Course 2 Book 2A: Numbers, Limits and the IntegralChapter 6

Sequences

Limits, lim sup and lim inf, monotone convergence and Bolzano–Weierstrass.

27 min read · Updated Oct 2, 2026

Read with Tao, Analysis I, chapter "Limits of sequences" (convergence and limit laws, the extended real number system, suprema and infima of sequences, lim sup, lim inf and limit points, some standard limits, subsequences, real exponentiation part II).

In this chapter · 7 sections

6.1Convergence
6.1.1Limit laws
6.1.2Some standard limits
6.2Monotone sequences
6.3The extended reals, suprema and infima
6.4lim sup and lim inf
6.5Subsequences and Bolzano–Weierstrass
6.5.1Completeness of the real numbers
6.6Real exponentiation
6.7Exercises

With the real numbers built and the logic of quantifiers in hand, we can finally say what it means for a sequence to converge, and prove the theorems that make limits usable. Three results from this chapter recur all the way to Perelman:

Completeness: a sequence of real numbers converges exactly when it is Cauchy. This is the property that the construction of 2A.4 The Real Numbers was designed to give.
Monotone convergence: a monotone, bounded sequence converges. A quantity that only ever increases, and can't increase forever, must settle down. Perelman's proof is driven by quantities of exactly this kind.
Bolzano–Weierstrass: every bounded sequence has a convergent subsequence. This is the first compactness theorem of the guidebook, and the engine of the contradiction–compactness template of 2A.5 Quantifiers and the Shape of a Proof.

Between them sit the tools for sequences that don't converge, $\limsup$ and $\liminf$ , which describe their long-run upper and lower envelopes.

By the end of this chapter you will be able to:

prove that a sequence converges, or that it doesn't, directly from the definition;
use the limit laws, the squeeze theorem and the standard limits;
prove and apply the monotone convergence theorem;
compute $\limsup$ and $\liminf$ , and use them to characterise convergence;
prove Bolzano–Weierstrass, and use it to show that Cauchy sequences of reals converge.

Convergence

Definition 6.1 Convergence

A sequence of real numbers $(a_n)$ converges to a real number $L$ if for every $\varepsilon > 0$ there is an $N$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$ . We then write $a_n \to L$ or $\lim_{n\to\infty} a_n = L$ , and call $L$ the limit. A sequence that converges to no real number diverges.

In quantifier form, $\forall \varepsilon > 0\ \exists N\ \forall n \geq N: |a_n - L| \leq \varepsilon$ . As a game (2A.5 Quantifiers and the Shape of a Proof): the adversary names a tolerance, you name a point beyond which every term stays within that tolerance of $L$ . The only difference from the Cauchy condition is that the terms are compared to a fixed $L$ instead of to each other.

Figure 6.1. Convergence to

L

: for every band of half-width

\varepsilon

around

L

, there is an index

N

after which all terms stay inside the band. A smaller

\varepsilon

may need a larger

N

In the world In use Settling time is N(ε)

Control engineers describe how quickly a system (a cruise control, a thermostat, a hard-disk head, a drone's attitude controller) reaches its target with a figure called the settling time. The usual definition is the time after which the response reaches and stays within 2% of its final value. That is the convergence definition with $\varepsilon$ equal to 2% of the final value. The settling time is the $N(\varepsilon)$ of the game, and the word "stays" is the $\forall n \geq N$ . A response that dips inside the band and then overshoots back out has not settled. Some specifications use 5% instead, which is the same idea with a different $\varepsilon$ .

Figure 6.2. A step response with a 2% band. The settling time is the last time the curve is outside the band; after it, every value lies within 2% of the target. That is

N(\varepsilon)

for

\varepsilon = 0.02

Proposition 6.2 Limits are unique

A sequence converges to at most one real number.

Proof. Suppose $a_n \to L$ and $a_n \to L'$ . Let $\varepsilon > 0$ . For $n$ large enough, both $|a_n - L| \leq \varepsilon/2$ and $|a_n - L'| \leq \varepsilon/2$ , so $|L - L'| \leq \varepsilon$ . As $\varepsilon$ was arbitrary, $L = L'$ (2A.5 Quantifiers and the Shape of a Proof).

Proposition 6.3 Convergent sequences are Cauchy and bounded

If $a_n \to L$ , then $(a_n)$ is Cauchy, and hence bounded.

Proof. Given $\varepsilon$ , choose $N$ with $|a_n - L| \leq \varepsilon/2$ for $n \geq N$ . Then for $n, m \geq N$ , $|a_n - a_m| \leq |a_n - L| + |L - a_m| \leq \varepsilon$ . Boundedness follows as in 2A.4 The Real Numbers.

The converse, that every Cauchy sequence of real numbers converges, is the completeness of $\mathbb{R}$ . We prove it at the end of the chapter with Bolzano–Weierstrass. Before that, one link to 2A.4 The Real Numbers: the formal limits used to build the reals are genuine limits.

Proposition 6.4 Formal limits are limits

If $(a_n)$ is a Cauchy sequence of rationals, then $a_n \to \operatorname{LIM}_{m\to\infty} a_m$ in $\mathbb{R}$ .

This is the exercise "terms approach their formal limit" in 2A.4 The Real Numbers. From now on we write $\lim$ throughout.

Limit laws

Theorem 6.5 Limit laws

Suppose $a_n \to a$ and $b_n \to b$ . Then:

$a_n + b_n \to a + b$ and $c\,a_n \to c\,a$ for every real $c$ ;
$a_n b_n \to ab$ ;
if $b \neq 0$ , then $b_n \neq 0$ for all large $n$ , and $a_n / b_n \to a/b$ ;
if $a_n \leq b_n$ for all large $n$ , then $a \leq b$ .

Proof (Parts 2 and 4). (2) Write $a_n b_n - ab = a_n(b_n - b) + b(a_n - a)$ . The sequence $(a_n)$ is bounded, say by $M$ , and we may take $M \geq |b|$ and $M > 0$ . Given $\varepsilon$ , take $N$ with $|a_n - a| \leq \varepsilon/(2M)$ and $|b_n - b| \leq \varepsilon/(2M)$ for $n \geq N$ . Then $|a_n b_n - ab| \leq M \cdot \frac{\varepsilon}{2M} + M \cdot \frac{\varepsilon}{2M} = \varepsilon$ .

(4) Suppose $a > b$ and let $\varepsilon = (a - b)/3$ . For large $n$ , $a_n \geq a - \varepsilon$ and $b_n \leq b + \varepsilon$ , so $a_n - b_n \geq a - b - 2\varepsilon = \varepsilon > 0$ , contradicting $a_n \leq b_n$ .

Part (4) says that limits preserve non-strict inequalities but not strict ones: $1/n > 0$ for every $n$ , but the limit is $0$ . In the many estimates of later courses, a strict inequality that holds along a sequence typically survives only as $\leq$ in the limit.

Corollary 6.6 Squeeze theorem

If $a_n \leq b_n \leq c_n$ for all large $n$ , and $a_n \to L$ and $c_n \to L$ , then $b_n \to L$ .

Some standard limits

These are used so often that they are worth having as named facts.

Proposition 6.7 Standard limits

$1/n^{\alpha} \to 0$ for every rational $\alpha > 0$ .
$x^n \to 0$ if $|x| < 1$ ; the sequence $x^n$ diverges if $|x| > 1$ or $x = -1$ .
$x^{1/n} \to 1$ for every $x > 0$ .

Proof (Part 2, for 0 < x < 1). Write $x = \tfrac{1}{1 + h}$ with $h > 0$ . By the binomial expansion (or Bernoulli's inequality, proved by induction in Exercise 6.19), $(1 + h)^n \geq 1 + nh > nh$ . So $0 < x^n < \frac{1}{nh}$ , and the squeeze theorem with part 1 gives $x^n \to 0$ .

The proof of (2) is a model for many estimates: to show something tends to zero, bound it by something you already know tends to zero.

Monotone sequences

Definition 6.8 Monotone sequences

A sequence is increasing if $a_{n+1} \geq a_n$ for every $n$ , decreasing if $a_{n+1} \leq a_n$ for every $n$ , and monotone if it is one or the other. It is bounded above if some $M$ has $a_n \leq M$ for all $n$ .

Theorem 6.9 Monotone convergence theorem

An increasing sequence that is bounded above converges, and its limit is $\sup_n a_n$ . Likewise, a decreasing sequence bounded below converges to $\inf_n a_n$ .

Proof. Let $S = \sup_n a_n$ , which exists by the least upper bound property (2A.4 The Real Numbers). Let $\varepsilon > 0$ . Since $S - \varepsilon$ is not an upper bound, some term $a_N > S - \varepsilon$ . Since the sequence is increasing, $a_n \geq a_N > S - \varepsilon$ for every $n \geq N$ ; and $a_n \leq S$ because $S$ is an upper bound. So $|a_n - S| < \varepsilon$ for all $n \geq N$ .

This is where the least upper bound property earns its keep: it supplies the limit. Over the rationals the theorem is false. The decimal truncations of $\sqrt 2$ are increasing and bounded, but have no rational limit.

Example 6.10 Compound interest and the number e

Lend $1$ at an annual interest rate of $100\%$ , compounded $n$ times a year: after one year you have $a_n = (1 + \tfrac1n)^n$ . Compounding more often helps: $a_1 = 2$ , $a_2 = 2.25$ , $a_{12} \approx 2.613$ , $a_{365} \approx 2.7146$ . Does it grow without bound? Jacob Bernoulli asked exactly this question in 1683, about continuous compounding.

The sequence increases. By the binomial theorem,

a_n = \sum_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n} \frac{1}{k!}\Big(1 - \frac1n\Big)\Big(1 - \frac2n\Big)\cdots\Big(1 - \frac{k-1}{n}\Big).

Going from $n$ to $n + 1$ , each bracket increases and an extra non-negative term is added, so $a_{n+1} \geq a_n$ .

It is bounded. Each bracket is at most $1$ , so $a_n \leq \sum_{k=0}^n \frac{1}{k!} \leq 1 + 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^{n-1}} < 3$ , using $k! \geq 2^{k-1}$ .

By Theorem 6.9 the sequence converges. Its limit is the number $e = 2.71828\ldots$ . Continuous compounding at $100\%$ multiplies money by $e$ , not by infinity.

Figure 6.3.

(1 + \tfrac1n)^n

increases, but stays below

3

. A monotone bounded sequence has nowhere to go but its supremum: here,

e = 2.71828\ldots

Where this goes Monotone quantities and Perelman

"Monotone and bounded, therefore convergent" is the logical skeleton of Perelman's approach. His $\mathcal{W}$ -entropy (12A.3 The 𝓦-Entropy) and reduced volume (12A.5 Reduced Distance and Reduced Volume) are quantities that can only move in one direction along a Ricci flow and are bounded, so they converge. The real power comes from a second fact. Each monotonicity formula says how fast the quantity changes, and the rate is a sum of squares that vanishes only on very special geometries (solitons). So in the limit, where the quantity has stopped changing, the geometry must be one of those special ones. Bounded monotone convergence identifies the limit; the equality case identifies what it looks like. You will see the same two-step argument in miniature for entropy along the heat equation (6A.10 Entropy, Information and Diffusion).

The extended reals, suprema and infima

Unbounded sequences are often best described as heading to $+\infty$ or $-\infty$ . To handle this cleanly, extend the real line by two symbols.

Definition 6.11 Extended reals

The extended real line $\mathbb{R}^* = \mathbb{R} \cup \{-\infty, +\infty\}$ is ordered by $-\infty < x < +\infty$ for every real $x$ . Every subset $E$ of $\mathbb{R}^*$ has a supremum and an infimum in $\mathbb{R}^*$ : $\sup E = +\infty$ if $E$ has no real upper bound, and $\sup \varnothing = -\infty$ , by convention.

The symbols $\pm\infty$ are not real numbers, and $\infty - \infty$ is left undefined. But they let statements such as " $a_n \to +\infty$ " (for every $M$ , eventually $a_n \geq M$ ) and " $\sup_n a_n = +\infty$ " be made without exceptions.

lim sup and lim inf

A sequence like $(-1)^n(1 + \tfrac1n)$ doesn't converge, but its long-run behaviour is easy to describe: the terms near the end come arbitrarily close to $1$ and to $-1$ , and to nothing else. The precise tool is the upper and lower envelope of the tail.

Definition 6.12 lim sup and lim inf

For a sequence $(a_n)$ , let $a_N^+ = \sup_{n \geq N} a_n$ and $a_N^- = \inf_{n \geq N} a_n$ , the largest and smallest values of the tail from $N$ on (in $\mathbb{R}^*$ ). Then

\limsup_{n\to\infty} a_n := \inf_N a_N^+ = \lim_{N\to\infty} a_N^+, \qquad \liminf_{n\to\infty} a_n := \sup_N a_N^- = \lim_{N\to\infty} a_N^-.

As $N$ grows, the tail gets smaller, so $a_N^+$ decreases and $a_N^-$ increases. By monotone convergence (in $\mathbb{R}^*$ ) both always have limits. So, unlike $\lim$ , $\limsup$ and $\liminf$ always exist.

Figure 6.4. For

a_n = (-1)^n(1 + \tfrac1n)

: the supremum of the tail from

N

on (upper staircase) decreases to

\limsup a_n = 1

, and the infimum of the tail (lower staircase) increases to

\liminf a_n = -1

. The sequence converges exactly when the two envelopes meet.

Proposition 6.13 Properties of lim sup and lim inf

Let $(a_n)$ be a sequence of reals.

$\liminf a_n \leq \limsup a_n$ .
If $L^+ = \limsup a_n$ is finite, then for every $\varepsilon > 0$ : eventually $a_n < L^+ + \varepsilon$ , and infinitely often $a_n > L^+ - \varepsilon$ . (Similarly for $\liminf$ .)
$(a_n)$ converges to $L$ if and only if $\limsup a_n = \liminf a_n = L$ (with $L$ finite).

Proof (Part 3). If $a_n \to L$ , then for every $\varepsilon$ the tails eventually lie in $[L - \varepsilon, L + \varepsilon]$ , so $L - \varepsilon \leq a_N^- \leq a_N^+ \leq L + \varepsilon$ for large $N$ , and both envelopes converge to $L$ . Conversely, if both envelopes converge to $L$ , then $a_N^- \leq a_n \leq a_N^+$ for $n \geq N$ , and the squeeze theorem gives $a_n \to L$ .

Part (2) gives the practical reading: $\limsup a_n$ is the smallest level that the sequence eventually stays below, even allowing any small margin. In engineering terms it is the long-run peak. For a damped vibration it is the amplitude the oscillation settles to, and for a noisy measurement it is the level that noise excursions keep approaching.

Definition 6.14 Limit points

A real $x$ is a limit point of $(a_n)$ if for every $\varepsilon > 0$ and every $N$ there is some $n \geq N$ with $|a_n - x| \leq \varepsilon$ : the sequence returns arbitrarily close to $x$ infinitely often.

Finite $\limsup$ and $\liminf$ are always limit points, the largest and the smallest (Exercise 6.21). For $(-1)^n(1 + \tfrac1n)$ the limit points are exactly $1$ and $-1$ .

Subsequences and Bolzano–Weierstrass

Definition 6.15 Subsequence

A subsequence of $(a_n)$ is a sequence $(a_{n_k})_{k=0}^\infty$ where $n_0 < n_1 < n_2 < \cdots$ . That is, it keeps infinitely many terms of the original sequence, in their original order, and drops the rest.

If $a_n \to L$ , every subsequence converges to $L$ too. The converse fails: $(-1)^n$ has the subsequences $1, 1, 1, \dots$ and $-1, -1, -1, \dots$ , which converge to different limits. In fact, $x$ is a limit point of $(a_n)$ exactly when some subsequence converges to $x$ (Exercise 6.21). The central theorem says that a bounded sequence always has at least one such subsequence.

Theorem 6.16 Bolzano–Weierstrass

Every bounded sequence of real numbers has a convergent subsequence.

Proof (By repeated halving). Let all terms lie in $[c_0, d_0]$ . Split the interval in half. At least one half contains $a_n$ for infinitely many $n$ (the two halves together contain every term, and there are infinitely many terms). Call that half $[c_1, d_1]$ . Split it again, and choose a half $[c_2, d_2]$ that contains infinitely many terms. Continuing, we get nested intervals with $d_k - c_k = (d_0 - c_0)/2^k$ , each containing $a_n$ for infinitely many $n$ .

Now choose indices: $n_0$ with $a_{n_0} \in [c_0, d_0]$ ; then $n_1 > n_0$ with $a_{n_1} \in [c_1, d_1]$ , which is possible since infinitely many indices qualify; and so on, $n_k > n_{k-1}$ with $a_{n_k} \in [c_k, d_k]$ . For $j, k \geq K$ , both $a_{n_j}$ and $a_{n_k}$ lie in $[c_K, d_K]$ , so they differ by at most $(d_0 - c_0)/2^K$ . The subsequence is therefore Cauchy.

To finish without assuming completeness of $\mathbb{R}$ , notice that the left endpoints $c_k$ increase and are bounded, so by Theorem 6.9 they converge to some $c$ . Since $c_k \leq a_{n_k} \leq d_k$ and $d_k - c_k \to 0$ , the squeeze theorem gives $a_{n_k} \to c$ .

Figure 6.5. Bolzano–Weierstrass by halving: keep the half that still contains infinitely many terms, and pick one term from each successive interval, later each time. The nested intervals shrink to a point, and the picked terms form a subsequence converging to it.

In the world In use Bisection: the same halving, computing a root

The proof of Bolzano–Weierstrass is also an algorithm. To solve $f(x) = 0$ for a continuous $f$ with $f(a) < 0 < f(b)$ , evaluate $f$ at the midpoint and keep the half on which $f$ changes sign. After $k$ steps the root is confined to an interval of length $(b - a)/2^k$ . For $f(x) = x^2 - 2$ on $[1, 2]$ , 52 halvings shrink the interval to $2^{-52}$ , about $2.2 \times 10^{-16}$ , which is the spacing of double-precision numbers in $[1, 2)$ (2A.4 The Real Numbers). By then the machine can't represent a smaller interval. Bisection is slow compared with Newton's method but cannot fail, which is why robust root-finders, such as Brent's method (1973), fall back on it whenever faster steps misbehave.

Figure 6.6. Bisection on

x^2 - 2

: each interval is half the previous one and still contains the root, because

f

changes sign across it. It is the halving of Figure 6.5, used to compute.

Completeness of the real numbers

Theorem 6.17 The real numbers are complete

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

Proof. Convergent implies Cauchy is Proposition 6.3. Conversely, let $(a_n)$ be Cauchy. It is bounded, so by Bolzano–Weierstrass some subsequence $a_{n_k}$ converges to some $L$ . We show the whole sequence converges to $L$ . Let $\varepsilon > 0$ . Choose $N$ with $|a_n - a_m| \leq \varepsilon/2$ for $n, m \geq N$ , and then some $k$ with $n_k \geq N$ and $|a_{n_k} - L| \leq \varepsilon/2$ . For every $n \geq N$ , $|a_n - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| \leq \varepsilon$ .

The proof follows a pattern worth recognising: compactness produces a convergent subsequence, and an extra property (here, being Cauchy) upgrades it to convergence of the whole sequence. The same two-step pattern recurs later. Compactness theorems for Ricci flows (11B.3 Compactness of Ricci Flows) produce a convergent subsequence of rescaled flows, and a separate uniqueness argument upgrades it to convergence of the whole family.

The idea Compactness, for the first time

Bolzano–Weierstrass is the first compactness theorem in the guidebook. It says that bounded sets of reals can't "escape": any sequence in them has a part that settles down. It is the ingredient behind step 3 of the contradiction–compactness template (2A.5 Quantifiers and the Shape of a Proof). The first use comes in 2A.9 Continuous Functions, where it shows that a continuous function on a closed interval attains its maximum. Each later course needs a version of it for bigger spaces: of points (2B.3 Compactness), of functions (2B.5 Uniform Convergence and Arzelà–Ascoli), of manifolds (9B.4 Convergence of Manifolds) and of Ricci flows (11B.3 Compactness of Ricci Flows). In infinite-dimensional spaces it fails in its naive form, and recovering some version of it is one of the main themes of functional analysis (4A.1 Banach Spaces and Bounded Operators, 4A.6 Weak Convergence and the Direct Method).

Real exponentiation

With limits available, powers $x^\alpha$ for irrational $\alpha$ can be defined. For $x > 0$ and real $\alpha$ , choose any sequence of rationals $q_n \to \alpha$ , and set

x^\alpha := \lim_{n\to\infty} x^{q_n}.

For this to be a definition, two things must hold: the limit must exist, and it must not depend on the sequence $(q_n)$ chosen. Both follow from the estimate $|x^q - x^r| \leq C\,|q - r|$ for rationals $q, r$ in a bounded range, where $C$ depends on $x$ and the range (Exercise 6.24). The first gives Cauchy, hence convergent; the second gives well-definedness, by interleaving two sequences. The familiar laws $x^{\alpha+\beta} = x^\alpha x^\beta$ and $(x^\alpha)^\beta = x^{\alpha\beta}$ pass to the limit. The exponential and logarithm functions, properly constructed from power series, come in 2B.6 Power Series, Exponentials and Bump Functions.

Recall Where we stand

Convergent sequences have unique limits and obey the limit laws. Monotone bounded sequences converge, to their supremum or infimum. Every sequence has a $\limsup$ and $\liminf$ , and converges exactly when they agree. Every bounded sequence has a convergent subsequence, and as a consequence $\mathbb{R}$ is complete. The next chapter, 2A.7 Series, applies all of this to infinite sums.

Exercises

Exercise 6.18 Limits from the definition

Prove directly from Definition 6.1 that $\frac{3n + 1}{n + 2} \to 3$ . Find an explicit $N(\varepsilon)$ .

Solution

$\left|\frac{3n+1}{n+2} - 3\right| = \frac{5}{n+2} < \frac{5}{n}$ . Given $\varepsilon$ , take $N \geq 5/\varepsilon$ .

Exercise 6.19 Bernoulli's inequality

Prove by induction that $(1 + h)^n \geq 1 + nh$ for every real $h \geq -1$ and every natural number $n$ . Where is $h \geq -1$ needed?

Solution

Base case: $(1+h)^0 = 1 = 1 + 0$ . Step: if $(1+h)^n \geq 1 + nh$ , multiply by $1 + h \geq 0$ (this is where $h \geq -1$ is used, so the inequality doesn't reverse): $(1+h)^{n+1} \geq (1 + nh)(1 + h) = 1 + (n+1)h + nh^2 \geq 1 + (n+1)h$ .

Exercise 6.20 The n-th root of n

Show that $n^{1/n} \to 1$ .

Hint

Write $n^{1/n} = 1 + h_n$ with $h_n \geq 0$ . Then $n = (1 + h_n)^n \geq \binom{n}{2}h_n^2$ for $n \geq 2$ , so $h_n^2 \leq \frac{2}{n - 1}$ .

Exercise 6.21 Limit points and subsequences

Show that (a) $x$ is a limit point of $(a_n)$ if and only if some subsequence converges to $x$ ; (b) a finite $\limsup a_n$ is a limit point, and no limit point is larger.

Exercise 6.22 Every sequence has a monotone subsequence

Call $n$ a peak of $(a_n)$ if $a_n \geq a_m$ for every $m > n$ . Show that if there are infinitely many peaks they give a decreasing subsequence, and if there are finitely many, one can build an increasing subsequence. Deduce Bolzano–Weierstrass again, from Theorem 6.9.

Exercise 6.23 Computing envelopes

Find $\limsup$ and $\liminf$ of: (a) $a_n = \sin(n\pi/3)$ ; (b) $a_n = n\,(-1)^n$ ; (c) $a_n = \big(1 + (-1)^n\big)/n$ .

Solution

(a) The values cycle through $0, \tfrac{\sqrt3}{2}, \tfrac{\sqrt3}{2}, 0, -\tfrac{\sqrt3}{2}, -\tfrac{\sqrt3}{2}$ , so $\limsup = \tfrac{\sqrt3}{2}$ and $\liminf = -\tfrac{\sqrt3}{2}$ . (b) $\limsup = +\infty$ , $\liminf = -\infty$ . (c) Both are $0$ : the terms are $0$ for odd $n$ and $2/n$ for even $n$ , so the sequence converges to $0$ .

Exercise 6.24 Rational powers are Lipschitz on bounded ranges

Let $x > 1$ and let $q < r$ be rationals in $[-K, K]$ . Show that $0 < x^r - x^q \leq x^K(x^{r-q} - 1)$ , and that $x^{s} - 1 \leq s(x - 1)$ for rational $0 \leq s \leq 1$ (hint: for $s = p/m$ , apply the AM–GM inequality to $p$ copies of $x$ and $m - p$ copies of $1$ ). Deduce that $x^\alpha$ is well defined for real $\alpha$ .

Exercise 6.25 Rehearsal: a first contradiction–compactness argument

Let $(a_n)$ be a bounded sequence such that every convergent subsequence has limit $L$ . Prove that $a_n \to L$ , by contradiction: if not, there is $\varepsilon > 0$ and a subsequence staying at least $\varepsilon$ away from $L$ ; apply Bolzano–Weierstrass to it. Identify the five steps of the template of 2A.5 Quantifiers and the Shape of a Proof in your proof. This "every convergent subsequence has the same limit, so the whole sequence converges" step is exactly how uniqueness of a limit object is used later in the route.

Solution

Suppose $a_n \not\to L$ . By negation (2A.5 Quantifiers and the Shape of a Proof) there is $\varepsilon > 0$ such that $|a_n - L| > \varepsilon$ for infinitely many $n$ : these form a subsequence (steps 1–2). It is bounded, so by Bolzano–Weierstrass it has a further subsequence converging to some $L'$ (step 4; no normalisation is needed here, so step 3 is empty). That further subsequence is a convergent subsequence of $(a_n)$ , so $L' = L$ . But all its terms satisfy $|a_n - L| > \varepsilon$ , so $|L' - L| \geq \varepsilon$ by the limit laws (step 5): contradiction.

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6.1Convergence

6.1.1Limit laws

6.1.2Some standard limits

6.2Monotone sequences

6.3The extended reals, suprema and infima

6.4lim sup and lim inf

6.5Subsequences and Bolzano–Weierstrass

6.5.1Completeness of the real numbers

6.6Real exponentiation

6.7Exercises

Convergence

Limit laws

Some standard limits

Monotone sequences

The extended reals, suprema and infima

lim sup and lim inf

Subsequences and Bolzano–Weierstrass

Completeness of the real numbers

Real exponentiation

Exercises