6. Sequences, the Number e, and Continuity Theorems

Learning goals and preparation

We assume real-number order, finite sums, binomial coefficients, and elementary functions. This chapter develops sequence and continuity foundations. You will define sequence convergence, justify the sequence construction of Euler's number, and distinguish existence theorems from algorithms or uniqueness claims. Later calculus may verify identities established by these definitions, but is not silently substituted for them.

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

6.1 Sequence convergence and elementary laws

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Worked example 1: an explicit threshold. For $a_n=3+2/n$, choose any integer $N>2/\varepsilon$. Then $n\ge N$ gives $|a_n-3|=2/n\le2/N<\varepsilon$, proving convergence to three. Saying only “$n$ gets large” does not specify the required tolerance relationship.

6.2 Monotone sequences and simple series

Monotone sequences

Theorem

Monotone convergence theorem

Every increasing real sequence bounded above converges to the supremum of its set of terms. Every decreasing sequence bounded below converges to the infimum.

Proof

From supremum to eventual closeness

Let $s=\sup\{x_n:n\in\mathbb N\}$. Given $\varepsilon>0$, the supremum criterion gives an index $N$ with $s-\varepsilon<x_N\le s$. Monotonicity gives $s-\varepsilon<x_N\le x_n\le s$ for all $n\ge N$, proving $x_n\to s$.

Worked example

A recursive sequence

Let $x_1=1$ and $x_{n+1}=(x_n+2)/2$. Induction gives $x_n<2$, and $x_{n+1}-x_n=(2-x_n)/2>0$, so the sequence increases and is bounded above. It converges to some $L$. Taking limits in the recurrence yields $L=(L+2)/2$, hence $L=2$. Solving the fixed-point equation without first proving convergence would not by itself show the sequence has a limit.

For real $r$ with $|r|<1$, $r^n\to0$. For $0<q=|r|<1$, put $h=q^{-1}-1>0$. The elementary inequality $(1+h)^n\ge1+nh$ gives

$$|r^n|=q^n\le\frac1{1+nh}\to0.$$

The inequality follows by multiplying $1+nh$ by $1+h$ and retaining the extra nonnegative $nh^2$ at each step, starting at $n=1$. For $r=0$, all positive-index powers are zero. At $r=1$ the sequence is one; at $r=-1$ it oscillates; for $|r|>1$ its magnitude grows without bound. These boundary cases must not be hidden in the formula for $|r|<1$.

6.4 An infinite series is a limit of finite totals

Definition

For terms $a_1,a_2,\ldots$, define the $N$th partial sum

$$S_N=\sum_{n=1}^{N}a_n.$$

The infinite series $\sum_{n=1}^{\infty}a_n$ converges to $S$ when the sequence $(S_N)$ converges to $S$. Its sum is that limit. If the partial sums do not have a finite limit, the series diverges. The notation is a question about a limit, not permission to finish an infinite number of operations.

A necessary condition for convergence is $a_n\to0$. If $S_N\to S$, then for $n\ge2$,

$$a_n=S_n-S_{n-1}\longrightarrow S-S=0.$$

Common mistake

Thus terms that fail to approach zero prove divergence. Terms that approach zero do not prove convergence.

Worked example

Worked example 4: distinguish terms from totals. The sequence $a_n=1/n$ converges to zero. Nevertheless its series diverges. Group a partial sum ending at $2^m$ as

$$1+\frac12+\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.$$

The block from $2^{j-1}+1$ through $2^j$ has $2^{j-1}$ terms, each at least $1/2^j$, so each block contributes at least $1/2$. Consequently $S_{2^m}\ge1+m/2$, which is unbounded. These are finite groupings of selected partial sums, not an unexplained rearrangement of an infinite series. This proves divergence of the harmonic series.

Consider $a+ar+ar^2+\cdots$, with partial sum $S_N=\sum_{k=0}^{N-1}ar^k$. If $a=0$, every term is zero and the series converges to zero for every real $r$. Suppose now that $a\ne0$.

A geometric progression has $a_1=a$ and $a_{k+1}=ra_k$. For $k\ge2$, $a_k=ar^{k-1}$. Describing $a_1$ separately also avoids an unnecessary $0^0$ expression when $r=0$. Its finite sum is derived by shifting and subtracting:

$$S_n=a+ar+\cdots+ar^{n-1},\quad rS_n=ar+ar^2+\cdots+ar^n.$$

Consequently $(1-r)S_n=a(1-r^n)$, giving

$$S_n=\frac{a(1-r^n)}{1-r}\quad(r\ne1),\qquad S_n=na\quad(r=1).$$

This is a finite identity for any real $a,r$ and positive integer $n$; no convergence assumption is required. Negative $r$ causes alternating signs. The exceptional ratio $1$ is not handled by division by zero.

When $|r|<1$, $r^N\to0$, and therefore

$$\sum_{k=0}^{\infty}ar^k=\frac a{1-r}.$$

When $|r|\ge1$, the terms $ar^k$ do not approach zero, so the series diverges. This includes $r=1$ and $r=-1$. Writing $a/(1-r)$ at $r=2$ gives an ordinary number but does not make it the sum of the divergent series.

For $|r|<1$, the exact remainder after the first $N$ terms is

$$R_N=S-S_N=\frac{ar^N}{1-r},\qquad |R_N|=\frac{|a|\,|r|^N}{|1-r|}.$$

The simpler estimate $|R_N|\le |a||r|^N/(1-|r|)$ also holds. The exact expression retains the sign and accounts correctly for negative ratios.

6.3 Defining Euler's number through a sequence

Let $e_n=(1+1/n)^n$. The binomial theorem gives

$$e_n=\sum_{k=0}^n\frac1{k!}\prod_{j=0}^{k-1}\left(1-\frac jn\right).$$

For fixed $k\le n$, each factor is nonnegative and nondecreasing when $n$ increases, and a new positive term appears in $e_{n+1}$. Hence $e_n$ is increasing. Also every product is at most one, so

$$e_n\le\sum_{k=0}^n\frac1{k!}<3.$$

For example, $k!\ge2^{k-1}$ for $k\ge1$ bounds the infinite tail by a geometric sum, with strict inequality for sufficiently large $k$. Since $e_1=2$, the monotone bounded theorem supplies a finite limit. Define the Euler number $e=\lim_ne_n$, so $2\le e\le3$ (sharper bounds are possible).

This sequence arises when a unit proportional increase is split into $n$ equal multiplicative substeps.

Once exponential and logarithmic calculus has been established, consistency with the natural exponential is checked by $x/(1+x)\le\ln(1+x)\le x$ for $x>0$. Setting $x=1/n$ gives $n/(n+1)\le n\ln(1+1/n)\le1$. Squeezing and exponentiating show $e_n\to\exp(1)$. Here that is a later verification, not a circular substitute for the sequence existence proof.

6.4 Function limits and continuity

Limits and domain points

Definition

Function limit

Let $A\subset\mathbb R$, and let $c$ be an accumulation point of $A$ (every punctured neighbourhood of $c$ meets $A$). For $f:A\to\mathbb R$, we say $\lim_{x\to c,\,x\in A}f(x)=L$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $x\in A$ and $0<|x-c|<\delta$ imply $|f(x)-L|<\varepsilon$.

The value $f(c)$, if defined, is irrelevant to this punctured limit. The accumulation-point condition prevents the definition from being vacuously true for every $L$ at an isolated point. Domain restrictions matter: endpoint limits on a closed interval are one-sided relative to the domain.

Worked example

An epsilon–delta estimate

For $f(x)=x^2$ at $c=2$, target $L=4$. If $|x-2|<1$, then $|x+2|<5$. Thus $|x^2-4|=|x-2||x+2|<5|x-2|$. Given $\varepsilon>0$, choose $\delta=\min(1,\varepsilon/5)$. Then $0<|x-2|<\delta$ gives $|x^2-4|<\varepsilon$.

Theorem

Sequential criterion for a function limit

For an accumulation point $c$ of $A$, the limit of $f$ at $c$ equals $L$ if and only if $f(x_n)\to L$ for every sequence $x_n\in A\setminus\{c\}$ with $x_n\to c$.

Proof

Proving the criterion

Given the epsilon–delta limit and any such sequence, eventually $0<|x_n-c|<\delta$, so $|f(x_n)-L|<\varepsilon$. Conversely, if the limit fails, some $\varepsilon_0>0$ defeats every $\delta>0$. For each $n$, choose $x_n\in A$ with $0<|x_n-c|<1/n$ and $|f(x_n)-L|\ge\varepsilon_0$. Then $x_n\to c$ but $f(x_n)$ does not converge to $L$, contradicting the sequential hypothesis.

Worked example

Two approaches can disprove a limit

For $f(x)=\sin(1/x)$ on $\mathbb R\setminus\{0\}$, the sequences $x_n=(\pi/2+2\pi n)^{-1}$ and $y_n=(3\pi/2+2\pi n)^{-1}$ both approach $0$, but $f(x_n)=1$ and $f(y_n)=-1$. No common limit exists. To prove a limit, by contrast, a few selected paths or sequences do not suffice: the criterion requires every qualifying sequence.

When the domain contains points approaching $a$ from both sides, a two-sided finite limit exists if and only if both one-sided finite limits exist and agree.

At a domain endpoint use the appropriate one-sided or relative-domain limit, not an unavailable opposite side.

A limit at infinity replaces the near-$a$ condition by $x>M$ for a sufficiently large threshold. For example,

$$\frac{2x^2+3}{x^2+1}=2+\frac1{x^2+1}\longrightarrow2\quad(x\to\infty).$$

Definition

Continuity relative to a domain

For $c\in A$, $f:A\to\mathbb R$ is continuous at $c$ if for every $\varepsilon>0$ there is $\delta>0$ such that $x\in A$ and $|x-c|<\delta$ imply $|f(x)-f(c)|<\varepsilon$. At an isolated point of $A$ this is automatic. At an accumulation point it is equivalent to the limit equaling $f(c)$.

The sequential criterion for continuity also includes isolated points: $f$ is continuous at $a$ relative to its domain if and only if every domain sequence $x_n\to a$ satisfies $f(x_n)\to f(a)$. The epsilon–delta condition immediately controls any convergent input sequence. Conversely, if continuity fails, a fixed output discrepancy can be found within distance $1/n$ for every $n$, constructing a sequence that violates this criterion. At an isolated point relative continuity is automatic; equality with the punctured limit is an equivalent formulation only when $a$ is an accumulation point.

Worked example 2: a discontinuity versus a removable hole. Let $f(x)=x+1$ for $x\ne2$ and $f(2)=0$. The nearby limit is three, so the function is not continuous at two. Defining $f(2)=3$ would repair this point. In contrast, no value at zero repairs $|x|/x$, since its two side limits differ.

6.5 Intermediate and extreme value theorems

Theorem

Intermediate value theorem

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, there is $c\in[a,b]$ with $f(c)=y$.

Proof

Completeness locates a crossing

If $y=f(a)$ or $y=f(b)$, take the corresponding endpoint. Otherwise, assume $f(a)<y<f(b)$; the reverse order follows by replacing both $f$ by $-f$ and $y$ by $-y$. The set $S=\{x\in[a,b]:f(x)\le y\}$ is nonempty and bounded above. Let $c=\sup S$. If $f(c)<y$, continuity gives nearby points to the right with value below $y$; since $c<b$, this contradicts the upper-bound property. If $f(c)>y$, continuity gives a neighbourhood of $c$ with value above $y$; the supremum approximation property supplies an element of $S$ in that neighbourhood, a contradiction. Thus $f(c)=y$.

This proves existence but not uniqueness. A monotonicity argument or another condition is needed before a bisection bracket can be called the location of a unique root. Bisection's numerical error bounds are developed in Chapter 1.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Worked example 3: hypotheses determine attainment. The function $f(x)=x$ on $(0,1)$ has infimum zero and supremum one but attains neither. On $[0,1]$, both are attained. On an unbounded closed interval, continuity still does not guarantee a maximum: $f(x)=x$ on $[0,\infty)$ is unbounded. The closedness and boundedness assumptions are not decorative.

6.6 Graded exercises

Foundation

1. Prove $a_n=5-3/n$ converges to five using an explicit $N(\varepsilon)$.
2. Give a bounded divergent sequence and identify the failed convergence condition.
3. Evaluate $\lim_n(4n+1)/(2n-3)$ by dividing by $n$ and checking the denominator limit.

Core

4. Find the sum of $\sum_{k=0}^{\infty}3(1/4)^k$ and justify convergence.
5. Explain why $a_n\to0$ does not establish convergence of $\sum a_n$.
6. Compute $e_1,e_2,e_3$ and compare them with the proved bounds and monotonicity.
7. Choose the value at $x=1$ making $(x^2-1)/(x-1)$ continuous after extension.
8. Show $x^3+x-1=0$ has at least one root in $[0,1]$ and state why the IVT alone does not assert uniqueness.

Synthesis

9. Let $a_n\ge0$ be increasing and bounded above by ten. Prove convergence using the supremum directly. Does the information determine its numerical limit?
10. Give examples showing why continuity, closedness and boundedness cannot each be dropped from the extreme-value theorem. Explain the failure in each example.

6.7 Full solutions

1. For $\varepsilon>0$, choose an integer $N>3/\varepsilon$. Then $n\ge N$ gives $|a_n-5|=3/n\le3/N<\varepsilon$.
2. Use $(-1)^n$. Even and odd subsequences stay at one and negative one, so sufficiently late terms cannot all lie in a small interval about one common limit.
3. The quotient becomes $(4+1/n)/(2-3/n)\to4/2=2$. The denominator tends to the nonzero number two, so the quotient law applies for sufficiently large indices.
4. Ratio magnitude $1/4<1$ makes the geometric remainder vanish. Sum is $3/(1-1/4)=4$.
5. The harmonic-series block argument supplies a counterexample: terms $1/n$ tend to zero but partial sums grow without bound. Vanishing terms are necessary, not sufficient.
6. $e_1=2$, $e_2=9/4$, $e_3=64/27$. They increase and all lie below three, consistent with the proof. A few terms alone would not prove monotonicity or convergence.
7. For $x\ne1$ the quotient is $x+1$, so its nearby limit is two. Defining the extended value to be two makes it continuous.
8. The polynomial is continuous, equals negative one at zero and one at one, so the IVT gives a root in the interior. A continuous function can cross zero repeatedly, so uniqueness needs further information; here a positive derivative can provide it once derivative methods are available.
9. Let $L=\sup\{a_n:n\ge1\}$. For every $\varepsilon>0$, some $a_N>L-\varepsilon$ by the defining supremum property. Increase gives $L-\varepsilon<a_N\le a_n\le L$ for $n\ge N$, proving $a_n\to L$. The bound ten does not determine $L$; constant sequences at different values meet the assumptions.
10. Without continuity, on $[0,1]$ define $f(x)=x$ for $x<1$ and $f(1)=0$; its supremum one is not attained. Without closedness, $f(x)=x$ on $(0,1)$ attains neither endpoint bound. Without boundedness, $f(x)=x$ on $[0,\infty)$ has no maximum. Each example preserves the other indicated properties while violating one needed hypothesis.

Summary and onward map

Convergence is a quantified accuracy statement. Monotone bounded sequences converge by completeness; a bounded sequence need not converge, and a vanishing series term need not give a finite sum. Euler's number has a sequence-based construction. Continuity transfers limits; IVT supplies intermediate values, while EVT supplies attained extrema only under its full interval hypotheses.