Real-notes Frédéric Latrémolière

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TREAL ANALYSIS

FRDRIC LATRMOLIRE

ABSTRACT. We present raw notes for the topological section of real analysis. Westrive for a minimal presentation, not for the greater generality nor to transmitmore than the strict proofs of the theorems of analysis regarding continuity. Thisdocument is a reference manual rather than a user manual; it is self-containedassuming the usual axiomatic of R. The scope of these notes are the topologi-cal theorems of calculus from the axiomatic of R in a manner suitable for a realanalysis class.

1. SEQUENCES

We work in R, the (unique, up to ordered-field isomorphism) Archimedeanlinearly ordered field with the property that every nonempty bounded subset of R hasa supremum.

1.1. Sequences and subsequences.

Definition 1.1.1. A sequence in R is a function fromN to R.

Lemma 1.1.2. If : N N is a strictly increasing function, then for all n N we have(n) > n.Proof. By definition, (0) N so (0) > 0. Assume (n) > n for some n N.Then (n+ 1) > (n) > n so (n+ 1) > n+ 1 (as n+ 1 is the smallest element ofN strictly greater than n). Definition 1.1.3. A subsequence of a sequence (xn)nN is a sequence of the form(x(n))nN for some strictly increasing function : N N.1.2. Convergence.

Definition 1.2.1. A sequence (xn)nN in R converges to l Rwhen: > 0 N N n N (n > N) = |xn l| < .

Theorem 1.2.2. If a sequence (xn)nN in R converges to l R and l R then l = l.Proof. Assume l 6= l. Let = 12 |l l|. Then there exists N N such that for alln > N, we have |xn l| < and there exists N N such that for all n > N wehave |xn l| < . Then, for n = max{N, N} we have:

|l l| 6 |l xn|+ |xn l| < 2 = |l l|.We have reached a contradiction, so l = l.

Date: November 7, 2013.2000 Mathematics Subject Classification. Primary: 46L89, 46L30, 58B34.Key words and phrases. Noncommutative metric geometry, Gromov-Hausdorff convergence,

Monge-Kantorovich distance, non-unital C*-algebras, Quantum Metric Spaces, Lip-norms.1

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Notation 1.2.3. If (xn)nN is a sequence in Rwhich converges to l, we denote l bylimn xn. We will also say that (xn)nN has a limit, or has limit l (at infinity).

Theorem 1.2.4. Let r N, r > 0 and M R. The sequence(

MNr

)nN

converges to 0.

Proof. Let > 0. Let = sup{x R : xr 6 }, which exists as the supremum of anonempty bounded subset ofR. SinceR is Archimedean, there exists N N suchthat N > |M| . Thus, for all n > N we have:Mnr

6 |M|Nr < r 6 .This concludes our proof. Theorem 1.2.5. Let (xn)nN and (yn)nN be two sequences. If there exists N N suchthat for all n > N we have xn = yn, then (xn)nN converges to l R if and only if(yn)nN converges to l.

Proof. Assume that (xn)nN converges to l. Let > 0. There exists N N suchthat for all n > N we have |xn l| < . Let n > max{N, N}. Then yn = xn andthus |yn l| < . Thus (yn)nN converges to l. The result is symmetric in the roleof (xn)nN and (yn)nN. Remark 1.2.6. We can thus use Theorem (1.2.5) to work with truncated sequences,i.e. functions from {n N : n > N} to R for an arbitrary N N: such truncatedsequences can be regarded as sequences by completing them with 0 for the first Nentries, without changing the behavior at infinity.

Theorem 1.2.7. A sequence (xn)nN converges to l R if and only if (|xn l|)nNconverges to 0.

Proof. This is by definition since ||x|| = |x| for all x R. Theorem 1.2.8. Let (xn)nN be a sequence in R. If (xn)nN converges, then everysubsequence of (xn)nN converges to the same limit.

Proof. Let l = limn xn. Let : N N be strictly increasing. Let > 0. Thereexists N N such that for all n > N we have |xn l| < . Since (n) > n forall n N by Lemma (1.1.2), we conclude that for all n > N we have |x(n) l| 0such that for all N N, there exists n > N such that |xn l| > . Thus, for anyN N, the set {n N : n > N and |xn l| > } is not empty.

Let (0) = min{n N : |xn l| > }. Assume that for some n N we haveconstructed (0) < (1) < . . . < (n) such that |x(n) l| > . Let:

(n + 1) = min {n N : n > (n) + 1 and |xn l| > } .

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We thus have constructed : N N strictly increasing such that for all n N, wehave |x(n) l| > . The subsequence (x(n))nN has no subsequence convergingto l, thus proving our theorem.

1.3. Order and Limit.

Theorem 1.3.1. If a sequence (xn)nN in R converges, then it is bounded.

Proof. Let l = limn xn. There exists N N such that for all n > N, we have|xn l| < 1 so |xn| < |l|+ 1. Let M = max{|l|+ 1, |xn| : n N, n 6 N}. Then forall n Nwe have |xn| 6 M. Theorem 1.3.2. Let (xn)nN and (yn)nN be two convergent sequences in R. If thereexists N N such that for all n > N we have xn 6 yn, then limn xn 6 limn yn.Proof. Let lx = limn xn and ly = limn yn. If lx > ly then let =

lxly2 > 0.

There exists Nx N such that for all n N with n > Nx, we have lx < xn

lx+ly2 for all n > Nx.

There exists Ny N such that for all n > Ny we have ly < yn < ly + , so inparticular lx+ly2 > yn.

For n > max{Nx,Ny}, we have xn > yn, which proves our theorem by contrap-position.

Theorem 1.3.3. If (xn)nN converges to l > 0 then there exists N N such that for alln > N we have xn > l2 .

Proof. Let = 12 l. There exists N N such that for all n > N we have xn > l =l2 .

Theorem 1.3.4 (Squeeze Theorem). Let (xn)nN, (yn)nN and (zn)nN be three se-quences in R such that:

(1) there exists N N such that for all n > N we have xn 6 yn 6 zn,(2) the sequences (xn)nN and (zn)nN converge and limn xn = limn zn.

Then (yn)nN converges and limn yn = limn xn.

Proof. Let l = limn xn = limn zn. Let > 0. There exists Nx N such thatfor all n > Nx we have |xn l| < so l < xn. There exists Ny N such thatfor all n > Ny we have |zn l| < so zn < l + . Let n > max{N, Nx, Ny}. Then:

l < xn 6 yn 6 zn < l + .This concludes our proof.

Corollary 1.3.5. If (xn)nN converges to l R then (|xn|)nN converges to |l|.Proof. For all n N we have 0 6 ||xn| |l|| 6 |xn l|. Conclude by applyingTheorem (1.3.4).

1.4. Limits and Algebra.

Theorem 1.4.1. If (xn)nN and (yn)nN are two convergent sequences and t R then(xn + tyn)nN converges to limn xn + t limn yn.

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Proof. Let lx = limn xn and ly = limn yn. Let > 0. There exists Nx Nsuch that for all n > Nx we have |xn lx| < . There exists Ny N such that forall n > Ny we have |yn ly| < (|t|+ 1)1.

Let n > max{Nx, Ny}. Then:|(xn + tyn) (lx + tly)| 6 |xn lx|+ |t||yn ly| < .

Theorem 1.4.2. If (xn)nN and (yn)nN are two convergent sequences and t R then(xnyn)nN converges to limn xn limn yn.Proof. Let lx = limn xn and ly = limn yn. Since (xn)nN converges, it isbounded: there exists M R such that for all n N we have |xn| 6 M. For alln Nwe have:

0 6 |xnyn lxly| = |xnyn xnly + xnly lxly|6 |xn lx||ly|+ |yn ly||xn|6 |xn lx||ly|+ |yn ly|M.

By assumption, (|xn lx|)nN and (|yn ly|)nN converge to 0. By Theorem(1.4.1), (|xn lx||ly|+ |yn ly|M)nN converges to 0 as well. By Theorem (1.3.4),we conclude (|xnyn lxly|)nN converges to 0. Hence (xnyn)nN converges tolxly.

Theorem 1.4.3. If (xn)nN is a sequence in R converging to some l R \ {0}, thenthere exists N N such that for all n > N we have xn 6= 0, and moreover (x1n )n>Nconverges to l1.

Proof. If l < 0, then replace (xn)nN by (xn)nN, so that we may assume l > 0.By Theorem (1.3.3), there exists N N such that for all n > N we have xn > l2 > 0.Now let > 0. There exists P N such that for all n > P we have |xn l| < 2l2 .Thus, for all n > max{P, N} we have: 1xn 1l

= |l xn|l|xn|6 |xn l|1

2 l2

< .

This completes our proof.

1.5. Monotone Sequences.

Definition 1.5.1. A sequence (xn)nN is increasing when for all n 6 m N wehave xn 6 xm.

Definition 1.5.2. A sequence (xn)nN is decreasing when (xn)nN is increasing.Definition 1.5.3. A sequence is monotone if it is either decreasing or increasing.

Definition 1.5.4. A sequence (xn)nN is strictly increasing (resp. strictly decreas-ing) when for all n 6 m Nwe have xn < xm (resp. xm < xn).

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Theorem 1.5.5 (Monotone Subsequence Theorem). If (xn)nN is a sequence in Rthen there exists a subsequence (x(n))nN which is monotone.

Proof. Let:P = {n N : k N k > n = xk 6 xn}.

If P is infinite, then let (0) = minP . Assume that for some n N wehave constructed (0) < . . . < (n) P . The set P \ {(j) : 0 6 j 6n} is not empty since P is infinite. Let (n + 1) be its smallest element.Now : N N is strictly increasing, and by construction, if n 6 m thenx(m) 6 x(n). Thus (x(n))nN is decreasing.

IfP is finite, let (0) N such that if n > M then n 6P . Now assume wehave constructed (0) < . . . (n) for some n N with x(j) 6 x(j+1) forall 0 6 j 6 n 1. Since (n) 6 P there exists (n + 1) > (n) such thatx(n+1) > x(n). The sequence (x(n))nN thus constructed by induction isincreasing.

Our theorem is thus proven.

Theorem 1.5.6 (Monotone Convergence Theorem). A monotone sequence (xn)nNis convergent if and only if it is bounded.

Proof. Any convergent sequence is bounded by Theorem (1.3.1). Assume now that(xn)nN is a bounded increasing sequence. The set X = {xn : n N} is thusnonempty and bounded, so it has a supremum l. Let > 0. Since l is notan upper bound for X, there exists N N such that xN > l . Since (xn)nNis increasing and l is an upper bound of the sequence, we get for all n > N thatl > xn > xN > l . Hence for all n N with n > N we have |xn l| < , so(xn)nN converges to l.

If (xn)nN is bounded decreasing, then (xn)nN is bounded increasing, andwe conclude with the method above and Theorem (1.4.1).

Theorem 1.5.7. Let a R. The sequence (an)nN (with the convention that a0 = 1 forthis theorem, even if a = 0) converges if and only if a (1, 1]; moreover if |a| < 1 thenlimn an = 0.

Proof. For all n Nwe note that xn+1 = axn. We note that if (an)nN converges tol, then:

l al = limn a

n a limn a

n

= limn a

n limn a

n+1

= 0 since (an+1)nN is a subsequence of (an)nN.

(1.5.1)

Thus l = al. If a 6= 1 then this equation is solved if and only if l = 0.The sequence (1)nN converges to 1; we shall henceforth assume a 6= 1.Assume first a > 0. We have x0 = 1 > 0 and if, for some n N we have

xn > 0 then xn+1 = axn > 0. By induction xn > 0 for all n N. Moreoverxn+1 xn = (1 a)xn for all n N.

Therefore, if a [0, 1) then (xn)nN is a decreasing sequence, bounded belowby 0, so it converges by Theorem (1.5.5), and its limit must be 0 by Equation (1.5.1).

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Now, if a > 1 then (xn)nN is increasing. If it did converge, then it would havelimit 0; yet xn > x0 > 1 for all n N, so by Theorem (1.3.2), (xn)nN may notconverge to 0; hence it does not converge.

Last, assume a 6 0. If a (1, 0], then (|a|n)nN = (|an|)nN converges to 0, so(an)nN does as well. If a < 1 then a2 (1,) so (a2n)nN does not converge,and thus (an)nN does not converge by Theorem (1.2.8). Last, ((1)n)nN admitsa subsequence convergent to 1 and a subsequence convergent to 1, so it can notconverge, by Theorem (1.2.8). Remark 1.5.8. For a > 1 we have shown that (an)nN is not bounded.

1.6. Cauchy Sequences.

Definition 1.6.1. A sequence (xn)nN in R is Cauchy when: > 0 N N p, q N (p, q > N) = |xp xq| < .

Theorem 1.6.2. If a sequence (xn)nN converges in R, then it is Cauchy.

Proof. Let l = limn xn. Let > 0. There exists N N such that for all n > N,we have |xn l| < 12 . Thus for all p, q > N, we have:

|xp xq| 6 |xp l|+ |l xq| < .

Theorem 1.6.3. If a sequence (xn)nN is Cauchy, then it is bounded.

Proof. There exists N N such that for all p, q > N we have |xp xq| < 1. Thusfor all p > N we have |xp| 6 |xN |+ 1. If M = max{|xj| : j 6 N} then for all p Nwe have |xp| 6 M. Theorem 1.6.4. If (xn)nN is a Cauchy sequence which admits a subsequence convergingto l R then (xn)nN converges to l.Proof. Let (x(n))nN be a subsequence of (xn)nN which converges to some l R.Let > 0. Since (x(n))nN converges to l, there exists N N such that for alln > N we have |x(n) l| < 12 . Since (xn)nN is Cauchy, there exists P N suchthat for all p, q > P we have |xp xq| < 12 . Let n > max{N, P} and note that, byLemma (1.1.2), we have (n) > n > max{N, P}. Then:

|xn l| 6 |xn x(n)|+ |x(n) l| N(k), we have

|xn l| < 12k . Now, there exists p(k) N such that for all p > p(k) we have|ynp xN | < 12k . Let zk = ykp(k). Pick z0 A arbitrarily.

By construction, zk A for all k N. Moreover, for all k N, k > 0 we have:

0 6 |zk l| 6 |zk xN(k)|+ |xN(k) l| n. Let(x(n))nN be a convergent subsequence of (xn)nN given by our assumption onD. Then (x(n))nN is bounded by Theorem (1.3.1). Yet for all n N we have|x(n)| > (n) > n, which is a contradiction. This concludes our theorem. 2.4. Open Sets.

Definition 2.4.1. A subset A of R is open when its complement A{ is closed.

Theorem 2.4.2. The sets and R are open.

Proof. Trivial. Theorem 2.4.3. If U, V are open subsets of R then U V is an open subset of R.Proof. By definition, U{ and V{ are closed, so (U V){ = U{ V{ is closed byTheorem (2.2.3). Theorem 2.4.4. If U is a nonempty collection of open subsets of R, then

U = {x R : U U x U}is open.

Proof. Let F = {U{ : U U }. By construction, F = (U ){. By Theorem(2.2.4),

F is closed, so

U is open.

Theorem 2.4.5. For all a < b the set (a, b) is open.

Proof. The complement of (a, b) is (, a] [b,), which is the union of twoclosed sets by Theorem (2.2.7), and thus is closed by Theorem (2.2.3). Theorem 2.4.6. A subset A of R is open if and only if for all x A there exists > 0such that (x , x + ) A.Proof. Let A R and assume that there exists x A such that for all > 0,there exists y (x , x + ) with y 6 A. Thus, for all n N there exists yn (

x 1n+1 , x + 1n+1) A{. By Theorem (1.3.4), the sequence (yn)nN converges to

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x. Since yn A{ for all n N, yet x A so x 6 A{, we conclude that A{ is notclosed, so A is not open.

Conversely, assume that for all x A there exists x > 0 such that (x x, x +x) A. Thus:

A =

xA(x x, x + x).

As a union of open sets, A is thus open. Theorem 2.4.7. Let A be a subset of R. Then:

A = {x R : V R (V is open and x V) = V A 6= } .Proof. Let x R such that, for some V R open with x V, we have V A = .Then A V{ and V{ is closed so A V{. Hence x 6 A.

Let x 6 A. Then x A{ and A{ is open, yet A A{ = . 2.5. Interior.

Definition 2.5.1. The interior A of a subset A of R is the largest open subset of Rcontained in A.

Theorem 2.5.2. Let A R. A point x R is in A if and only if there exists > 0 suchthat (x , x + ) A.Proof. By definition, A is open so if x A then there exists > 0 with (x , x +) A A.

If x R such that there exists > 0 with (x , x + ) A then, since (x , x + ) is an open set contained in A, we conclude (x , x + ) A.

3. CONTINUOUS FUNCTIONS

3.1. Limits of Functions.

Definition 3.1.1. Let D R, f : D R and a D. The function f convergesto l at a along D when for every sequence (xn)nN in D converging to a, we have( f (xn))nN converges to l.

Theorem 3.1.2. Let D R, f : D R and a D. If f converges to l R and l Rat a along D, then l = l.

Proof. Let (xn)nN be a sequence in D converging to a. Then ( f (xn))nN convergesto both l and l so l = l by Theorem (1.2.2). Notation 3.1.3. If f : D R converges to l along D at a D, then l is denoted bylimxa,xD f (x).

Theorem 3.1.4. If f : D R, with D R, converges to l along D at a D, thenl = f (a).

Proof. The sequence (a)nN lies in D and converges to a. Theorem 3.1.5. Let D R, a D and f : D R. The function f has limit l at aalong D if and only if:

> 0 > 0 t D |t a| < = | f (t) l| < .

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Proof. Assume that there exists > 0 such that, for all > 0 there exists t D suchthat |t a| < and | f (t) l| > . Thus for all n N there exists tn D such that|tn a| < 1n+1 and | f (tn) l| > . Now by Theorem (1.3.4), (tn)nN converges toa, yet ( f (tn))nN does not converge to l, so f does not have l for limit at a.

Conversely, assume that:

> 0 > 0 t D |t a| < = | f (t) l| < .Let (tn)nN be a sequence in D converging to a. Let > 0. There exists > 0 suchthat for all t D, if |t a| < then | f (t) l| < . Now there exists N N suchthat for all n > N we have |tn a| < . Thus for all n > N we have | f (tn) l| < .So limn f (tn) = l as desired. Remark 3.1.6. Thanks to Theorem (3.1.5), we have a characterization of conver-gence for functions which does not involve sequences. Every proof is the rest ofthis section can be done using either a sequence based method or an e deltabased method. We shall propose only one method, chosen solely to keep our ex-position minimal. It is very useful to prove all the following theorems using whichevermethod is not employed in these notes.

3.2. Limits and Order.

Theorem 3.2.1. Let D R, a D and f : D R. The following are equivalent:(1) the function f has limit l at a along D,(2) for all > 0, the function f has limit l at a along D (a , a + ),(3) there exists > 0 such that f has limit l at a along D (a , a + ).

Proof. The equivalences are obvious following Theorem (3.1.5). Convention 3.2.2. If f : D R is a given function and a D, then limxa f (x), orsimply lima f , designates, if it exists, the limit of f along a set (a , a) (a+ ) D for some > 0.

Moreover, lima+ f = limxa+ f (x) and lima f = limxa f (x) designate, ifthey exists, the limits of f along (a , a) and (a, a + ) for some > 0 such thatthese intervals are contained in D.

Last, limxa,P(x) f (x)where P is a predicate, is a short hand for limxa,x{tR:P(t)} f (x).

Theorem 3.2.3. Let D R, a D and f : D R. If f has a limit at a along Dthen there exists > 0 and M R such that for all t D (a , a + ) we have| f (t)| 6 M.Proof. By Theorem (3.1.5), there exists > 0 such that for all t D, if |t a| < then | f (t) l| < 1, so | f (t)| 6 1+ |l|. Theorem 3.2.4. Let D R, a D and f , g : D R such that, for some > 0 wehave f (x) > g(x) for all x D (a , a + ). If f and g have limits at a along Dthen:

limxa,xD

f (x) > limxa,xD

g(x).

Proof. Let (xn)nN be a sequence in D converging to a. There exists N N suchthat for all n > N we have f (xn) > g(xn). Thus by Theorem (1.3.2) we conclude:

limxa,xD

f (x) = limn f (xn) > limn g(xn) = limxa,xD g(x).

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Theorem 3.2.5. Let D R, a D and l 6= 0. If f : D R has limit l at a along D thenthere exists > 0 such that for all t (a , a + ) D we have | f (t)| > 12 |l| > 0.Proof. By Theorem (3.1.5), there exists > 0 such that for all t D, if |t a| < then | f (t) l| < 12 |l| since 12 |l| > 0. Thus for all t D (a , a + ) we have| f (t)| > 12 |l|. Theorem 3.2.6 (Squeeze Theorem). Let D R, a D and f , g, h : D R. If thetwo following conditions hold:

(1) there exists > 0 such that for all t D (a, a+) we have f (t) 6 g(t) 6h(t),

(2) there exists l R such that both f and h converge to l at a along Dthen g converges to l at a along D.

Proof. We employ Theorem (3.1.5) repeatedly.Let > 0. There exists f > 0 such that for all t D with |t a| < f we have

| f (t) l| < , so in particular l < f (t).There exists h > 0 such that for all t D with |t a| < h we have |h(t) l| 0 such that 1f is defined on D (a , a + ) and has limit 1l at aalong D (a , a + ).Proof. Let > 0 be given by Theorem (3.2.5) such that for all t D (a , a+ )we have | f (t)| > 0. Then 1f is defined on D (a , a + ). Let (xn)nN be asequence in D converging to a. Since ( f (xn))nN converges to l, we conclude byTheorem (1.4.3) that

(1

f (xn)

)nN

converges to 1l as desired.

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3.4. Continuity at a point.

Definition 3.4.1. Let D R and a D. A function f : D R is continuous at aon D when limxa,xD f (x) = f (a).

Theorem 3.4.2. Let D R and a D. If f , g are continuous at a on D then:(1) for all t R, the function f + tg is continuous at a on D,(2) the function f g is continuous at a on D,(3) if f (a) 6= 0 then there exists > 0 such that the function 1f is continuous at a on

D (a , a + ).Proof. Apply the limit theorems for algebra and limits of functions.

Theorem 3.4.3. Let D, E R, f : D R, g : E R and a D. If f is continuousat a on D, if g is continuous at f (a) on E, and if there exists > 0 such that if t D (a , a + ) then f (t) E, then g f is continuous at a on D (a , a + ).Proof. Let > 0. There exists g > 0 such that if t E and |t f (a)| < g then|g(t) g( f (a))| < . Now, there exists f > 0 such that if t D and |t a| < fthen | f (t) f (a)| < min{g,}. Thus for all t D (a , a + ), if |t a| < fthen |g f (t) g f (a)| < . Remark 3.4.4. The hypothesis a D and f (D) E is stronger than the assumption > 0 f (D (a , a + )) E.3.5. Continuity on a set.

Definition 3.5.1. Let D R and f : D R. The function f is continuous (on D)when for all x D, we have:

limtx,tD

f (t) = f (x).

Theorem 3.5.2. Let D R be a compact set. If f : D R is continuous, thenf (D) = { f (x) : x D} is bounded.Proof. Assume that for all M R there exists x D such that | f (x)| > M. In par-ticular, for all n N there exists xn D such that | f (xn)| > n. By Theorem (2.3.2),the sequence (xn)nN admits a convergent subsequence (x(n))nN with limit l D since D is closed and bounded. Since f is continuous at l D by assumption,we conclude that ( f (x(n)))nN converges to f (l) and in particular by Theorem(1.3.1), the sequence ( f (x(n)))nN is bounded. Yet | f (x(n))| > (n) > n for alln N, which is a contradiction. Hence our result is proven. Theorem 3.5.3 (Continuous Image of a Compact). Let D R be a compact set. Iff : D R is continuous, then f (D) = { f (x) : x D} is compact.Proof. By Theorem (3.5.2), the set f (D) is bounded. Let (yn)nN be a sequence inf (D). Then for all n N, there exists xn D such that f (xn) = yn. By Theo-rem (2.3.2), since D is closed and bounded, there exists a subsequence (x(n))nNwhich converges to some l D. Now f is continuous at l by assumption, sof (l) = limn yn. Thus f (D) is closed.

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Theorem 3.5.4. Let f : [a, b] R be a continuous function, with a 6 b. If f (a) 6 0and f (b) > 0 then there exists c [a, b] such that f (c) = 0.Proof. Let a0 = a and b0 = b. Let = b a. Assume that for some n N we haveconstructed a0 6 . . . 6 an and bn 6 . . . 6 b0 such that for all j {0, . . . , n} wehave f (aj) 6 0, f (bj) > 0 and bj ab 6 2j.

Let mn = 12 (bn an). If f (mn) 6 0 then let an+1 = mn and bn+1 = mn, and ob-serve that f (an+1) 6 0 while f (bn+1) > 0 while bn+1 an+1 = 12 2n = 2n1.

If f (mn) > 0 then set instead an+1 = an and bn+1 = mn; once again f (an+1) 6 0while f (bn+1) > 0 while bn+1 an+1 = 12 2n = 2n1.

This completes the construction of our sequences (an)nN and (bn)nN. Now,by construction, (an)nN is increasing and bounded (above by b) so it converges;since [a, b] is closed by Theorem (2.2.7), we conclude that c [a, b]. Moreover,since:

0 6 c bn 6 an bn 6 2nfor all n N, Theorem (1.3.4) implies that limn bn = c.

Since f is continuous at c [a, b], we conclude:lim n f (an) = f (c) = limn f (bn).

Now, since for all n N we have f (an) 6 0, we conclude that f (c) 6 0 byTheorem (1.3.2). Since f (bn) > 0 for all n N, we similarly conclude that f (c) > 0.Hence f (c) = 0.

Theorem 3.5.5 (Intermediate Value Theorem). Let I R be a nonempty interval ofR and f : I R be a continuous function. Let a 6 b I and let d [ f (a), f (b)] [ f (b), f (a)]. Then there exists c [a, b] I such that f (c) = d.Proof. Assume first that f (a) 6 f (b). Let g : x [a, b] 7 g(x) d. Then g iscontinuous on [a, b] by Theorem (3.4.2) and g(a) 6 0 while g(b) > 0. Thus byTheorem (3.5.4), there exists c [a, b] such that g(c) = 0, i.e. f (c) = d. Now sinceI is an interval and a, b I, we have [a, b] I, so c I.

Assume now that f (a) > f (b), so f (a) 6 f (b). Since f is continuouson [a, b] by Theorem (3.4.2), there exists c [a, b] such that f (c) = d by thereasoning above. Thus our theorem is completed.

Theorem 3.5.6. If I R is an interval and f : I R is continuous then f (I) is aninterval.

Proof. This is simply a rephrasing of Theorem (3.5.5).

Theorem 3.5.7 (Extreme Value Theorem). If f : [a, b] R is continuous and a 6b R then there exists m, M R such that f ([a, b]) = [m, M].Proof. By Theorem (3.5.5), f ([a, b]) is an interval. By Theorem (3.5.3), f ([a, b]) is acompact set. Thus I = f ([a, b]) is closed and bounded. It is nonempty by assump-tion.. Let m = inf I. Then for all n N, there exists xn I with m 6 xn 6 m+ 1n+1 .Thus (xn)nN converges to m by Theorem (1.3.4), and since I is closed, m I. Thesame reasoning holds for M I.

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REAL ANALYSIS 15

3.6. Uniform Continuity.

Definition 3.6.1. Let D R and f : D R. The function f is uniformly continu-ous on D when:

> 0 > 0 x, y D (|x y| < ) = | f (x) f (y)| < .Theorem 3.6.2. If f : D R is uniformly continuous on D then it is continuous on D.Proof. Follows from the definition of uniform continuity and Theorem (3.1.5). Theorem 3.6.3 (Heines Theorem). Let D R be a compact subset of R. A functionf : D R is continuous on D if and only if it is uniformly continuous on D.Proof. The condition is sufficient by Theorem (3.6.2).

Assume that f is continuous on D and suppose that f is not uniformly contin-uous. Thus there exists > 0 such that for all > 0, there exists x, y D such that|x y| < and | f (x) f (y)| > .

Let n N. There exists xn, yn D such that |xn yn| < 1n+1 and | f (xn) f (yn)| > . Since D is compact, there exists a convergent subsequence (x(n))nNwith limit in D which we denote by lx. Since D is compact again, there exists asubsequence (y(n))nN of (y(n))nN which converges in D to a limit we denoteby ly. Note that (x(n))nN converges to lx by Theorem (1.2.8).

Now, since for all n N, we have |x(n) y(n)| < 1(n)+1 < 1n+1 , weconclude by Theorem (1.3.4) that lx = ly.

Now, f is continuous at lx, so ( f (x(n)))nN and ( f (y(n)))nN converge tof (lx) = f (ly). Yet this is a contradiction since | f (xn) f (yn)| > for all n N.This concludes our proof.

4. COMPLEMENT OF TOPOLOGY

This section is optional.

4.1. Continuity.

Theorem 4.1.1. Let D R and f : D R. The following assertions are equivalent:(1) f is continuous on D,(2) for any open subset U of R, there exists V R open such that f1(U) = {y

R : f (y) U} = V D,(3) for any closed subset F of R, there exists G R closed such that f1(F) =

G D,Proof. Assume (1). Assertion (2) is trivial when U = , so assume U R is opneand nonempty. Let x f1(U). Since f (x) U and U is open, by Theorem (2.4.6)there exists > 0 such that ( f (x) , f (x) + ) U. By Theorem (3.1.5), thereexists > 0 such that for all t D with |t x| < we have f (t) ( f (x) , f (x) + ). Thus (x , x + ) f1(U). By Theorem (2.4.6) we conclude thatf1(U) is open.

Assume (2). Let F R closed. Then U = F{ is open, so by (2), there existsV R open such that f1(U) = D V. Thus f1(F) = D (V{) and V{ isclosed.

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Last, assume (3). Let x D and let (xn)nN be a sequence in D convergingto x. Suppose that ( f (xn))nN does not converge to f (x). By Theorem (1.2.9),there exists a subsequence (x(n))nN of (xn)nN such that ( f (x(n)))nN has nosubsequence converging to f (x). Thus f (x) 6 F = { f (x(n) : n N}. Now F isclosed so f1(F) = D G where G is closed.

Now, (x(n))nN G D and converges to x D, so x D G since Gis closed. Yet f (x) 6 F, so we have reached a contradiction. Hence, ( f (xn)nNconverges to f (x) as desired, and (1) holds. 4.2. Connectedness.

Definition 4.2.1. A subset D of R is connected when for any two open subsetsU, V of R such that U D, V D are not empty while I (U V) = , we haveI 6= (I U) (I V).Theorem 4.2.2. A subset I of R is connected if and only if every continuous function : I {0, 1} is constant on I.Proof. Assume there exists U, V open subsets of R such that I = (I U) (I V)with U I 6= and V I 6= and I U V = . Let:

: x I 7{

1 if x U,0 if x V.

Let R be an open subset of R. If 0, 1 6 1() is empty. If 1 , 0 6 then 1() = U I. If 0 , 1 6 then 1() = I V. If 0, 1 then1() = I = I R. By Theorem (4.1.1), the function is continuous on I, andsince U I and V I are not empty, the function is not constant.

Assume now that there exists a nonconstant continuous function : I {0, 1}.Then by Theorem (4.1.1), there exists U, V such that U I = 1

((12 , 2))

and

V I = 1((1, 12

)). Note that U I = 1({1}) and V I = 1({0}). Thus

since is not constant, U I and V I are not empty, and I (U V) = I byconstruction. Theorem 4.2.3. A subset I of R is connected if and only if it is a nonempty interval.

Proof. Let I be a nonempty interval. Let : I {0, 1} be continuous. If is notconstant, then there exists a, b I such that (a) = 0 and (b) = 1. By Theorem(3.5.5), there exists c I such that (c) = 12 , which is a contradiction. So I isconnected by Theorem (4.2.2).

Conversely, assume that I is not an interval. Let x < y < z R with x, z I.Let U = (, y) and V = (y,). Then U and V are open, x U I, z V Iand I = I \ {y} = (I U) (I V). 4.3. Compactness. In this section, collection will be meant as a synonim for set,to clarify the exposition.

Theorem 4.3.1 (Total Boundedness). Let D R be a compact set. For all > 0 thereexists a finite subset F D such that:

D xF

(x , x + ).

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REAL ANALYSIS 17

Proof. Let D R. Suppose that there exists > 0 such that for all finite subset Fof D, there exists x D with x 6 yF(y , y + ). Let x0 D. Assume that forsome n N, we have constructed x0, . . . , xn D such that xj 6 k N. Yet this is prohibited by construction(for instance if (p) > (q) then x(p) 6 (x(q) , x(q) + )). Thus D is notcompact. Theorem 4.3.2. If D R is compact andU is a collection of open subsets ofR such thatD U , then there exists > 0 such that for all x D, there exists U U such that(x , x + ) U.Proof. Let D R be a subset of R such that, there exists a collection U of opensubsets of R such that D U , yet for all > 0 there exists x D such that(x , x + ) 6 U for any U U .

Let n N. Let xn D be chosen so that(

xn 1n+1 , xn 1n+1)

is not a subset ofany U U . Assume now that (xn)nN thus built has a convergent subsequence(x(n))nN with limit in D, and let l be its limit. Then since l D, there existsU U such that l U. Since U is open, by Theorem (2.4.6), there exist > 0such that (l , l + ) U. On the other hand, there exists N N such that|x(n) l| < 12 for all n > N and tere exists N N such that 1(n)+1 6 12 for alln > N. Let n = max{N, N}.

Let t (x(n) 1(n)+1 , x(n) + 1(n)+1 ). Since n > N we conclude |t x(n)| N, we conclude |x(n) l| < 12 . Hence |t l| < , so t U. This isa contradiction.

Thus (xn)nN has no convergent subsequence with limit in D and thus, D is notcompact. Theorem 4.3.3. If D R is compact and if U is a collection of oppen subsets of R suchthat D

U then there exists a finite subset V of U such that D V .

Proof. Let > 0 be given by Theorem (4.3.2) for the compact set D and the opencoveringU . By Theorem (4.3.1), there exists F D finite such that D xF(x, x+ ). For each x F, let Ux U be chosen so that (X , x+ ) Ux, which ispossible with our choice of. Then D xF Ux as desired, so V = {UX : x F}answers the theorem. Theorem 4.3.4 (Heine-Borel Theorem). Let D R. The following are equivalent:

(1) D is compact,(2) for all collectionsU of open subsets ofR such that D U , there exists a fintie

sub-collection V U such that D V .(3) for all collectionsF of closed subsets of R such that D F = there exists a

finite subset G F such that D G = .

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(4) for all collectionsF of closed subsets ofR, if for any finite subset G ofF we haveD G 6= , then we have DF 6= .

Proof. By Theorem (4.3.3), we have (1) implies (2).Assume (2). LetF be a collection of closed subsets ofR such that DF = .Let U = {F{ : F F}. Then if x D, there exists F F such that x 6 F by

assumption, so x F{. Hence D U . Thus by (2), there exists V U finitesuch that D V . Let G = {U{ : U V } F . Now G is finite, and if x Dthen x U for some U V so x 6 U{; hence DG = as desired. This proves(3).

Assume (3). LetF of closed subsets of R such that for any finite subset G ofFwe have D G 6= . Then, if DF =, there exists by (3) a finite subset G Fwith D

G = , an obvious contradiction. Thus (4) holds.Now, assume (4). Let (xn)nN be a sequence in D. For all n N, let Xn =

{xk : k > n}. Note that Xn D 6= for all n N. Moreover, for all n > m we haveXn Xm. In particular, for any finite nonempty F Nwe havenF Xn = Xmax F.Thus F = {Xn : n N} satisfies assumption (4). Consequently, nN Xn D 6=. Let l nN Xn D.

Let (0) = min{n N : |xn l| < 1}, which exists since l X0 = {xn : n N}.Assume that for some n N, we have built (0) < . . . < (n) N suchthat |x(j) l| < 1j+1 for all j {0, . . . , n}. Then since l X(n)+1, the set{k > (n) + 1 : |xk l| < 1n+2} is not empty; let (n + 1) be its smallest ele-ment.

By induction, we have constructed a subsequence (x(n))nN of (xn)nN con-verging to l D. Thus D is compact, i.e. (1) holds.

5. DERIVATION

5.1. Derivation.

Definition 5.1.1. Let D R, f : D R and a D. If x D 7 f (x) f (a)xa has alimit l R at a then f is differentiable at a, and l is denoted by f (a), called thederivative number of f at a.

Definition 5.1.2. Let D R, f : D R. Let S be the set of points in D where fis differentiable (note: S D). The derivative function f of f is defined on S byx S 7 f (x).Theorem 5.1.3. Let D R, f : D R and a D. The function f is differentiableat a with derivative number l at a if and only if there exists a function o : D R withlima o = 0 and l R such that, for all x D:(5.1.1) f (x) = f (a) + (x a)l + (x a)o(x).Proof. If there exists o : D R with lima o = 0 and l R such that f (x) =f (a) + l(x a) + (x a)o(x) for all x D, then:

f (x) f (a)x a = l + o(x)

whose limit at a is l, as desired.

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Conversely, if f has derivative number l at a then for all x Dwe have:

f (x) = f (a) + (x a)l + (x a)o(x) where o(x) = f (x) f (a)x a l

and by definition of the derivative number, lima o = 0. This completes our proof.

Theorem 5.1.4. If f : D R is differentiable at a D then f is continuous at a.Proof. For all x D we have:

0 6 | f (x) f (a)| 6 f (x) f (a)x a

|x a|,and the right hand side converges to f (a) 0 at a, hence by Theorem (1.3.4), ourresult is proven.

5.2. Derivation and Algebra.

Theorem 5.2.1. If D R, f , g : D R are differentiable at a D, then for all t R,the function f + tg is differentiable at a and ( f + tg)(a) = f (a) + tg(a).

Proof. With the hypothesis of the theorem, for all x D \ {a}:( f + tg)(x) ( f + tg)(a)

x a =f (x) f (a)

x a + tg(x) g(a)

x awhose limit is f (a) + tg(a).

Theorem 5.2.2 (Leibniz). If D R, f , g : D R are differentiable at a D, then f gis differentiable at a and:

( f g)(a) = f (a)g(a) + f (a)g(a).

Proof. For all x D \ {a} we have:f g(x) f g(a)

x a =f (x)g(x) f (x)g(a) + f (x)g(a) f (a)g(a)

x a= f (x)

g(x) g(a)x a +

f (x) f (a)x a g(a).

The second term converges to f (a)g(a) at a. Moreover since f is differentiable ata, it is continuous at a, so f converges to f (a) at a; hence the first term convergesto f (a)g(a) as desired.

Theorem 5.2.3. Let f : D R and a D. If f is differentiable at a and f (a) 6= 0 then1f is differentiable at a and: (

1f

)= f (a)f 2(a)

.

Proof. Since f is continuous at a and f (a) 6= 0, there exists > 0 such that for allx D (a , a + ) we have f (x) 6= 0. Since a D, we may assume > 0 issmall enough so that (a , a + ) D. For all x (a , a + ), x 6= a:

1f (x) 1f (a)

x a =f (a) f (x)

f (x) f (a)(x a) =f (a) f (x)

x a1

f (x) f (a).

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Since f converges to f (a) at a since f is differentiable, hence continuous at a, ourtheorem is proven.

Theorem 5.2.4. Let f : D R, g : E R, a D. If f is differentiable at a, f (a) Eand g is differentiable at f (a) then g f is differentiable at a and:

(g f )(a) = g( f (a)) f (a).Proof. Since f is differentiable at a there exists o : D R such that lima o = 0 andfor all x D:

f (x) = f (a) + f (a)(x a) + (x a)o(x).Since g is differentiable at f (a) there exists e : E R such that lim f (a) e = 0 andfor all x E:(5.2.1) g(x) = g( f (a)) + g( f (a))(x f (a)) + (x f (a))e(x).Up to changing the value of e at f (a), we assume e( f (a)) = 0 without changes toEquation (5.2.1), and thus e is continuous at f (a). Thus, for all x D:g( f (x)) = g( f (a)) + g( f (a))( f (x) f (a)) + ( f (x) f (a))e( f (x))

= g( f (a)) + g( f (a))( f (a)(x a)) + [g( f (a))(x a)o(x) + (x a)( f (a) + o(x))e( f (x))] .Now, since e is continuous at f (a) and f is continuous at a, we conclude e fconverges to e( f (a)) = 0 at a. Thus ( f (a)+ o(x))e( f (x)) converges to f (a) 0 = 0at a. Morevoer, o converges to 0 at a. Thus, if we set for all x D:

r(x) = g( f (a))o(x) + ( f (a) + o(x))e( f (x))

then lima r = 0, and for all x D we have:g f (x) = g f (a) + g( f (a)) f (a)(x a) + (x a)r(x)

so our theorem is proven.

5.3. Mean Value Theorem.

Theorem 5.3.1. Let D R, f : D R. If there exists a D and > 0 such that forall t (a , a+ ), we have t D and f (t) 6 f (a), and if f is differentiable at a, thenf (a) = 0.

Proof. Let t (a , a). Then since f (t) f (a) 6 0, we have f (t) f (a)ta > 0. Thuswe have f (a) 6 0.

Let t (a, a + ). Then f (t) f (a)ta 6 0, and thus f (a) 6 0.So f (a) = 0.

Theorem 5.3.2 (Rolles Theorem). Let a < b. Let f : [a, b] R be continuous on[a, b], differentiable on (a, b). If f (a) = f (b) then there exists c (a, b) such thatf (c) = 0.

Proof. By Theorem (3.5.7), we have f ([a, b]) = [m, M]. Let x, y [a, b] such thatf (x) = m and f (y) = M. Now, if {x, y} = {a, b} then since f (a) = f (b), weconclude that m = f (x) = f (y) = M and thus f is constant on [a, b]. Thus for anyc (a, b) we have f (c) = 0.

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Assume now {x, y} 6= {a, b}. Without loss of generality, assume M 6 {a, b}.Then y (a, b). Since (a, b) is open and f (t) 6 f (y) = M for all t (a, b), weconclude by Theorem (5.3.1) that f (y) = 0. Theorem 5.3.3. Let a < b. If f : [a, b] R is continuous on [a, b] and differentiable on(a, b), then there exists c (a, b) such that:

f (b) f (a) = f (c)(b a).Proof. Let:

: x [a, b] 7 f (x) f (a) f (b) f (a)b a (x a).

Then is continuous on [a, b] and differentiable on (a, b) as the sum of two func-tions, both continuous on [a, b] and differentiable on (a, b).

Moreover, (a) = 0 = (b). Thus by Theorem (5.3.2), there exists c (a, b)such that (c) = 0. Now, (c) = f (c) f (b) f (a)ba . This completes our proof. Theorem 5.3.4. Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b).

(1) for all x (a, b), we have f (x) 6 0 if and only if f is decreasing on [a, b],(2) for all x (a, b), we have f (x) > 0 if and only if f is increasing on [a, b],(3) for all x (a, b), we have f (x) = 0 if and only if f is constant on [a, b].

Proof. Since f is increasing if and only if f is decreasing and since ( f ) = f wherever defined, it is sufficient to prove (1) to deduce (2) and (3).

Assume f is increasing on [a, b], and differentiable on (a, b). Let c (a, b). Thenfor all x (a, b), x 6= c we have:

f (x) f (c)x a > 0

since if x < c then f (x) 6 f (c) and x c < 0, while if x > c then f (x) > f (c)while x c > 0. Thus f (c) > 0.

Conversely, assume that f > 0 on (a, b). Let x, y [a, b] with x 6 y. Sincef is continuous on [x, y] and differentiable on (x, y), there exists c (x, y) suchthat f (y) f (x) = f (c)(y x). Since f (c) > 0 and y x > 0 we concludef (y) > f (x) as desired. 5.4. Higher Order Derivatives.

Definition 5.4.1. Let a < b and let f : (a, b) R. Set f (0) = f on (a, b). If, forsome n N, the function f (n) is differentiable on (a, b) then its derivative is the(n+ 1)-th derivative f (n+1) of f and f is said to be n-times differentiable on (a, b).

Notation 5.4.2. We are familiar with f (1) = f . We also may use f for f (2) andf for f (3), and technically continue with this scheme for all n N, though afterf , readability encourages to use the notation f (n).

Definition 5.4.3. Let f : [a, b] R be an n-times differentiable function on (a, b).The Taylor Polynomial of f at c (a, b) of order n is:

n

j=0

f (j)(c)j!

(x c)j.

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Theorem 5.4.4 (Taylor-Lagrange). Let f : [a, b] R be an n + 1-times differentiablefunction on (a, b) such that for all j {0, . . . , n} the function f (j) has a limit at a and balong [a, b]. Then there exists (a, b) such that:

f (b) =n

j=0

f (j)(a)j!

(b a)j + f(n+1)()

(n + 1)!(b a)n+1.

Proof. For all x [a, b] we set:

g(x) = f (b) f (x)n

j=1

f (j)(t)j!

(b t)j.

Since f (j) is continuous on (a, b) by assumption, since it is differentiable, and more-over since this function has limits at a and b, up to extending it by continuity, thefunction g is continuous on [a, b] and differentiable on (a, b).

Let h(x) = g(x) (bt)(n+1)(n+1)! for all x [a, b] where R is chosen so that

h(a) = 0. Note that by construction, h(b) = 0. Moreover, h is continuous on [a, b]and differentiable on (a, b). By Theorem (5.3.2), there exists (a, b) such thath() = 0. The key is to observe that g(x) = (bt)nn! f (n+1)(x) for all x (a, b), soh(x) = (bt)nn! f (n+1)(x) + (bt)

n

n! , and thus h() = 0 implies = f (n+1)(), as

desired. E-mail address: [email protected]: http://www.math.du.edu/~frederic

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF DENVER, DENVER CO 80208

1. Sequences1.1. Sequences and subsequences1.2. Convergence1.3. Order and Limit1.4. Limits and Algebra1.5. Monotone Sequences1.6. Cauchy Sequences

2. Basic Topology2.1. Closure2.2. Closed Sets2.3. Compact Sets2.4. Open Sets2.5. Interior

3. Continuous Functions3.1. Limits of Functions3.2. Limits and Order3.3. Limits and Algebra3.4. Continuity at a point3.5. Continuity on a set3.6. Uniform Continuity

4. Complement of Topology4.1. Continuity4.2. Connectedness4.3. Compactness

5. Derivation5.1. Derivation5.2. Derivation and Algebra5.3. Mean Value Theorem5.4. Higher Order Derivatives

Real-notes Frédéric Latrémolière

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