1. Algebra for Business and Social-Science Models

Learning goals and preparation

This chapter develops algebraic reasoning for readers who need mathematical tools before calculus. We assume signed arithmetic, fractions, and elementary percentages. You will learn to translate a verbal model into equations, identify feasible solutions, solve linear and quadratic problems, and use exponentials and logarithms to compare growth. The examples use hypothetical models.

Within a model's stated range, state its assumptions, solve the equations, and interpret the result in the original units.

1.1 Quantities, units, and percentages

A variable represents a quantity that can change. A parameter is held fixed within a calculation. A function assigns one output to each allowed input. For example, $C(q)=120+8q$ can model total cost in dollars for $q\ge0$ units of production. The number $120$ is a fixed cost; the coefficient $8$ has units dollars per unit. The domain may need further restriction, such as integer $q$ or a capacity limit.

A percentage change compares a difference with its starting value:

$$\text{relative change}=\frac{\text{new}-\text{old}}{\text{old}},\qquad\text{old}\ne0.$$

If a positive price $P$ rises by a proportion $r$, where $0<r<1$, the new price is $P(1+r)$. A subsequent decrease of proportion $r$ gives $P(1+r)(1-r)=P(1-r^2)$, not $P$. Percentage changes apply to different bases.

Worked example 1: reversing a discount. A price after a $20\%$ discount is $72$. If the original price is $P$, then $0.8P=72$, so $P=90$. Adding $20\%$ to $72$ would give $86.4$ and would not reverse the discount.

An index number is a ratio relative to a reference value. An index rising from $120$ to $132$ has increased by $12$ index points but by $10\%$. Distinguish a difference in levels from a relative change.

1.2 Linear equations and break-even analysis

An affine function has the form $f(x)=mx+c$; its graph is a straight line. The slope $m$ is the change in output per unit change in input. In mathematical terminology “linear” strictly means $c=0$, although elementary applications often call all straight-line models linear.

Suppose a product sells at a constant price $p$ per unit, costs $v$ per unit to make, and has fixed cost $F\ge0$. Revenue, cost and profit are

$$R(q)=pq,\qquad C(q)=F+vq,\qquad \Pi(q)=(p-v)q-F.$$

If $p>v$, the break-even quantity is $F/(p-v)$. If $p=v$, no positive fixed cost can be recovered; if $p<v$ and $F>0$, every nonnegative production level makes a loss. The familiar formula must not conceal these cases.

Worked example 2: integer feasibility. Let $p=18$, $v=11$, and $F=500$. Nonnegative profit requires $7q-500\ge0$, so $q\ge500/7$. If production is in whole units, the smallest break-even-or-better level is $72$. Its profit is $4$, while $71$ units gives a loss of $3$. Rounding to the nearest integer would give the wrong decision rule; we need the smallest feasible integer satisfying an inequality.

1.3 Systems and equilibrium

A pair of linear equations can represent two simultaneous conditions. Substitution or elimination is justified because each step is reversible.

Worked example 3: a simple market model. Demand and supply in units per week are

$$q_d=140-4p,\qquad q_s=20+2p.$$

Assume $p\ge0$ and $q\ge0$, and assume the model is applicable near equilibrium. Setting $q_d=q_s$ gives $120=6p$, so $p=20$ and $q=60$. Both are admissible.

For a general system $ax+by=u$, $cx+dy=v$, elimination produces $(ad-bc)y=av-cu$ after multiplying appropriately. If $ad-bc\ne0$, the solution is

$$x=\frac{du-bv}{ad-bc},\qquad y=\frac{av-cu}{ad-bc}.$$

Substitution gives $a(du-bv)+b(av-cu)=u(ad-bc)$ and $c(du-bv)+d(av-cu)=v(ad-bc)$, so both equations hold. Every solution satisfies these two formulas by elimination, proving uniqueness. If this coefficient vanishes, the two equations may be redundant or inconsistent; more information is needed. Counting two equations and two unknowns does not guarantee uniqueness.

1.4 Quadratics and feasible domains

A quadratic function is $f(x)=ax^2+bx+c$ with $a\ne0$. Completing the square gives

$$f(x)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.$$

This identity proves the vertex formula without calculus. If $a>0$, the square term is nonnegative and the vertex is a minimum on the real line. If $a<0$, it is a maximum. On a restricted interval, the vertex may be infeasible; compare the relevant endpoints as well.

The equation $ax^2+bx+c=0$ has roots

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

when the discriminant is nonnegative. A negative discriminant gives no real roots.

Worked example 4: revenue under a demand curve. Let price be $p(q)=30-q/2$ with feasible quantity $0\le q\le60$. Revenue is

$$R(q)=q\left(30-\frac q2\right)=450-\frac12(q-30)^2.$$

Therefore the maximum is $450$ at $q=30$, with price $15$. The quantity is inside the feasible domain. If capacity instead restricted $0\le q\le20$, then $(q-30)^2$ is smallest at $q=20$, so the maximum feasible revenue would be $400$, not $450$.

To solve a quadratic inequality, factor or locate its real roots and determine signs between roots. For example, $(q-2)(q-5)\le0$ holds on $[2,5]$. The sign is not determined by treating the two factors independently as both nonpositive: a product can be negative when one factor is positive.

1.5 Powers, exponentials, and logarithms

A geometric growth model is $A_n=A_0(1+r)^n$, where $A_0>0$, $1+r>0$, and $n$ counts equal periods. A continuous-time exponential model is $A(t)=A_0e^{kt}$. These formulas describe assumptions about growth; their parameters have different meanings. If time is measured in years, $k$ has units year$^{-1}$.

The natural logarithm $\ln$ is the inverse of $e^x$ and is defined only for positive inputs. It obeys $\ln(uv)=\ln u+\ln v$ for $u,v>0$ and $\ln(u^a)=a\ln u$ for $u>0$. There is no rule splitting the logarithm of a sum.

Worked example 5: a threshold time. A hypothetical programme's participation is modeled by $N(t)=200e^{0.08t}$ until capacity is reached, with time in years. The time to reach $300$ is

$$e^{0.08t}=\frac32\implies t=\frac{\ln(3/2)}{0.08}.$$

This is an exact answer, approximately $5.07$ years. It is meaningful only if the exponential approximation remains valid until that time.

When comparing positive quantities, taking logarithms preserves inequalities because $\ln$ is increasing. If $0<a<1$, then $a^n\le r$ for $0<r<1$ becomes $n\ln a\le\ln r$; dividing by the negative number $\ln a$ reverses the inequality.

1.6 Piecewise rules and accounting for thresholds

A tariff can charge different rates in different bands. Suppose the first $100$ units cost $2$ each, and each unit above $100$ costs $3$. For $q\ge0$,

$$T(q)=\begin{cases}2q,&0\le q\le100,\\200+3(q-100),&q>100.\end{cases}$$

The second formula includes the first band's cost. It is not $3q$. Both branches approach the same bill at $100$, so the rule has no jump there. In contrast, a fixed charge triggered by crossing a threshold would produce a jump. The model should say which interpretation is intended.

Worked example 6: invert a piecewise bill. A bill of $290$ exceeds the $200$ first-band total, so use the second branch: $200+3(q-100)=290$, yielding $q=130$. The answer satisfies $q>100$. Solving a branch and ignoring its condition can produce an inadmissible candidate.

1.7 Graded exercises

Foundation

1. A positive price rises by $25\%$ and then falls by $20\%$. What is the overall percentage change? Explain using multipliers.
2. A discounted price of $136$ is $15\%$ below the original price. Find the original price.
3. With fixed cost $360$, unit cost $9$, and unit price $15$, find the minimum whole-number quantity with nonnegative profit.
4. Solve the system $2x+y=11$, $x-y=1$ and verify both equations.

Core

5. Demand is $q=100-2p$ and supply is $q=10+p$. Find the model's equilibrium and test nonnegativity.
6. On $0\le q\le40$, maximize $R(q)=40q-q^2$ without calculus. Repeat if capacity is $q\le15$.
7. Solve $q^2-7q+10\le0$, and interpret the answer if $q$ must be a whole number.
8. Let $N(t)=500e^{-0.1t}$, $t\ge0$. Find the exact time at which $N=200$ and explain why it is positive.

Synthesis

9. A service charges $5$ per unit for the first $20$ units and $3$ per additional unit, with no fixed charge. Write its bill function, calculate the bill at $35$ units, and invert a bill of $160$.
10. A quantity is multiplied by $0.9$ each month. Determine the smallest nonnegative integer $n$ for which it is no more than half its original positive value. Explain the inequality reversal and verify the two adjacent integers.

1.8 Full solutions

1. The multiplier is $1.25(0.8)=1$, so the overall change is $0\%$. This cancellation is special to these unequal rates; equal nonzero percentage increases and decreases do not cancel.
2. If the original is $P$, then $0.85P=136$. Thus $P=160$.
3. Profit is $(15-9)q-360=6q-360$. It is nonnegative when $q\ge60$. The minimum whole-number quantity is $60$.
4. Adding the equations gives $3x=12$, so $x=4$, and $y=3$. Check: $2(4)+3=11$ and $4-3=1$.
5. Equate $100-2p=10+p$, so $p=30$ and $q=40$. Both are nonnegative and the demand and supply formulas both give $40$.
6. $R(q)=400-(q-20)^2$, so the maximum on $[0,40]$ is $400$ at $20$. On $[0,15]$, the distance to $20$ is smallest at $15$; the maximum is $400-25=375$ there.
7. Factor as $(q-2)(q-5)\le0$. The leading coefficient is positive and the expression is nonpositive between the roots, including them: $q\in[2,5]$. If $q$ is a whole number, the allowed values are $2,3,4,5$.
8. $e^{-0.1t}=2/5$, hence $t=-10\ln(2/5)=10\ln(5/2)$. It is positive because $5/2>1$, so its logarithm is positive. Time has the model's stated time unit.
9. For $0\le q\le20$, $T(q)=5q$. For $q>20$, $T(q)=100+3(q-20)$. At $35$, the bill is $145$. A bill of $160$ exceeds $100$, so use the upper branch: $3(q-20)=60$, giving $q=40$, which satisfies the branch condition.
10. Require $0.9^n\le0.5$. Logarithms give $n\ln0.9\le\ln0.5$. Since $\ln0.9<0$, divide to get $n\ge\ln0.5/\ln0.9\approx6.579$. The smallest integer is $7$. Directly, $0.9^6=0.531441>0.5$ and $0.9^7=0.4782969\le0.5$.

Summary and onward map

Use units to distinguish quantities from coefficients. A feasible domain is part of a model. Linear systems impose simultaneous conditions; quadratics can be optimized by completing the square; logarithms invert positive exponential models. Percentage changes compound through multiplication, and piecewise formulas must retain their branch conditions.

Prerequisite mapping. Algebraic functions and logarithms prepare for differentiation and integration. Systems prepare for vectors and matrices. The distinction between a continuous model and integer admissibility remains useful in later optimization.

Proofs supporting the methods used in this note

Finite real limit laws and geometric sequences: the missing proofs

Work over the ordered real numbers, with the Archimedean property, positive-integer induction and the usual absolute-value triangle inequality. A sequence converges to a finite real $L$ when every $\varepsilon>0$ admits an $N$ that works for every $n\ge N$.

Uniqueness, boundedness and subsequences. If $a_n\to L,M$ and $L\ne M$, require both errors below $|L-M|/3$ at the same sufficiently large index. The triangle inequality gives $|L-M|<2|L-M|/3$, a contradiction. For boundedness, require $|a_n-L|<1$ in the tail, and take a maximum of $|L|+1$ and the finitely many initial absolute values. For a subsequence $n_1<n_2<\cdots$, induction gives $n_k\ge k$; hence every tail threshold for the original sequence also works eventually for the subsequence.

Sum, scalar and product. If $a_n\to a$, $b_n\to b$, choose both input errors below $\varepsilon/2$; then

$$|(a_n+b_n)-(a+b)|\le |a_n-a|+|b_n-b|<\varepsilon.$$

For a fixed scalar $c\ne0$ use input tolerance $\varepsilon/|c|$; $c=0$ is immediate. Choose $B\ge1$ bounding all $|b_n|$. Require

$$|a_n-a|<\frac{\varepsilon}{2B},\qquad |b_n-b|<\frac{\varepsilon}{2(|a|+1)}.$$

The common later threshold gives $|a_nb_n-ab|\le B|a_n-a|+|a||b_n-b|<\varepsilon$.

Reciprocal and quotient. If $b\ne0$, first require $|b_n-b|<|b|/2$, ensuring $|b_n|>|b|/2$ and nonzero denominators. Then

$$\left|\frac1{b_n}-\frac1b\right|\le\frac{2|b_n-b|}{|b|^2}.$$

Requiring also $|b_n-b|<\varepsilon|b|^2/2$ proves reciprocal convergence. The product law now proves $a_n/b_n\to a/b$. Any undefined initial quotients may be omitted; convergence is a tail assertion.

Order and squeeze. If $a_n\le b_n$ eventually and $a_n\to a$, $b_n\to b$, then $a\le b$: otherwise errors below $(a-b)/3$ would force $a_n>b_n$. If $a_n\le z_n\le b_n$ eventually and both outer limits are $L$, require both errors below $\varepsilon$. Then $L-\varepsilon<z_n<L+\varepsilon$, proving $z_n\to L$. This proof does not require a previously known limit of $z_n$.

For the corresponding function-limit laws, fix a common allowed approach domain $D\subseteq\mathbb R$ and an accumulation point $a$ of $D$: every punctured neighborhood of $a$ contains a point of $D$. Let $f,g:D\to\mathbb R$ have the stated finite relative-domain limits at $a$. Replace each tail threshold in the displayed estimates by a punctured-domain radius, and take the minimum of finitely many positive radii. For the product proof only local boundedness is needed: $g(x)\to b$ gives $|g(x)|\le|b|+1$ on a sufficiently small punctured $D$-neighborhood, even if $g$ is unbounded elsewhere on $D$. The same estimates prove sum, scalar, product, reciprocal and quotient laws with their stated nonzero limiting-denominator condition; the same order and squeeze estimates apply on the common approach domain. Accumulation supplies actual points when proving uniqueness or excluding an incompatible order of limiting values.

For composition let $B\subseteq\mathbb R$, let $b\in B$, and let $F:B\to\mathbb R$ be continuous at $b$ relative to $B$. Suppose $g:D\to\mathbb R$ tends to $b$ and $g(x)\in B$ throughout some sufficiently small punctured $D$-neighborhood of $a$. Given $\varepsilon>0$, continuity supplies $\eta>0$ such that $u\in B$ and $|u-b|<\eta$ imply $|F(u)-F(b)|<\varepsilon$. The limit of $g$ and the composition-domain condition supply a common positive radius forcing both conditions for every allowed approaching $x$. Thus $F(g(x))\to F(b)$. This includes inputs with $g(x)=b$ and does not divide by an inner increment.

Geometric classification. Bernoulli's bound $(1+t)^n\ge1+nt$ for $t\ge0$ follows by induction: multiply the case $n$ by the positive $1+t$ and retain the nonnegative remainder $nt^2$. If $0<q<1$, set $t=q^{-1}-1>0$. Thus

$$0<q^n\le\frac1{1+nt}\longrightarrow0.$$

For any requested tolerance, the Archimedean property supplies $N$ making the right side smaller. Hence $r^n\to0$ for $0<|r|<1$ by $|r^n|=|r|^n$; $r=0$ gives zero for every positive index. At $r=1$ the sequence is constantly one. At $r=-1$, the even and odd subsequences are one and minus one, so uniqueness and subsequence inheritance exclude convergence. If $r>1$, Bernoulli with $t=r-1$ gives $r^n\ge1+nt\to+\infty$. If $r<-1$, $|r^n|\to+\infty$ by the same argument, contradicting boundedness of any convergent real sequence. These cases exhaust all real ratios.

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.

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Proof

Why continuity suffices

A continuous function on a compact interval is uniformly continuous. Given $\varepsilon>0$ and $b>a$, choose $\delta$ so that points less than $\delta$ apart have output difference less than $\varepsilon/(2(b-a))$. A partition with mesh below $\delta$ has oscillation on every subinterval at most that bound, hence $U-L<\varepsilon$. The Darboux criterion yields integrability. For $a=b$ the integral is zero by convention.

Bounded proper Riemann integrals: mesh equivalence and elementary properties

Let $a<b$ be finite, and let $f:[a,b]\to\mathbb R$ be bounded. For a partition $P$, write $L(f,P),U(f,P)$ for the lower and upper Darboux sums. We use real completeness and the proved Darboux criterion: equality of lower and upper integrals is equivalent to arbitrarily small $U-L$. This packet does not assume continuity of $f$.

Darboux implies every fine tagged sum converges. Let the Darboux integral be $I$. Given $\varepsilon>0$, choose a fixed partition $P_0$ with gap below $\varepsilon/2$. Choose $M>0$ with $|f|\le M$, and let $k$ be the number of interior partition points of $P_0$. For any other partition $P$, mark as bad every cell whose closure contains one of those $k$ points. Their total length is at most $2k\|P\|$, because each point belongs to at most two cells. Every remaining cell lies within a single $P_0$ cell. The contribution of all good cells to the oscillation sum $U(f,P)-L(f,P)$ is at most the gap for $P_0$; the bad cells contribute at most $2M(2k\|P\|)$. Thus

$$U(f,P)-L(f,P)<\varepsilon/2+4Mk\|P\|.$$

Take $\|P\|<\varepsilon/[8M\max(1,k)]$. Every tagged sum $S(P)$ and $I$ lie between $L(f,P)$ and $U(f,P)$, so $|S(P)-I|<\varepsilon$, independently of tags.

The converse. Suppose all sufficiently fine tagged sums are within $\eta$ of a common finite $I$. Fix one such fine partition. Each cell supremum and infimum can be approximated by actual function values with error below $\eta/(b-a)$. The resulting upper-approximating tagged sum exceeds $U(f,P)-\eta$, and the lower-approximating sum is below $L(f,P)+\eta$. Therefore $U(f,P)<I+2\eta$ and $L(f,P)>I-2\eta$. Since $\eta$ can be arbitrarily small, the Darboux criterion gives integrability. Lower and upper bounds on tagged sums then identify its integral as $I$. Suprema need not be attained for this argument.

Linearity, constants and order. For integrable $f,g$ and real $\alpha,\beta$, on any common tagged partition the exact finite identity is $S(\alpha f+\beta g)=\alpha S(f)+\beta S(g)$. Its limit exists, so the equivalence just proved gives integrability and the corresponding linear integral identity. The constant function $c$ has every sum equal to $c(b-a)$. If $f\le g$, their tagged sums with the same tags obey the same order; their limits do too. Consequently $|\int f|\le M(b-a)$ whenever $|f|\le M$.

Absolute values. The reverse triangle inequality gives $\operatorname{osc}_J|f|\le\operatorname{osc}_J f$ on every cell $J$: apply $||f(x)|-|f(y)||\le|f(x)-f(y)|$ and approximate the suprema/infima. Thus every Darboux gap for $|f|$ is at most the gap for $f$, and $|f|$ is integrable. Since $-|f|\le f\le|f|$, order and linearity give $|\int_a^b f|\le\int_a^b|f|$.

Restriction and interval addition. For $a<c<b$, refine any small-gap partition by $c$. The nonnegative gaps on $[a,c]$ and $[c,b]$ sum to the refined whole gap; each can therefore be arbitrarily small. Both restrictions are integrable. Join partitions of the two pieces with meshes tending to zero. The exact sum on the joined partition is the sum of the two tagged sums, so passage to their three limits gives $\int_a^b f=\int_a^c f+\int_c^b f$. Conversely, if both restrictions are bounded integrable, join two partitions with gaps below $\varepsilon/2$; the resulting whole gap is below $\varepsilon$, proving whole integrability and the same identity. At a shared endpoint both restrictions use the same value of $f$; no endpoint term has positive interval length. Set equal-bound integrals to zero and reversed-bound integrals to the negative. Splitting a finite interval and cancellation now extend addition to every ordering of its three endpoints.

These proofs justify the estimates used in accumulation, continuous FTC, integration by parts and proper substitution. They assert bounded proper Riemann statements only, and do not replace the separate convergence definitions at singularities or infinite endpoints.

Definition

Function limit and continuity

Suppose $f$ is defined at points arbitrarily close to $a$ other than $a$. We write $\lim_{x\to a}f(x)=L$ if for every $\varepsilon>0$ there is $\delta>0$ such that $0<|x-a|<\delta$ and $x$ in the domain imply $|f(x)-L|<\varepsilon$. The function is continuous at $a$ when it is defined there and this limit equals $f(a)$. For endpoints use the corresponding one-sided limit.

When the domain contains points arbitrarily close to $a$ on both sides, the two-sided limit exists exactly when the left- and right-hand limits both exist and agree. Polynomial functions are continuous everywhere, rational functions where the denominator is nonzero, and sums, products, quotients with nonzero denominator, and compositions of continuous functions are continuous where defined. In radian measure the standard limit is $\lim_{x\to0}\sin x/x=1$; it supports the derivative of sine later. Limits at $+∞$ and $-∞$ describe the two directions separately; neither infinity is a real input.

Theorem

Intermediate and extreme values

If $f$ is continuous on $[a,b]$, it attains a maximum and a minimum there. If $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in[a,b]$. Closedness, boundedness and continuity matter for the extrema assertion; continuity on an open interval alone need not give extrema.

Proof

Proof idea for the intermediate value theorem

Assume $f(a)<y<f(b)$ and let $S=\{x\in[a,b]:f(x)\le y\}$. The set is nonempty and bounded above, so $c=\sup S$ exists. Continuity rules out $f(c)<y$ because then nearby points to the right would also be in $S$; it rules out $f(c)>y$ because points of $S$ arbitrarily close to $c$ would then have values above $y$. Therefore $f(c)=y$. Endpoint equality is immediate, and reversing the inequalities handles the other order.

Learning goals and setting

Read differentiability as linear approximation, handle zero inner increments in the chain rule, construct a differentiable function with discontinuous derivative, and state sufficient hypotheses for a local differentiable inverse. Unless an endpoint is explicitly mentioned, the function is defined on an open neighborhood of the point.

Differentiability at $a$ means a finite limit $f'(a)$ of the difference quotient. Equivalently,

$$f(a+h)=f(a)+f'(a)h+h\varepsilon(h),\qquad\varepsilon(h)\to0.$$

Define the error quotient for nonzero $h$ and give it value zero at zero. This form is the precise content of a tangent approximation. It implies continuity by letting the increment tend to zero, but does not by itself supply a numerical error constant or a bound on the second derivative.

A vertical or one-sided infinite slope is not a finite derivative under this definition. At a domain endpoint, a one-sided derivative may be a separately named object; it must not be silently identified with an unavailable two-sided derivative.

1. A chain-rule proof that handles zero inner changes

Suppose $g$ is differentiable at $a$, $F$ is differentiable at $b=g(a)$, and the nearby composition is defined. Write

$$F(b+k)-F(b)=F'(b)k+k\eta(k),\qquad\eta(k)\to0,$$

with $\eta(0)=0$. Let $k=g(a+h)-g(a)=g'(a)h+h\varepsilon(h)$. Continuity of $g$ ensures $k\to0$. Substitution gives

$$\frac{F(g(a+h))-F(g(a))}{h} =[F'(b)+\eta(k)][g'(a)+\varepsilon(h)]\longrightarrow F'(b)g'(a).$$

This proof never divides by $k$, which may be zero for infinitely many increments. The familiar product of two difference quotients needs an extra case when its proposed denominator is zero; the error-function proof includes that case automatically.

For a quotient $f/g$, the requirement $g(a)\ne0$ plus continuity of $g$ gives a neighborhood where $|g(x)|>|g(a)|/2$. Thus the quotient is locally defined and its derivative formula can be justified. Merely writing a squared denominator in the final answer does not prove this local domain fact.

From a proper area logarithm to the natural exponential

Use proper integrals of continuous functions on finite closed intervals, their proved additivity/order/linearity, the completeness and Archimedean properties, and the intermediate value theorem. Define

$$L(x)=\int_1^x\frac{dt}{t}\qquad(x>0),$$

with oriented bounds. Every integral here is proper: its interval stays a positive distance from zero.

Continuity and strict increase. For $0<u<v$, additivity and order give $L(v)-L(u)=\int_u^vdt/t\ge(v-u)/v>0$. Near any fixed $a>0$, restrict $x>a/2$. Then $|L(x)-L(a)|\le(2/a)|x-a|$, proving continuity without using a logarithmic derivative.

The product law from actual sums. Fix $u>0$. Scaling a tagged partition of the interval between $1$ and $v>0$ by $u$ scales its mesh by $u$ and its widths by $u$. The corresponding sums satisfy

$$\sum_i\frac{u\Delta s_i}{u\xi_i}=\sum_i\frac{\Delta s_i}{\xi_i}.$$

Both are sums of continuous proper integrands, so their all-mesh limits are equal. For $v<1$ reverse orientation; $v=1$ is zero directly. Hence $\int_u^{uv}dt/t=L(v)$ and

$$L(uv)=L(u)+L(v).$$

In particular $L(1)=0$, $L(1/u)=-L(u)$ and $L(2^n)=nL(2)$. As $L(2)>0$, the latter values are unbounded above by the Archimedean property; the reciprocal values are unbounded below. For any real $y$, choose positive arguments with $L$ below and above $y$. Continuity and the intermediate value theorem give an $x>0$ with $L(x)=y$. Strict increase makes that $x$ unique.

Construct and control the inverse. Define $E:\mathbb R\to(0,\infty)$ as this inverse. It is strictly increasing and continuous. To prove continuity at $y_0=L(a)$, choose $0<d<\min(a,\varepsilon)$. The numbers $L(a-d)<y_0<L(a+d)$ have positive gaps. If $|y-y_0|$ is smaller than both gaps, monotonicity forces $a-d<E(y)<a+d$, so $|E(y)-a|<\varepsilon$. The product law and uniqueness yield $E(s+t)=E(s)E(t)$, $E(0)=1$ and $E(-t)=1/E(t)$. Define $e=E(1)>1$ and $a^t=E(tL(a))$ for $a>0$. It follows that $a^{s+t}=a^sa^t$ and $(a^s)^t=a^{st}$, since $L(E(z))=z$. For integer exponents these equal repeated products; for rational $m/n$, $n>0$, the definition is the unique positive $n$th root of $a^m$. Thus it agrees with the elementary positive rational powers and gives their continuous real extension. In particular $E(t)=e^t$, and the area logarithm is its inverse $\ln=L$.

Derivatives without a cycle. For $a>0$, the average of $1/t$ on the interval between $a$ and $a+h$, with the sign of $h$ and orientation both retained, lies between its smallest and largest values on that interval. Both bounds tend to $1/a$. Thus

$$\frac{L(a+h)-L(a)}h\longrightarrow\frac1a.$$

For $y_0=L(a)$ and $h\ne0$, put $v=E(y_0+h)\ne a$. Inverse continuity gives $v\to a$, and

$$\frac{E(y_0+h)-E(y_0)}h =\left(\frac{L(v)-L(a)}{v-a}\right)^{-1}\longrightarrow a=E(y_0).$$

The denominator limit is nonzero, so the reciprocal is justified. This proves $L'=1/x$ and $E'=E$ from proper area and inverse continuity, with neither l'Hopital nor a previously assumed exponential derivative. The chain rule then gives $(x^r)'=rx^{r-1}$ on $x>0$ by $x^r=E(rL(x))$, and $(b^x)'=(\ln b)b^x$ for $b>0$. The case $b=1$ is the constant one. This supplies the normalization and identity bridge between the area logarithm, the inverse exponential and the real-power notation.