Real Analysis

Real analysis is calculus done honestly. Every limit, derivative, and integral you met before rested on one unstated assumption: that the real line has no gaps. This course makes that assumption explicit and derives everything from it. We begin with the algebraic and order structure the rationals and reals have in common, then locate the exact property the rationals lack.

The field axioms

A field is a set F with two operations, addition and multiplication, satisfying the familiar rules: both operations are commutative and associative, multiplication distributes over addition, there are distinct identities 0 and 1, every element has an additive inverse, and every nonzero element has a multiplicative inverse. The rationals Q and the reals R are both fields; the integers Z are not, because 2 has no integer reciprocal.

The order axioms

An ordered field also carries a relation < that is compatible with the operations: for all a, b, c, exactly one of a < b, a = b, b < a holds (trichotomy); if a < b then a + c < b + c; and if a < b and c > 0 then ac < bc. From these few rules every inequality you know can be proved. For instance, squares are never negative.

Proposition. In any ordered field, a^2 >= 0 for every a, and in particular 1 > 0.

Proof. By trichotomy either a = 0, a > 0, or a < 0. If a = 0 then a^2 = 0 >= 0. If a > 0, multiplying a > 0 by the positive number a gives a^2 > 0. If a < 0 then -a > 0, so (-a)^2 > 0; but (-a)^2 = a^2, so again a^2 > 0. In every case a^2 >= 0. Since 1 = 1^2 and 1 != 0, we get 1 > 0. This completes the proof.

The hole the rationals cannot fill

Everything above holds for Q. Yet the rationals are riddled with gaps, and here is the cleanest example.

Theorem. There is no rational number x with x^2 = 2.

Proof. Suppose, for contradiction, that x = p/q in lowest terms with x^2 = 2. Then p^2 = 2q^2, so p^2 is even, which forces p to be even, say p = 2k. Substituting, 4k^2 = 2q^2, so q^2 = 2k^2, making q even too. But then p and q share the factor 2, contradicting lowest terms. Hence no such rational exists.

The number we would call sqrt(2) is missing from Q, even though rationals get arbitrarily close to it from both sides. The reals are precisely the ordered field with no such holes. The property that seals the gaps is completeness, and stating it exactly is the goal of the next lesson.

Completeness is the axiom that distinguishes R from Q. To state it we need the language of bounds. Let S be a nonempty subset of R.

Bounds and least upper bounds

A number M is an upper bound of S if s <= M for every s in S. The number M is the supremum (or least upper bound), written sup S, if it is an upper bound and no smaller number is an upper bound. Symmetrically, m is a lower bound if m <= s for all s, and the infimum inf S is the greatest lower bound. The supremum need not belong to S: for the open interval (0, 1) we have sup = 1 and inf = 0, though neither lies in the set.

The completeness axiom

Completeness Axiom. Every nonempty subset of R that is bounded above has a supremum in R.

This single sentence is what Q lacks. The set S = {x in Q : x^2 < 2} is bounded above, but within Q it has no least upper bound; in R its supremum is sqrt(2). By applying the axiom to -S one gets the companion fact: every nonempty set bounded below has an infimum.

The epsilon characterization of the supremum

The following restatement is the workhorse of nearly every proof. It says the supremum is approached arbitrarily closely from inside the set.

Lemma. Let S be nonempty and bounded above. Then M = sup S if and only if (i) s <= M for all s in S, and (ii) for every epsilon > 0 there exists s in S with s > M - epsilon.

Proof. Suppose M = sup S. Then (i) holds because M is an upper bound. For (ii), given epsilon > 0, the number M - epsilon is smaller than M, so it is not an upper bound; hence some s in S exceeds it, giving s > M - epsilon. Conversely, suppose (i) and (ii) hold. Condition (i) makes M an upper bound. If some M' < M were also an upper bound, set epsilon = M - M' > 0; by (ii) there is s > M - epsilon = M', contradicting that M' bounds S. So M is the least upper bound. This completes the proof.

Read condition (ii) as a promise: no matter how tight a tolerance epsilon you name, the set contains a point within epsilon of the supremum. That is the mechanism by which suprema power the limit arguments to come.

Completeness is not just an abstract guarantee; it has concrete consequences that we use constantly. The first is that the natural numbers are unbounded, a fact that seems obvious but must be proved from the axioms.

The Archimedean property

Theorem (Archimedean property). For every real number x there is a natural number n with n > x.

Proof. Suppose not. Then some x is an upper bound for the set N of natural numbers. Since N is nonempty and bounded above, completeness gives a supremum M = sup N. Because M - 1 < M, it is not an upper bound, so there is a natural number n with n > M - 1. But then n + 1 > M, and n + 1 is also a natural number, contradicting that M bounds N. Hence N is unbounded, which is exactly the claim. This completes the proof.

Corollary. For every epsilon > 0 there is a natural number n with 1/n < epsilon. Indeed, applying the theorem to x = 1/epsilon gives n > 1/epsilon, and rearranging yields 1/n < epsilon. This corollary is the engine of countless convergence proofs, since it lets us shrink 1/n below any target.

Density of the rationals

The rationals have gaps, yet they are sprinkled everywhere along the line. Between any two distinct reals sits a rational.

Theorem (Density of Q). If a < b are real numbers, there is a rational p/q with a < p/q < b.

Proof. Since b - a > 0, the Archimedean corollary gives a natural number q with 1/q < b - a, so 1 < q(b - a). Let p be the smallest integer greater than qa (such an integer exists by the Archimedean property applied to qa and -qa). Then p - 1 <= qa < p. The right inequality gives qa < p, hence a < p/q. From p - 1 <= qa we get p <= qa + 1 < qa + q(b - a) = qb, so p/q < b. Therefore a < p/q < b. This completes the proof.

The idea is simple once seen: choose a denominator q fine enough that steps of size 1/q cannot jump over the interval (a, b), then take the first such step past a. The same reasoning shows the irrationals are dense too, so both kinds of numbers are woven tightly together throughout the line.

A sequence is a function from the natural numbers to the reals, written (a_n). The heart of analysis is making the phrase "the terms get close to L" completely precise. Here is the definition on which the entire subject rests.

The definition

Definition. A sequence (a_n) converges to L, written lim a_n = L, if for every epsilon > 0 there exists a natural number N such that for all n >= N we have |a_n - L| < epsilon.

Unpack the logic slowly. The challenger names a tolerance epsilon > 0, however small. You must respond with a threshold N so that from the N-th term onward, every term lies within epsilon of L. The quantity |a_n - L| is the distance from the term to the proposed limit, so the condition says that distance is eventually smaller than any positive number. Crucially N is allowed to depend on epsilon: tighter tolerances generally need larger thresholds.

A first direct proof

Claim. lim (1/n) = 0.

Proof. Let epsilon > 0 be given. By the Archimedean corollary choose a natural number N with 1/N < epsilon. Then for every n >= N we have |1/n - 0| = 1/n <= 1/N < epsilon. Since epsilon was arbitrary, the definition is satisfied and lim (1/n) = 0. This completes the proof.

Notice the shape of the argument, which every direct proof repeats: let epsilon be given, produce N (often by solving the target inequality for n), then verify the bound for all n >= N.

A second direct proof

Claim. lim ((2n + 1)/n) = 2.

Proof. Here |(2n + 1)/n - 2| = |1/n| = 1/n. Given epsilon > 0, choose N with 1/N < epsilon. For n >= N, 1/n <= 1/N < epsilon, so the terms are within epsilon of 2. Hence the limit is 2. This completes the proof.

Limits are unique

Theorem. A convergent sequence has exactly one limit.

Proof. Suppose a_n to L and a_n to L' with L != L'. Set epsilon = |L - L'|/2 > 0. There is N_1 with |a_n - L| < epsilon for n >= N_1, and N_2 with |a_n - L'| < epsilon for n >= N_2. For n >= max(N_1, N_2), the triangle inequality gives |L - L'| <= |L - a_n| + |a_n - L'| < epsilon + epsilon = |L - L'|, so |L - L'| < |L - L'|, an absurdity. Therefore L = L'. This completes the proof.

Proving every limit from scratch would be exhausting. This lesson builds a toolkit: a few theorems that let you compute limits by combining known ones, each justified by an epsilon argument.

Convergent implies bounded

Theorem. Every convergent sequence is bounded, meaning there is B with |a_n| <= B for all n.

Proof. Let a_n to L. Apply the definition with epsilon = 1: there is N with |a_n - L| < 1 for all n >= N, hence |a_n| < |L| + 1 for those n by the triangle inequality. The finitely many earlier terms a_1, ..., a_{N-1} are bounded by their maximum absolute value. Taking B = max(|a_1|, ..., |a_{N-1}|, |L| + 1) bounds the whole sequence. This completes the proof.

The converse fails: the bounded sequence a_n = (-1)^n does not converge, because its terms alternate between -1 and 1 and never settle.

The algebraic limit theorems

Suppose a_n to A and b_n to B. Then limits respect arithmetic:

• Sum: a_n + b_n to A + B.
• Product: a_n b_n to A B.
• Scalar: c a_n to c A for any constant c.
• Quotient: a_n / b_n to A / B, provided B != 0.

Proof of the sum rule. Let epsilon > 0. Choose N_1 with |a_n - A| < epsilon/2 for n >= N_1, and N_2 with |b_n - B| < epsilon/2 for n >= N_2. For n >= max(N_1, N_2), the triangle inequality gives |(a_n + b_n) - (A + B)| <= |a_n - A| + |b_n - B| < epsilon/2 + epsilon/2 = epsilon. Hence the sum converges to A + B. This completes the proof.

The trick of splitting epsilon into epsilon/2 + epsilon/2 is a recurring move; splitting into thirds appears in the product and quotient proofs.

The squeeze theorem

Theorem (Squeeze). If x_n <= a_n <= y_n for all large n, and x_n to L and y_n to L, then a_n to L.

For example, since -1/n <= (sin n)/n <= 1/n and both outer sequences tend to 0, the squeeze theorem gives lim ((sin n)/n) = 0 even though sin n itself does not converge. Trapping an awkward sequence between two tame ones is often the easiest route to its limit.

The definition of a limit lets you verify a proposed value, but it does not tell you a limit exists when you have no candidate. Completeness fills that gap for a large and useful class of sequences.

Monotone sequences

A sequence (a_n) is increasing if a_n <= a_{n+1} for all n, and decreasing if a_n >= a_{n+1} for all n. A sequence that is either is called monotone. Monotone sequences are the well-behaved ones: they can only head in a single direction.

The monotone convergence theorem

Theorem. A monotone increasing sequence that is bounded above converges, and its limit is sup {a_n}. Likewise a decreasing sequence bounded below converges to its infimum.

Proof (increasing case). Let (a_n) be increasing and bounded above. The set of terms {a_n} is nonempty and bounded above, so by completeness it has a supremum L = sup {a_n}. We show a_n to L. Let epsilon > 0. By the epsilon characterization of the supremum, L - epsilon is not an upper bound, so there is an index N with a_N > L - epsilon. Since the sequence is increasing, for every n >= N we have a_n >= a_N > L - epsilon. Also a_n <= L because L is an upper bound. Combining, L - epsilon < a_n <= L, so |a_n - L| < epsilon. Hence a_n to L. This completes the proof.

This is the first existence theorem for limits: it produces a limit from boundedness and monotonicity alone, with no formula in sight.

Worked application: a recursive sequence

Define a_1 = 1 and a_{n+1} = sqrt(2 + a_n). Does it converge?

1. Bounded above by 2. Check by induction: a_1 = 1 < 2, and if a_n < 2 then a_{n+1} = sqrt(2 + a_n) < sqrt(4) = 2.
2. Increasing. Since 0 <= a_n < 2, we have a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = -(a_n - 2)(a_n + 1) > 0, so a_{n+1} > a_n.
3. Conclusion. The sequence is increasing and bounded above, so by the theorem it converges to some L.

To find L, take limits in a_{n+1} = sqrt(2 + a_n): since both sides tend to L, we get L = sqrt(2 + L), so L^2 = 2 + L, giving L^2 - L - 2 = 0 and (L - 2)(L + 1) = 0. The limit must be positive, so L = 2. The theorem told us the limit exists; the recursion then pinned its value.

Not every bounded sequence converges, but something almost as good is always true: a bounded sequence must contain a convergent piece. Making that precise requires the idea of a subsequence.

Subsequences

A subsequence of (a_n) is obtained by selecting an infinite, strictly increasing list of indices n_1 < n_2 < n_3 < ... and keeping only those terms, giving (a_{n_k}). For instance, the even-indexed terms a_2, a_4, a_6, ... form a subsequence.

Fact. If a_n to L, then every subsequence also converges to L. The contrapositive is a handy divergence test: if two subsequences head to different limits, the sequence diverges. The sequence (-1)^n diverges precisely because its even terms tend to 1 while its odd terms tend to -1.

Every sequence has a monotone subsequence

Lemma. Every sequence of reals has a monotone subsequence.

Proof sketch. Call an index m a peak if a_m >= a_n for all n > m. If there are infinitely many peaks, listing them in order gives a decreasing subsequence. If there are only finitely many peaks, then past the last peak every index fails to be a peak, so each such term has a strictly larger term further along; chaining these choices produces a strictly increasing subsequence. Either way a monotone subsequence exists. This completes the sketch.

The Bolzano-Weierstrass theorem

Theorem (Bolzano-Weierstrass). Every bounded sequence of real numbers has a convergent subsequence.

Proof. Let (a_n) be bounded. By the lemma it has a monotone subsequence (a_{n_k}). This subsequence inherits the bound of the original sequence, so it is a bounded monotone sequence, and by the monotone convergence theorem it converges. Hence (a_n) has a convergent subsequence. This completes the proof.

This theorem is a cornerstone. It is the reason closed bounded intervals are compact, and compactness is what makes continuous functions on such intervals attain their maxima and behave uniformly, results you will prove in Module 4. A useful picture: no matter how the terms of a bounded sequence bounce around inside an interval, infinitely many of them must cluster near some single point, and that cluster point is the limit of the subsequence.

The limit definition requires you to know the limit L in advance. Cauchy's insight was to characterize convergence internally, using only how the terms relate to each other, with no reference to any external value.

The definition

Definition. A sequence (a_n) is Cauchy if for every epsilon > 0 there exists N such that for all m, n >= N we have |a_m - a_n| < epsilon.

The terms of a Cauchy sequence get arbitrarily close to one another, not to a named target. Intuitively the sequence bunches up. Note the two indices m and n: both must lie past N, so the whole tail is squeezed into an interval of width epsilon.

Convergent sequences are Cauchy

Theorem. Every convergent sequence is Cauchy.

Proof. Suppose a_n to L and let epsilon > 0. Choose N with |a_n - L| < epsilon/2 for all n >= N. Then for m, n >= N, the triangle inequality gives |a_m - a_n| <= |a_m - L| + |L - a_n| < epsilon/2 + epsilon/2 = epsilon. Hence the sequence is Cauchy. This completes the proof.

Cauchy sequences converge: the completeness of R

The deep direction is the converse, and it is equivalent to the completeness axiom.

Theorem (Cauchy criterion). Every Cauchy sequence of real numbers converges.

Proof. Let (a_n) be Cauchy. First it is bounded: taking epsilon = 1, there is N with |a_n - a_N| < 1 for n >= N, so those terms lie within 1 of a_N, and the finitely many earlier terms are bounded as well. By Bolzano-Weierstrass the bounded sequence has a subsequence a_{n_k} to L. We claim the whole sequence converges to that same L. Let epsilon > 0. Since the sequence is Cauchy, choose N_1 with |a_m - a_n| < epsilon/2 for m, n >= N_1. Since the subsequence converges, choose an index n_k >= N_1 with |a_{n_k} - L| < epsilon/2. Then for any n >= N_1, |a_n - L| <= |a_n - a_{n_k}| + |a_{n_k} - L| < epsilon/2 + epsilon/2 = epsilon. Hence a_n to L. This completes the proof.

The two theorems together give the Cauchy criterion: a sequence of reals converges if and only if it is Cauchy. This is enormously useful because it lets you prove convergence without guessing the limit, and it is the notion of completeness used to build the reals and, later, to work in more general spaces.

An infinite series is an attempt to add infinitely many numbers. We give it meaning through sequences, so nothing new is assumed; convergence of a series is just convergence of an associated sequence.

Partial sums

Given a sequence (a_k), its partial sums are s_n = a_1 + a_2 + ... + a_n. The series sum a_k converges to S if the sequence of partial sums converges to S, and then we write sum a_k = S. If the partial sums diverge, the series diverges. So every question about series is secretly a question about the sequence (s_n).

The geometric series

The most important example is the geometric series sum r^k starting at k = 0. Its partial sum has a closed form: s_n = 1 + r + r^2 + ... + r^{n-1} = (1 - r^n)/(1 - r) for r != 1. If |r| < 1, then r^n to 0, so s_n to 1/(1 - r). If |r| >= 1, the terms do not shrink and the series diverges.

Result. For |r| < 1, sum (from k=0 to infinity) r^k = 1/(1 - r). For example sum (1/2)^k = 1/(1 - 1/2) = 2.

A telescoping series

Example. Evaluate sum (from k=1 to infinity) 1/(k(k+1)). Using partial fractions, 1/(k(k+1)) = 1/k - 1/(k+1). The partial sum collapses:

s_n = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1)) = 1 - 1/(n+1).

As n to infinity, s_n to 1, so the series sums to 1. When consecutive terms cancel like this, the series is called telescoping, and its sum is read off the surviving endpoints.

The n-th term test

Theorem (Divergence test). If sum a_k converges, then a_k to 0. Equivalently, if a_k does not tend to 0, the series diverges.

Proof. Suppose sum a_k = S, so s_n to S and also s_{n-1} to S. Then a_n = s_n - s_{n-1} to S - S = 0. This completes the proof.

Warning: the converse is false. Terms tending to 0 do not guarantee convergence, as the harmonic series in the next lesson shows. The n-th term test can only prove divergence, never convergence.

Most series have no closed-form partial sum, so we judge convergence indirectly with tests. This lesson collects the essential ones, each provable from the tools of Module 2.

The harmonic series diverges

Theorem. The harmonic series sum 1/k diverges, even though 1/k to 0.

Proof (grouping). Group the terms in blocks whose lengths are powers of two: (1/2) + (1/3 + 1/4) + (1/5 + ... + 1/8) + .... Each parenthesized block is at least 1/2, since for example 1/3 + 1/4 > 1/4 + 1/4 = 1/2 and 1/5 + 1/6 + 1/7 + 1/8 > 4 * (1/8) = 1/2. Adding enough blocks makes the partial sums exceed any bound, so the series diverges. This completes the proof.

More generally the p-series sum 1/k^p converges if and only if p > 1. So sum 1/k^2 converges while sum 1/k and sum 1/sqrt(k) diverge.

The comparison test

Comparison test. Suppose 0 <= a_k <= b_k for all large k. If sum b_k converges then sum a_k converges; if sum a_k diverges then sum b_k diverges. The proof rests on the monotone convergence theorem: partial sums of nonnegative terms are increasing, so they converge exactly when bounded, and the larger series bounds the smaller.

Example. sum 1/(k^2 + 1) converges because 1/(k^2 + 1) < 1/k^2 and the p-series sum 1/k^2 converges.

The ratio test

Ratio test. For a series with nonzero terms, suppose |a_{k+1}/a_k| to L. If L < 1 the series converges absolutely; if L > 1 it diverges; if L = 1 the test is inconclusive. It works by comparison with a geometric series of ratio near L.

Example. sum 1/k! has ratio a_{k+1}/a_k = 1/(k+1) to 0 < 1, so it converges. (Its sum is the number e - 1.)

The integral test

Integral test. If f is positive, continuous, and decreasing with f(k) = a_k, then sum a_k and the improper integral of f from 1 to infinity either both converge or both diverge. This is the cleanest way to settle p-series.

Absolute versus conditional convergence

A series converges absolutely if sum |a_k| converges. Absolute convergence implies convergence. A series that converges but not absolutely converges conditionally. The alternating harmonic series sum (-1)^{k+1}/k is the classic example: it converges (to ln 2) by the alternating series test, yet sum 1/k diverges, so its convergence is only conditional.

We now move from sequences to functions. The limit of a function at a point is defined with two tolerances instead of one: an output tolerance epsilon and an input tolerance delta.

The definition

Definition. Let f be defined near c (except possibly at c). We say lim (x to c) f(x) = L if for every epsilon > 0 there exists delta > 0 such that whenever 0 < |x - c| < delta, it follows that |f(x) - L| < epsilon.

Read it as a game. The challenger fixes an output band of half-width epsilon around L. You must find an input band of half-width delta around c so that every x in your band (except c itself) is sent into the output band. The clause 0 < |x - c| excludes x = c, because the limit concerns behavior near c, not the value there.

A linear example

Claim. lim (x to 3) (2x + 1) = 7.

Proof. Let epsilon > 0. We need |(2x + 1) - 7| < epsilon, that is |2x - 6| = 2|x - 3| < epsilon, i.e. |x - 3| < epsilon/2. So choose delta = epsilon/2. Then 0 < |x - 3| < delta gives |(2x + 1) - 7| = 2|x - 3| < 2 delta = epsilon. This completes the proof.

The strategy is standard: start from the desired output inequality, solve it for |x - c|, and let that bound define delta.

A quadratic example

Claim. lim (x to 2) x^2 = 4.

Proof. We estimate |x^2 - 4| = |x - 2||x + 2|. To control the factor |x + 2|, first restrict |x - 2| < 1, which forces 1 < x < 3 and hence |x + 2| < 5. Then |x^2 - 4| < 5|x - 2|. Given epsilon > 0, choose delta = min(1, epsilon/5). For 0 < |x - 2| < delta, both restrictions hold, so |x^2 - 4| < 5 |x - 2| < 5 (epsilon/5) = epsilon. This completes the proof.

The min trick handles the extra factor: one part of delta tames the coefficient, the other delivers the final epsilon.

The sequential criterion

Theorem. lim (x to c) f(x) = L if and only if for every sequence x_n to c with x_n != c, we have f(x_n) to L. This bridge lets you import every sequence theorem, and it gives a quick way to prove a limit does not exist: exhibit two sequences approaching c whose images head to different values.

Continuity is the property that the value of a function agrees with its limit. It is deceptively simple to state and astonishingly powerful in its consequences.

The definition

Definition. A function f is continuous at c if lim (x to c) f(x) = f(c). Spelled out with epsilon and delta: for every epsilon > 0 there is delta > 0 such that |x - c| < delta implies |f(x) - f(c)| < epsilon. Here x = c is allowed, and the inequality holds trivially there. A function is continuous on a set if it is continuous at each point of the set.

By the algebraic limit theorems, sums, products, quotients (with nonzero denominator), and compositions of continuous functions are continuous. Polynomials are continuous everywhere; rational, root, exponential, logarithmic, and trigonometric functions are continuous throughout their domains.

The intermediate value theorem

Theorem (IVT). If f is continuous on [a, b] and y lies strictly between f(a) and f(b), then there exists c in (a, b) with f(c) = y.

Proof. Assume f(a) < y < f(b) (the other case is symmetric). Let S = {x in [a, b] : f(x) < y}. This set is nonempty (it contains a) and bounded above by b, so by completeness c = sup S exists. We claim f(c) = y. If f(c) < y, continuity provides a small interval around c on which f stays below y, so points slightly larger than c lie in S, contradicting that c is an upper bound. If f(c) > y, continuity keeps f above y near c, so c is not the least upper bound. By trichotomy the only possibility left is f(c) = y. This completes the proof.

The IVT is why a continuous function cannot jump over a value: to pass from below y to above it, the graph must cross the level y. This guarantees roots, as when a continuous function changes sign.

The extreme value theorem

Theorem (EVT). If f is continuous on a closed bounded interval [a, b], then f attains a maximum and a minimum on [a, b].

Proof idea. First, f is bounded on [a, b]; otherwise some sequence x_n would have f(x_n) to infinity, but Bolzano-Weierstrass gives a subsequence x_{n_k} to p in [a, b], and continuity forces f(x_{n_k}) to f(p), a finite number, a contradiction. Let M = sup f([a, b]). Choose x_n with f(x_n) to M; a convergent subsequence x_{n_k} to p yields f(p) = M by continuity, so the maximum is attained. The minimum follows by applying the result to -f. This completes the sketch.

Both hypotheses are essential. On the open interval (0, 1) the continuous function 1/x is unbounded, and f(x) = x attains no maximum. Closedness and boundedness together, that is compactness, are what make the extreme value theorem work.

Ordinary continuity lets delta depend on both epsilon and the point c. Uniform continuity demands a single delta that works everywhere at once. The distinction is subtle but decisive, especially for integration.

The definition

Definition. A function f is uniformly continuous on a set D if for every epsilon > 0 there exists delta > 0 such that for all x, u in D with |x - u| < delta, we have |f(x) - f(u)| < epsilon.

Compare the quantifiers. In pointwise continuity the order is: fix c, fix epsilon, then find delta, so delta may shrink from point to point. In uniform continuity delta is chosen before any point is named, so the same delta controls the whole domain. Uniform continuity is the stronger property, and it always implies ordinary continuity.

Continuous but not uniformly continuous

Example. The function f(x) = 1/x is continuous on (0, 1) but not uniformly continuous there. Near 0 the graph steepens without bound, so no single delta can keep the outputs within epsilon. Concretely, take the sequences x_n = 1/n and u_n = 1/(n+1). Then |x_n - u_n| = 1/(n(n+1)) to 0, yet |f(x_n) - f(u_n)| = |n - (n+1)| = 1 for every n. Since inputs get arbitrarily close while outputs stay a full unit apart, no delta works for epsilon = 1. Hence f is not uniformly continuous.

A similar failure afflicts f(x) = x^2 on the whole line R: as x grows, a fixed input gap produces an ever larger output gap, since (x + h)^2 - x^2 = 2xh + h^2 grows with x.

The saving theorem

Theorem. If f is continuous on a closed bounded interval [a, b], then f is uniformly continuous on [a, b].

Proof idea. Suppose not. Then for some epsilon > 0 and every n there are points x_n, u_n in [a, b] with |x_n - u_n| < 1/n but |f(x_n) - f(u_n)| >= epsilon. By Bolzano-Weierstrass a subsequence x_{n_k} to p in [a, b]; since |x_n - u_n| to 0, also u_{n_k} to p. Continuity at p forces both f(x_{n_k}) and f(u_{n_k}) to converge to f(p), so their difference tends to 0, contradicting |f(x_{n_k}) - f(u_{n_k})| >= epsilon. Hence f is uniformly continuous. This completes the sketch.

Once again compactness of [a, b] does the heavy lifting. This theorem is exactly what we will need in Module 6 to prove that every continuous function on a closed interval is Riemann integrable.

With limits of functions now on firm footing, we can define the derivative precisely and prove its basic properties rather than merely assert them.

The definition

Definition. A function f is differentiable at c if the limit

f'(c) = lim (h to 0) [f(c + h) - f(c)] / h

exists; the value of the limit is the derivative at c. An equivalent form replaces c + h by x: f'(c) = lim (x to c) [f(x) - f(c)]/(x - c). The quotient inside is the slope of the secant line through (c, f(c)) and a nearby point; the limit is the slope of the tangent line, the instantaneous rate of change. Every derivative computation is therefore an epsilon-delta limit in disguise.

A derivative from the definition

Claim. If f(x) = x^2, then f'(c) = 2c.

Proof. The difference quotient is [(c + h)^2 - c^2]/h = [2ch + h^2]/h = 2c + h for h != 0. As h to 0, this tends to 2c by the algebraic limit theorems. Hence f'(c) = 2c. This completes the proof.

Claim. If f(x) = 1/x with c != 0, then f'(c) = -1/c^2.

Proof. The difference quotient is [1/(c + h) - 1/c]/h = [(c - (c + h))/(c(c + h))]/h = [-h/(c(c + h))]/h = -1/(c(c + h)). Letting h to 0 gives -1/c^2. This completes the proof.

Differentiability implies continuity

Theorem. If f is differentiable at c, then f is continuous at c.

Proof. For x != c write f(x) - f(c) = [ (f(x) - f(c))/(x - c) ] (x - c). As x to c, the first factor tends to f'(c) and the second tends to 0, so by the product rule for limits f(x) - f(c) to f'(c) * 0 = 0. Hence f(x) to f(c), which is continuity at c. This completes the proof.

The converse fails. The function f(x) = |x| is continuous at 0 but not differentiable there: the difference quotient |h|/h equals +1 for h > 0 and -1 for h < 0, so the one-sided limits disagree and the two-sided limit does not exist. A sharp corner has no single tangent slope.

The mean value theorem is the central theorem of differential calculus. Nearly every qualitative fact about derivatives, monotonicity, constancy, error bounds, flows from it. We build up to it in three steps.

Fermat's theorem on interior extrema

Theorem (Fermat). If f has a local maximum or minimum at an interior point c and is differentiable there, then f'(c) = 0.

Proof. Suppose f has a local maximum at c. For small h > 0, f(c + h) - f(c) <= 0, so the quotient [f(c + h) - f(c)]/h <= 0, giving f'(c) <= 0 in the limit. For small h < 0, the same numerator is <= 0 but h < 0, so the quotient is >= 0, giving f'(c) >= 0. Both hold only if f'(c) = 0. This completes the proof.

Rolle's theorem

Theorem (Rolle). If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is c in (a, b) with f'(c) = 0.

Proof. By the extreme value theorem f attains a maximum and a minimum on [a, b]. If both occur at the endpoints, then since f(a) = f(b) the maximum equals the minimum, so f is constant and f' = 0 throughout. Otherwise an extremum occurs at an interior point c, and Fermat's theorem gives f'(c) = 0. This completes the proof.

The mean value theorem

Theorem (Mean Value Theorem). If f is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) with

f'(c) = [f(b) - f(a)] / (b - a).

Proof. Define the auxiliary function g(x) = f(x) - [f(a) + ((f(b) - f(a))/(b - a))(x - a)], which subtracts from f the straight secant line joining the endpoints. Then g is continuous on [a, b], differentiable on (a, b), and g(a) = g(b) = 0. By Rolle's theorem there is c with g'(c) = 0. But g'(x) = f'(x) - (f(b) - f(a))/(b - a), so g'(c) = 0 gives exactly f'(c) = (f(b) - f(a))/(b - a). This completes the proof.

Geometrically, somewhere on the curve the tangent line is parallel to the secant joining the endpoints: the instantaneous rate equals the average rate at least once.

A fundamental consequence

Corollary. If f'(x) = 0 for all x in an interval, then f is constant there.

Proof. Take any two points x_1 < x_2 in the interval. By the mean value theorem there is c between them with f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0, so f(x_2) = f(x_1). Since the two points were arbitrary, f is constant. This completes the proof.

This is the rigorous reason two antiderivatives of the same function differ by a constant, the fact that makes the fundamental theorem of calculus, and integration itself, well defined.

Integration measures accumulated area. We make this rigorous by trapping the region under a graph between over-estimates and under-estimates built from rectangles, then squeezing them together.

Partitions and Darboux sums

Let f be bounded on [a, b]. A partition P is a finite set of points a = x_0 < x_1 < ... < x_n = b, cutting [a, b] into subintervals. On the i-th subinterval let M_i be the supremum of f and m_i the infimum. The upper sum and lower sum are

U(f, P) = sum M_i (x_i - x_{i-1}),   L(f, P) = sum m_i (x_i - x_{i-1}).

The upper sum uses the tallest rectangle on each piece and overestimates the area; the lower sum uses the shortest and underestimates it. Always L(f, P) <= U(f, P). Refining a partition by adding points can only lower U and raise L, tightening the trap.

The upper and lower integrals

The upper integral is the infimum of U(f, P) over all partitions, and the lower integral is the supremum of L(f, P). These always exist for a bounded function, by completeness, and the lower integral never exceeds the upper. We say f is Riemann integrable on [a, b] when the two are equal, and their common value is the integral, written integral (from a to b) f.

The Riemann criterion

Theorem (Riemann criterion). A bounded function f is integrable on [a, b] if and only if for every epsilon > 0 there exists a partition P with U(f, P) - L(f, P) < epsilon.

The criterion turns integrability into a checkable condition: the gap between the over- and under-estimates can be made arbitrarily small by choosing a fine enough partition. This is the practical test used in almost every integrability proof.

Continuous functions are integrable

Theorem. Every function continuous on [a, b] is Riemann integrable.

Proof. Since [a, b] is closed and bounded, f is uniformly continuous there (Module 4). Let epsilon > 0. Uniform continuity gives delta > 0 so that |x - u| < delta implies |f(x) - f(u)| < epsilon/(b - a). Choose a partition with every subinterval shorter than delta. On each subinterval f attains its max and min (by the extreme value theorem), and these points are within delta, so M_i - m_i < epsilon/(b - a). Therefore U(f, P) - L(f, P) = sum (M_i - m_i)(x_i - x_{i-1}) < (epsilon/(b - a)) sum (x_i - x_{i-1}) = (epsilon/(b - a))(b - a) = epsilon. By the Riemann criterion, f is integrable. This completes the proof.

Notice how three pillars converge here: completeness gives the upper and lower integrals, the extreme value theorem supplies the max and min on each piece, and uniform continuity guarantees the oscillation shrinks uniformly. That is the payoff of the whole course.

We close by letting whole functions vary. Given functions f_1, f_2, f_3, ..., in what sense can they approach a limit function f? There are two natural answers, and the difference between them is one of the most important distinctions in analysis.

Pointwise convergence

Definition. A sequence (f_n) converges pointwise to f on a set D if for each fixed x in D, the number sequence f_n(x) converges to f(x). Spelled out: for each x and each epsilon > 0 there is N (which may depend on x) with |f_n(x) - f(x)| < epsilon for n >= N.

The catch is that N can vary wildly from point to point, and this lets the limit misbehave.

Pointwise convergence can destroy continuity

Example. Let f_n(x) = x^n on [0, 1]. For 0 <= x < 1, x^n to 0; at x = 1, x^n = 1 for all n, so the limit is 1. The pointwise limit is

f(x) = 0 for 0 <= x < 1, and f(1) = 1.

Each f_n is continuous, yet the limit f has a jump at x = 1 and is discontinuous. Pointwise convergence alone does not preserve continuity. The trouble is that near x = 1 you need larger and larger n to push x^n close to 0, so no single N works across the interval.

Uniform convergence

Definition. A sequence (f_n) converges uniformly to f on D if for every epsilon > 0 there is a single N such that for all n >= N and all x in D, |f_n(x) - f(x)| < epsilon.

The quantifier order is the whole story, exactly as with uniform continuity: here N is chosen before x, so the same threshold works everywhere. Equivalently, the maximum gap sup_x |f_n(x) - f(x)| tends to 0. Picture an epsilon-tube of vertical half-width epsilon drawn around the graph of f; uniform convergence means that from some N on, the entire graph of f_n lies inside the tube.

Uniform convergence preserves continuity

Theorem. If each f_n is continuous on D and f_n to f uniformly, then f is continuous on D.

Proof. Fix c in D and epsilon > 0. By uniform convergence choose N with |f_N(x) - f(x)| < epsilon/3 for all x. Since f_N is continuous at c, choose delta > 0 so that |x - c| < delta implies |f_N(x) - f_N(c)| < epsilon/3. Then for such x, the triangle inequality gives

|f(x) - f(c)| <= |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| < epsilon/3 + epsilon/3 + epsilon/3 = epsilon.

Hence f is continuous at c, and since c was arbitrary, on all of D. This completes the proof.

This famous "epsilon/3 argument" is the reason uniform convergence is the right notion for interchanging limits with continuity, and similar theorems let you interchange uniform limits with integration. You have now traced the logical thread from the completeness of the real line all the way to the analysis of function spaces, the doorway to advanced analysis. Well done.