a single limit.
\[ a_n = (-1)^n \qquad\qquad b_n = (-1)^n\!\left(1 + \tfrac{1}{n}\right) \]
Split the first by parity of the index
Why: Even indices give plus one and odd indices give minus one, so the even subsequence and odd subsequence are each constant.
\[ a_{2k} = 1 \to 1, \qquad a_{2k+1} = -1 \to -1 \]
Do the same for the second
Why: The extra factor tends to 1, so the even terms approach plus one and the odd terms approach minus one from outside.
\[ b_{2k} = 1 + \tfrac{1}{2k} \to 1, \qquad b_{2k+1} \to -1 \]
Read off the cluster values
Why: Both sequences have exactly the two subsequential limits plus one and minus one, and no others.
\[ \text{subsequential limits} = \{\, -1,\; +1 \,\} \]
Verify neither sequence converges
Why: A convergent sequence forces every subsequence to the same value. Here two subsequences head to different values, so no overall limit exists. Check: plus one is not minus one.
\[ 1 \ne -1 \;\Rightarrow\; \lim_n a_n \text{ and } \lim_n b_n \text{ do not exist} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "subsequential limits of oscillating sequences", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A convergent sequence forces every subsequence to the same value. Here two subsequences head to different values, so no overall limit exists. Check: plus one is not minus one.
Trap
The tempting error: 'I found a subsequence converging to 1, therefore the sequence converges to 1.'
\[ a_n = (-1)^n, \qquad a_{2k} = 1 \to 1 \]
The even subsequence really does go to 1. But the odd subsequence goes to minus one. A single subsequence tells you about a cluster value, not about the sequence as a whole.
A subsequence pins down the limit only when you ALSO know the sequence converges. Then all subsequences share that one value.
\[ (a_n)\text{ converges} \;\Rightarrow\; \lim_k a_{n_k} = \lim_n a_n \text{ for every subsequence} \]
So to conclude the sequence has limit L, first establish convergence (often via Cauchy or monotone), then a convenient subsequence identifies L.
Concept
The second headline theorem drops the monotonicity assumption entirely. Boundedness alone rescues a piece of the sequence.
\[ (a_n)\text{ bounded} \;\Rightarrow\; (a_n)\text{ has a convergent subsequence} \]
It does not promise the whole sequence converges. It promises you can always extract an infinite, convergent piece from any bounded sequence, no matter how wildly it jumps.
Picture it
Figure (svg): A bounded interval with many dots, denser near one location marked as a cluster point.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Put infinitely many points into an interval of finite length. They cannot all stay far apart, because there is only so much room. Somewhere they must pile up.
Intuition
Put infinitely many points into an interval of finite length. They cannot all stay far apart, because there is only so much room. Somewhere they must pile up.
Figure (svg): A bounded interval with many dots, denser near one location marked as a cluster point.
That pile-up spot is where a convergent subsequence lives. The proof just makes 'pile up' precise by trapping infinitely many terms in shrinking intervals.
Concept
The engine of the proof is repeated halving. Cut the interval in two. Since the whole contains infinitely many terms, at least one half must also contain infinitely many.
\[ \text{infinite in } [a,b] \;\Rightarrow\; \text{infinite in } \big[a,\tfrac{a+b}{2}\big] \text{ or } \big[\tfrac{a+b}{2},b\big] \]
Keep a half that still holds infinitely many terms and repeat forever. The intervals nest and their lengths halve toward zero.
Hypothesis
Predict first
Worked example: Bolzano-Weierstrass by bisection is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Halve and keep an infinite half
Why: The starting interval contains all infinitely many terms. Splitting it, at least one half still contains infinitely many. Name that half and repeat, generating nested intervals.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Let the sequence be bounded, so every term lives in some interval from negative M to M. We extract a convergent subsequence.
\[ -M \le a_n \le M \quad \text{for all } n \]
Halve and keep an infinite half
Why: The starting interval contains all infinitely many terms. Splitting it, at least one half still contains infinitely many. Name that half and repeat, generating nested intervals.
\[ I_0 = [-M, M] \supseteq I_1 \supseteq I_2 \supseteq \cdots, \qquad |I_k| = \tfrac{2M}{2^{k}} \]
Choose one index from each interval, in order
Why: Because each interval holds infinitely many terms, we can always pick a NEW index larger than the last, keeping the choice strictly increasing so it is a genuine subsequence.
\[ n_1 < n_2 < \cdots, \qquad a_{n_k} \in I_k \]
Identify the limit point
Why: The nested intervals have lengths shrinking to zero, so by the nested interval property they share exactly one common point L, and completeness is what guarantees the intersection is nonempty.
\[ \bigcap_{k} I_k = \{L\} \]
Verify the chosen subsequence converges to L
Why: Both the k-th term and L sit in the same interval of length 2M over two-to-the-k, so their distance is at most that length, which tends to zero. Check: pick k large enough that the interval is shorter than any target tolerance.
\[ |a_{n_k} - L| \le |I_k| = \tfrac{2M}{2^{k}} \to 0 \;\Rightarrow\; a_{n_k} \to L \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Bolzano-Weierstrass by bisection", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both the k-th term and L sit in the same interval of length 2M over two-to-the-k, so their distance is at most that length, which tends to zero. Check: pick k large enough that the interval is shorter than any target tolerance.
Concept
There is a slicker route to Bolzano-Weierstrass through a surprising fact about EVERY sequence, bounded or not.
\[ \text{Every real sequence has a monotone subsequence.} \]
peak — An index m is a peak of the sequence if no later term ever exceeds the term at m. That is, a_m is at least a_n for every n greater than m. Peaks are the pivot of the lemma's proof.
Definition probe
Sort into buckets
Every line below is part of the definition of subsequence or of peak — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: every sequence has a monotone subsequence into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. List the peaks in increasing index order.
Worked example
We prove the lemma by splitting on how many peaks the sequence has. Exactly one of two cases must hold.
Case 1: infinitely many peaks
Why: List the peaks in increasing index order. By the defining property of a peak, each peak term dominates all later terms, in particular the next peak. So the peak terms form a decreasing subsequence.
\[ m_1 < m_2 < \cdots \;\Rightarrow\; a_{m_1} \ge a_{m_2} \ge a_{m_3} \ge \cdots \]
Case 2: only finitely many peaks
Why: Go past the last peak. Any index there is not a peak, so by definition some later term is strictly larger; jump to it. That index is also not a peak, so repeat, building a strictly increasing subsequence.
\[ n_1 < n_2 < \cdots \;\text{with}\; a_{n_1} < a_{n_2} < a_{n_3} < \cdots \]
Combine the cases
Why: Case 1 gives a decreasing subsequence, case 2 an increasing one. Every sequence falls into one case, so a monotone subsequence always exists.
Either way we obtained a monotone subsequence.
Verify the two cases are exhaustive
Why: The number of peaks is either infinite or finite, with no third option, and each possibility was handled. Check: a sequence with, say, three peaks lands in case 2 and yields an increasing subsequence beyond the third peak.
\[ \#\{\text{peaks}\} = \infty \;\text{ or }\; \#\{\text{peaks}\} < \infty \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "every sequence has a monotone subsequence", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The number of peaks is either infinite or finite, with no third option, and each possibility was handled. Check: a sequence with, say, three peaks lands in case 2 and yields an increasing subsequence beyond the third peak.
Concept
Now Bolzano-Weierstrass falls out in one line. Take any bounded sequence.
By the lemma it has a monotone subsequence. That subsequence inherits the bound, so it is monotone and bounded, and the Monotone Convergence Theorem makes it converge.
\[ \text{bounded} \;\xrightarrow{\text{lemma}}\; \text{monotone subsequence} \;\xrightarrow{\text{MCT}}\; \text{convergent subsequence} \]
Trap
The tempting overreach: 'the sequence is bounded, and Bolzano-Weierstrass gives a convergent subsequence, so the sequence converges.'
\[ a_n = (-1)^n : \quad -1, 1, -1, 1, \ldots \]
This is bounded, and it does have convergent subsequences. But the sequence itself oscillates forever and has no limit. Bolzano-Weierstrass promises a subsequence, never the whole.
A bounded sequence converges exactly when all its subsequential limits coincide. One cluster value means convergence; two or more means oscillation.
\[ \text{bounded} + \text{unique subsequential limit} \;\Rightarrow\; \text{convergent} \]
So Bolzano-Weierstrass is an existence result about pieces. To conclude the whole sequence converges you need an extra structural hypothesis, such as monotone or Cauchy.
Break the constraint
Discussion prompt
The rule this trap just fixed:
A bounded sequence converges exactly when all its subsequential limits coincide. One cluster value means convergence; two or more means oscillation.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
We used this inside the bisection proof; here it is on its own. A shrinking tower of closed intervals never closes down to nothing.
\[ I_1 \supseteq I_2 \supseteq \cdots, \;\; |I_n| \to 0 \;\Rightarrow\; \bigcap_n I_n = \{ x \} \text{ for a unique } x \]
This is another face of completeness. In the rationals a nested tower can squeeze down on a gap where no rational lives, and the intersection is empty.
Estimation
Predict first
We prove the nested interval property from monotone convergence, tying the two together.
Commit before you compute: what does Worked example: nested intervals pin down a point come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the point lies in every interval and is unique
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The common value sits at or above every left endpoint and at or below every right endpoint, so it belongs to each interval.
Worked example
We prove the nested interval property from monotone convergence, tying the two together.
\[ I_n = [a_n, b_n], \quad I_{n+1} \subseteq I_n, \quad b_n - a_n \to 0 \]
The left endpoints climb, the right endpoints fall
Why: Nesting means each new interval sits inside the last, so left endpoints increase and right endpoints decrease. Each left endpoint stays below the fixed right endpoint b_1, so the left endpoints are bounded above.
\[ a_1 \le a_2 \le \cdots \le b_1, \qquad b_1 \ge b_2 \ge \cdots \ge a_1 \]
Both endpoint sequences converge
Why: Increasing-and-bounded and decreasing-and-bounded each converge by the Monotone Convergence Theorem, to a supremum and an infimum respectively.
\[ a_n \uparrow a = \sup_n a_n, \qquad b_n \downarrow b = \inf_n b_n \]
The two limits coincide
Why: The gap between them is the limit of b_n minus a_n, which is zero by hypothesis, so the two limits are equal and the common point lies in every interval.
\[ b - a = \lim_n (b_n - a_n) = 0 \;\Rightarrow\; a = b \]
Verify the point lies in every interval and is unique
Why: The common value sits at or above every left endpoint and at or below every right endpoint, so it belongs to each interval. Any two points of the intersection differ by at most the shrinking lengths, hence by zero. Check: the intersection is exactly one point.
\[ a = b \in \bigcap_n I_n, \qquad \bigcap_n I_n = \{a\} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "nested intervals pin down a point", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The common value sits at or above every left endpoint and at or below every right endpoint, so it belongs to each interval. Any two points of the intersection differ by at most the shrinking lengths, hence by zero. Check: the intersection is exactly one point.
Concept
The third headline idea changes the question. Instead of asking whether terms approach a known target, ask whether the terms bunch together among THEMSELVES.
\[ \forall \varepsilon > 0 \;\; \exists N \;\; \forall m, n \ge N \;\; |a_n - a_m| < \varepsilon \]
Cauchy sequence — A sequence whose terms eventually all lie within any prescribed tolerance of each other. The definition never mentions a limit value, only the mutual closeness of late terms.
Intuition
Convergence measures distance to an external point L. Cauchy measures distance between the terms themselves. The beauty is you can test Cauchy without knowing, or even having, a limit.
This is exactly what you want for an algorithm: you can certify that its iterates are settling down purely by watching successive outputs stop changing, before you know the answer they are settling on.
\[ \text{convergent: close to } L \qquad\quad \text{Cauchy: close to each other} \]
Concept
The easy direction holds in any setting: if a sequence has a limit, its terms must also bunch together.
\[ a_n \to L \;\Rightarrow\; (a_n) \text{ is Cauchy} \]
The idea is the triangle inequality: if two terms are each close to L, they are close to each other. We make it precise next.
Step zero
Discussion prompt
Worked example: convergent implies Cauchy — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Turn the tolerance into a half-tolerance
Answer:
Worked example
Assume the sequence converges to L. We show it is Cauchy directly from the definition.
Turn the tolerance into a half-tolerance
Why: Given a target closeness epsilon, apply convergence with half of it, getting an index past which every term is within epsilon over two of L.
\[ \exists N \;\; \forall n \ge N \;\; |a_n - L| < \tfrac{\varepsilon}{2} \]
Add two closeness bounds via the triangle inequality
Why: For any two indices past N, insert L in the middle. Each piece is under epsilon over two, so the total is under epsilon.
\[ |a_n - a_m| \le |a_n - L| + |L - a_m| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \]
Verify the Cauchy condition holds
Why: The same N witnesses the Cauchy definition for the original epsilon. Check: with epsilon equal to one hundredth, the terms past N are within one hundredth of each other because each is within one two-hundredth of L.
\[ \forall \varepsilon>0 \;\exists N \;\forall m,n\ge N \;\; |a_n - a_m| < \varepsilon \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "convergent implies Cauchy", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The same N witnesses the Cauchy definition for the original epsilon. Check: with epsilon equal to one hundredth, the terms past N are within one hundredth of each other because each is within one two-hundredth of L.
Concept
Before proving the hard direction we need a stepping stone: a Cauchy sequence cannot run off to infinity.
\[ (a_n)\text{ Cauchy} \;\Rightarrow\; (a_n)\text{ bounded} \]
This is what will let us feed a Cauchy sequence into Bolzano-Weierstrass. Only finitely many early terms sit outside a fixed band, and finitely many things are always bounded.
Missing information
Discussion prompt
We bound a Cauchy sequence by pinning down its tail and handling the finitely many early terms separately.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Apply the Cauchy definition with tolerance one to get an index N past which every term is within one of the anchor term a_N.
Worked example
We bound a Cauchy sequence by pinning down its tail and handling the finitely many early terms separately.
Fix the tolerance at one
Why: Apply the Cauchy definition with tolerance one to get an index N past which every term is within one of the anchor term a_N.
\[ \exists N \;\; \forall n \ge N \;\; |a_n - a_N| < 1 \;\Rightarrow\; |a_n| < |a_N| + 1 \]
Bundle the finite head with the tail bound
Why: The terms before N are finitely many, so they have a largest absolute value. The maximum of that and the tail bound dominates every term.
\[ B = \max\{\, |a_1|, \ldots, |a_{N-1}|, \; |a_N| + 1 \,\} \]
Verify the bound covers all terms
Why: Early terms are below B because B is their maximum; late terms are below the tail piece of B. Check: every single term, head or tail, satisfies the bound.
\[ |a_n| \le B \quad \text{for all } n \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Cauchy implies bounded", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Early terms are below B because B is their maximum; late terms are below the tail piece of B. Check: every single term, head or tail, satisfies the bound.
Concept
The final piece: if a Cauchy sequence has even one convergent subsequence, the whole sequence is dragged to the same limit.
\[ (a_n)\text{ Cauchy},\;\; a_{n_k} \to L \;\Rightarrow\; a_n \to L \]
Cauchy says all late terms are near each other; the subsequence says some late terms are near L; combine them and all late terms are near L.
Estimation
Predict first
Assume the sequence is Cauchy and some subsequence converges to L. We show the full sequence converges to L.
Commit before you compute: what does Worked example: the drag lemma come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the whole sequence converges to L
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The bound holds for every index past N and every epsilon, which is precisely convergence to L.
Worked example
Assume the sequence is Cauchy and some subsequence converges to L. We show the full sequence converges to L.
Use Cauchy with a half-tolerance
Why: Given epsilon, Cauchy gives an index N past which any two terms are within epsilon over two.
\[ \exists N \;\; \forall m,n \ge N \;\; |a_n - a_m| < \tfrac{\varepsilon}{2} \]
Grab one subsequence term deep enough
Why: The subsequence converges to L, so choose an index in it that is both past N and within epsilon over two of L. Such an index exists because the subsequence indices grow without bound.
\[ \exists\, n_k \ge N \;\; \text{with}\;\; |a_{n_k} - L| < \tfrac{\varepsilon}{2} \]
Chain the two bounds
Why: For any index past N, it is within epsilon over two of that subsequence term, which is within epsilon over two of L. The triangle inequality gives within epsilon of L.
\[ |a_n - L| \le |a_n - a_{n_k}| + |a_{n_k} - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \]
Verify the whole sequence converges to L
Why: The bound holds for every index past N and every epsilon, which is precisely convergence to L. Check: the limit equals the subsequential limit, as it must.
\[ n \ge N \;\Rightarrow\; |a_n - L| < \varepsilon \;\Rightarrow\; a_n \to L \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the drag lemma", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The bound holds for every index past N and every epsilon, which is precisely convergence to L. Check: the limit equals the subsequential limit, as it must.
Concept
Now assemble the pieces. In the real numbers, every Cauchy sequence converges. This property is called completeness.
The chain: Cauchy makes the sequence bounded; Bolzano-Weierstrass extracts a convergent subsequence; the drag lemma pulls the whole sequence to that same limit.
\[ \text{Cauchy} \;\xrightarrow{\;}\; \text{bounded} \;\xrightarrow{\text{B-W}}\; \text{conv. subseq.} \;\xrightarrow{\text{drag}}\; \text{convergent} \]
Intuition
Cauchy alone cannot produce a limit out of thin air; it needs the ambient space to actually contain the point the terms are crowding toward.
Bolzano-Weierstrass is where completeness of the reals enters, since its proof used nested intervals. That is the exact step that fails over the rationals, as we are about to see.
\[ \text{Cauchy} + \text{completeness} \;\Longleftrightarrow\; \text{convergent} \]
Concept
The reverse implication depends on the space. Over the rationals a sequence can be Cauchy yet converge to nothing, because its target is an irrational hole.
\[ \text{in } \mathbb{Q}: \quad \text{Cauchy} \;\not\Rightarrow\; \text{convergent} \]
This is the precise sense in which the rationals are incomplete and the reals are their completion: the reals are what you get by filling every such gap.
Step zero
Discussion prompt
Worked example: decimal truncations of root two — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Show the sequence is Cauchy
Answer:
Worked example
Let the sequence be the successive decimal truncations of the square root of 2, each a perfectly good rational number.
\[ x_1 = 1.4,\; x_2 = 1.41,\; x_3 = 1.414,\; x_4 = 1.4142,\; \ldots \]
Show the sequence is Cauchy
Why: Two truncations agree on all leading digits up to the shorter length, so their difference is bounded by a power of ten that shrinks to zero.
\[ |x_n - x_m| \le 10^{-\min(n,m)+1} \to 0 \]
Suppose it converged to a rational r
Why: In the reals these truncations converge to the square root of 2. A limit is unique, so any rational limit would have to equal the square root of 2.
\[ x_n \to r \in \mathbb{Q} \;\Rightarrow\; r = \sqrt{2} \]
Reach the contradiction
Why: The square root of 2 is irrational, so no rational r can be the limit. The sequence is Cauchy in the rationals but has no rational limit.
\[ \sqrt{2} \notin \mathbb{Q} \;\Rightarrow\; \text{no rational limit exists} \]
Verify via the squares
Why: The squares of the truncations approach 2, so any rational limit would have to square to 2, which is impossible in the rationals. Check: 1.414 squared is about 1.9994, marching toward 2 but never landing on a rational whose square is exactly 2.
\[ x_n^2 \to 2, \quad \text{but } r^2 = 2 \text{ has no solution in } \mathbb{Q} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "decimal truncations of root two", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The squares of the truncations approach 2, so any rational limit would have to square to 2, which is impossible in the rationals. Check: 1.414 squared is about 1.9994, marching toward 2 but never landing on a rational whose square is exactly 2.
Trap
The tempting overgeneralization: 'Cauchy sequences always converge, so this rational Cauchy sequence has a limit.'
\[ x_n \in \mathbb{Q}, \;\; x_n \to \sqrt{2} \notin \mathbb{Q} \]
Inside the rationals this Cauchy sequence has no limit at all, because the point it aims for is not a rational number. Completeness is a property of the SPACE, not of the sequence.
The correct statement carries its space with it: in a COMPLETE space, Cauchy and convergent coincide; the reals are complete, the rationals are not.
\[ \text{in } \mathbb{R}: \;\text{Cauchy} \iff \text{convergent} \qquad \text{in } \mathbb{Q}: \;\text{Cauchy} \;\not\Rightarrow\; \text{convergent} \]
So always name the space. The same list of numbers converges in the reals and fails to converge in the rationals.
Concept
Everything in this deck flows from one axiom wearing three costumes. Over an Archimedean ordered field, these are logically equivalent.
\[ \textbf{(1)}\;\; \text{every nonempty bounded-above set has a supremum (LUB)} \]
\[ \textbf{(2)}\;\; \text{every bounded monotone sequence converges (MCT)} \]
\[ \textbf{(3)}\;\; \text{every Cauchy sequence converges (Cauchy completeness)} \]
Picture it
Figure (svg): A number line with a small dashed gap between two arrows, one labelled rationals with a hole, one labelled reals filled in.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Each formulation says the line has no holes, but through a different door. The least-upper-bound version speaks of sets, the monotone version of climbing sequences, the Cauchy version of self-bunching sequences.
Intuition
Each formulation says the line has no holes, but through a different door. The least-upper-bound version speaks of sets, the monotone version of climbing sequences, the Cauchy version of self-bunching sequences.
Figure (svg): A number line with a small dashed gap between two arrows, one labelled rationals with a hole, one labelled reals filled in.
The rationals satisfy none of the three; a bounded-above set of rationals can miss its supremum, a bounded monotone rational sequence can climb toward an irrational, and a rational Cauchy sequence can aim at a hole.
Concept
These theorems are the rigor behind 'the algorithm's iterates settle down.' Fixed-point iteration, gradient descent, and Newton's method all produce recursively defined sequences.
The Cauchy criterion is the practical one: you certify convergence by watching successive iterates stop changing, without knowing the answer. Completeness of the reals guarantees the answer they approach actually exists.
Floating-point numbers, by contrast, form an incomplete and even finite world, which is exactly why numerical convergence needs the real-number theory standing behind it.
Concept
Underlying every 'THE limit' statement is a fact worth isolating: a sequence cannot converge to two different values.
If it did, pick a tolerance smaller than half the gap between the two candidates. Past some index the terms would have to sit within that tolerance of both at once, which is impossible.
\[ a_n \to L \;\text{and}\; a_n \to L' \;\Rightarrow\; L = L' \]
Explain it
Discussion prompt
Explain Limits are unique to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Underlying every 'THE limit' statement is a fact worth isolating: a sequence cannot converge to two different values.
Concept
One small fact powers nearly every subsequence proof: because the chosen indices strictly increase through the whole numbers, the k-th index is at least k.
\[ n_1 < n_2 < \cdots \;\Rightarrow\; n_k \ge k \quad \text{for all } k \]
So 'deep into the subsequence' automatically means 'deep into the original sequence,' which is exactly what lets a Cauchy bound past some index N reach every subsequence term past that point.
Analogy
Discussion prompt
Explain How far a subsequence has travelled by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
One small fact powers nearly every subsequence proof: because the chosen indices strictly increase through the whole numbers, the k-th index is at least k.
Ranking
Put in order
Put the moves of Worked example: an averaging recursion into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. If a term is below 6 then averaging it with 6 stays below 6.
Worked example
One more recursive convergence, to drill the pattern on a fresh rule that averages toward a target.
\[ a_1 = 0, \qquad a_{n+1} = \tfrac{1}{2}(a_n + 6) \]
Bound it above by 6 (induction)
Why: If a term is below 6 then averaging it with 6 stays below 6. The base term 0 is below 6, so every term is.
\[ a_n < 6 \;\Rightarrow\; a_{n+1} = \tfrac{1}{2}(a_n + 6) < \tfrac{1}{2}(6+6) = 6 \]
Show it is increasing
Why: The difference between consecutive terms is half the distance from the current term to 6, which is positive while the term is below 6.
\[ a_{n+1} - a_n = \tfrac{6 - a_n}{2} > 0 \]
Invoke monotone convergence and solve
Why: Increasing and bounded above by 6, so it converges; the limit is the fixed point of the averaging rule.
\[ L = \tfrac{1}{2}(L + 6) \;\Rightarrow\; 2L = L + 6 \;\Rightarrow\; L = 6 \]
Verify the fixed point
Why: Averaging 6 with 6 returns 6, so the limit is consistent with the rule. Check: the terms 0, 3, 4.5, 5.25 climb steadily toward 6 without ever passing it.
\[ \tfrac{1}{2}(6 + 6) = 6 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "an averaging recursion", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Averaging 6 with 6 returns 6, so the limit is consistent with the rule. Check: the terms 0, 3, 4.5, 5.25 climb steadily toward 6 without ever passing it.
Trap
The tempting slip: 'the sequence converges to its supremum, so some term actually equals the supremum.'
\[ a_n = 1 - \tfrac{1}{n}, \qquad \sup_n a_n = 1 \]
The supremum is 1, yet no term ever equals 1; every term falls short. A supremum that a monotone sequence converges to is a limit, not necessarily a maximum.
Distinguish the least upper bound from a largest element. The Monotone Convergence Theorem promises the terms approach the supremum, and says nothing about reaching it.
\[ a_n \to 1, \qquad a_n < 1 \;\text{ for every } n \]
A term equals the supremum only in special cases, such as an eventually constant sequence. For a strictly increasing bounded sequence the limit is always just out of reach.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Intuition
Think of each term as an estimate of an unknown answer. The Cauchy condition says the estimates eventually stop disagreeing with one another by more than any tolerance you name.
You never had to know the true answer to check this; you only compared estimates to estimates. Completeness is the promise that such self-consistent estimates really are closing in on something that exists.
\[ m, n \ge N \;\Rightarrow\; |a_n - a_m| < \varepsilon \]
Counterexample
Discussion prompt
Think of each term as an estimate of an unknown answer. The Cauchy condition says the estimates eventually stop disagreeing with one another by more than any tolerance you name.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Pattern
1. Guess and prove a bound by induction
Why: Show the update rule keeps terms inside a fixed interval; the base case is the starting value, the step reuses the rule.
2. Prove monotonicity
Why: Compare consecutive terms, usually by sign-analysing a_{n+1} minus a_n; a bound from step 1 often supplies the needed inequality.
3. Invoke the Monotone Convergence Theorem
Why: Bounded and monotone forces a finite limit to exist, before you know its value.
4. Solve the fixed-point equation and verify
Why: Pass to the limit in the recursion, solve for L, discard roots that violate the bound, and check the survivor satisfies the equation.
Pattern
1. Establish boundedness
Why: Find a fixed interval containing every term; Cauchy sequences qualify automatically, since Cauchy implies bounded.
2. Extract a convergent subsequence
Why: Bolzano-Weierstrass guarantees one exists; via bisection or via the monotone subsequence lemma plus the Monotone Convergence Theorem.
3. Do not overclaim
Why: The subsequence converges, not necessarily the whole sequence; upgrade to full convergence only with monotonicity, Cauchy, or a unique cluster value.
Pattern
1. Split the tolerance
Why: To land inside epsilon, aim each partial estimate inside epsilon over two so the triangle inequality closes the gap.
2. Bound distances between terms, not to a limit
Why: The whole point of Cauchy is to avoid naming a limit; estimate the difference of two indices directly.
3. Invoke completeness to get the limit
Why: In the reals, once the sequence is Cauchy, a limit exists; in the rationals it need not, so always confirm the space is complete.
Real world
Discussion prompt
Outside this lesson: where does Monotone Convergence, Bolzano-Weierstrass & Cauchy actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: run a Cauchy argument is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers the three theorems that make "the iterates settle down" rigorous: the Monotone Convergence Theorem, the Bolzano-Weierstrass theorem, and the Cauchy criterion. It targets the misconceptions that monotonicity alone gives convergence, that every bounded sequence converges, that Cauchy implies convergent in any space, and that a subsequential limit is THE limit.
Elimination
Eliminate the wrong options
Using only the Monotone Convergence Theorem, which of these sequences is guaranteed to converge?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: B
Why: The sequence 1 minus 1 over n is strictly increasing and bounded above by 1, so the Monotone Convergence Theorem guarantees it converges to its supremum, which is 1. It is the only option that is both monotone and bounded.
Check
Check your understanding
Using only the Monotone Convergence Theorem, which of these sequences is guaranteed to converge?
Answer: B
Why: The sequence 1 minus 1 over n is strictly increasing and bounded above by 1, so the Monotone Convergence Theorem guarantees it converges to its supremum, which is 1. It is the only option that is both monotone and bounded.
Check
Recall the recursion from the worked example, which is increasing and bounded above by 2.
\[ a_1 = 1, \qquad a_{n+1} = \sqrt{2 + a_n} \]
Check your understanding
What is the limit of this sequence?
Answer: A
Why: Passing to the limit gives L equal to the square root of 2 plus L, so L squared minus L minus 2 equals 0, which factors as L minus 2 times L plus 1. The positive root is 2, and indeed the square root of 2 plus 2 equals 2.
Prediction
Predict first
What does Bolzano-Weierstrass guarantee about this sequence?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Some subsequence converges
Why: Bolzano-Weierstrass says a bounded sequence has a convergent subsequence, and here the even-indexed subsequence converges to 1. It never claims the full sequence converges, which this one does not.
Check
Consider a sequence that is bounded but oscillates, taking the values minus one and one alternately forever.
\[ a_n = (-1)^n \]
Check your understanding
What does Bolzano-Weierstrass guarantee about this sequence?
Answer: B
Why: Bolzano-Weierstrass says a bounded sequence has a convergent subsequence, and here the even-indexed subsequence converges to 1. It never claims the full sequence converges, which this one does not.
Commit first
Predict first
In which number system does this sequence converge?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: It converges in the reals but not in the rationals
Why: The limit is the square root of 2, which is real but not rational, so the sequence converges in the complete reals yet has no limit inside the incomplete rationals. This is the standard witness that the rationals are incomplete.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Take the decimal truncations of the square root of 2, each a rational number, forming a Cauchy sequence.
\[ 1.4,\; 1.41,\; 1.414,\; 1.4142,\; \ldots \]
Check your understanding
In which number system does this sequence converge?
Answer: B
Why: The limit is the square root of 2, which is real but not rational, so the sequence converges in the complete reals yet has no limit inside the incomplete rationals. This is the standard witness that the rationals are incomplete.
Prediction
Predict first
What is the set of subsequential limits of this sequence?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The set containing 1 and minus 1
Why: The even-indexed terms 1 plus 1 over 2k approach 1 and the odd-indexed terms approach minus 1, so the subsequential limits are exactly 1 and minus 1. No other value is a cluster point.
Check
Consider the sequence below, whose sign alternates while its size shrinks toward 1 from above.
\[ a_n = (-1)^n\!\left(1 + \tfrac{1}{n}\right) \]
Check your understanding
What is the set of subsequential limits of this sequence?
Answer: A
Why: The even-indexed terms 1 plus 1 over 2k approach 1 and the odd-indexed terms approach minus 1, so the subsequential limits are exactly 1 and minus 1. No other value is a cluster point.
Elimination
Eliminate the wrong options
Which statement correctly defines a Cauchy sequence?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The Cauchy condition is about mutual closeness of the terms, with no reference to any external limit: past some index, any two terms are within the given tolerance of each other. That is exactly option A.
Check
Sort the true definition of Cauchy from three plausible impostors.
Check your understanding
Which statement correctly defines a Cauchy sequence?
Answer: A
Why: The Cauchy condition is about mutual closeness of the terms, with no reference to any external limit: past some index, any two terms are within the given tolerance of each other. That is exactly option A.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: prove a recursive sequence converges · Pattern: apply Bolzano-Weierstrass · Pattern: run a Cauchy argument · Monotone sequences · Strict versus weak, and the umbrella word. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can prove a bounded monotone sequence converges, and use that to force recursive sequences to a limit you find by solving the fixed-point equation.
You can prove Bolzano-Weierstrass two ways, by bisection or via the monotone subsequence lemma, and you know it delivers a subsequence, never the whole sequence.
You can define a Cauchy sequence, prove Cauchy is equivalent to convergent in the reals, and explain why the same sequence fails in the rationals.
| Theorem | Hypothesis | Conclusion |
|---|---|---|
| Monotone Convergence | bounded and monotone | converges to sup or inf |
| Bolzano-Weierstrass | bounded | a convergent subsequence exists |
| Cauchy criterion | Cauchy, in the reals | the sequence converges |
These three, together with the least-upper-bound axiom, are the same fact of completeness seen from three sides: no gaps in the real line.
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