Boot Camp: Real Analysis Lecture Notes - UCLA Mathematics

Corollary 1.1.19. R is a dense linear ordering.Proof. This is clear, since Q R!Proposition 1.1.20. R has no endpoints.1.2Lecture 2 - Uniqueness of R and Basic General TopologyToday, we will talk about the properties which characterize R, as well as some generaltopology.Definition 1.2.1. An order (L; ) is Dedekind complete if(i) Every A L which is bounded above has a supremum (i.e., a -least upper bond)(ii) Every A L which is bounded below has an infimum (i.e., a -greatest lower bound)Proposition 1.2.2. R is Dedekind complete.Proof. Let A R be bounded above. Let f : N Q be onto, i.e. enumerate the rationals.Put An {f (i) i 6 n, x A such that f (i) 6 x}, and let (an ) be the sequence defined byan max{An }. Clearly, an 6 an 1 , since An An 1 . Colloquially, we are building (an ) tobe a sequence of nondecreasing elements of Q which are less than some element of A, withthe goal of showing that [an ] is the sup of A. We break into two cases:Case 1: Suppose (an ) n 0 is eventually constant. Say that an p Q for all n k for somek N. Can check that p is a sup for A. Case 2: Otherwise, we can thin (an ) n 1 to a subsequence (ank )k 1 which is strictly increasing. Can check that [ank ] is a sup for A.Example 1.2.3. Q is not Dedekind complete. Let A {q Q q 2 2}. Then A has noleast upper bound in Q. Let z sup A; we will see z 2 2.Definition 1.2.4. A linear order (L; ) is separable if there is a countable D L which isdense in L.Let’s recap. So far, we know that (R, ) is:(1) a dense linear order with no endpoints(2) Dedekind complete(3) separableProposition 1.2.5. (1) (2) (3) characterizes (R; ) uniquely up to isomorphism.Proof. Let (L; ) satisfy (1) (2) (3). Using (3), pick a countable dense subset D of L. Dnaturally inherits the linear order from L, and cannot have any endpoints. Indeed, if Dhas a (say) right endpoint α, then since L has no endpoints, there are β1 , β2 L such thatα β1 β2 ; but then L has no point of D between β1 and β2 , contradicting the densenessof D in L. Hence, D is a countable dense linear order with no endpoints, so (D; ) is7

isomorphic to Q. Let f : D Q witness this. We will extend f to an isomorphism from Lto R.For every x L, let Ωx {u u D, u 6 x} L. Since Ωx is a bounded set in L, andsince D has no endpoints, there exists d D such that d x. Since f is order preserving,f (d) f (u) for every u Ωx , so f (u) must be an upper bound for f (Ωx ) in R. Thus, wemay put f (x) supR (f (Ωx )) by the Dedekind completeness of R.Note that if x y L, then there exist z1 , z2 D such that x z1 z2 y by thedensity of D in L. Since f is order-preserving on the elements of D, f (z2 ) f (z1 ) f (u)for every u Ωx , so f (z2 ) f (z1 ) f (x). However, since z2 Ωy , f (z1 ) f (z2 ) 6 f (y), sof (x) f (y). This estalishes that f is order-preserving and injective on L.Fix z R and let A {p Q p 6 z}. Let x be the sup in L of B f 1 (A) D; x existssince L is Dedekind complete. Note B Ωx , since B D, and b 6 x for each b B, soz supR (f (B)) 6 supR (f (Ωx )) f (x). Suppose z f (x); then there is an element u Ωxsuch that z f (u) 6 f (x). Take q Q such that z q f (u); then f 1 (q) u 6 x, sincef 1 is order preserving on Q, and f 1 (q) b for each b B. So f 1 (q) is an upper boundfor B, but f 1 (q) x, a contradiction. So f (x) z, and hence f is surjective.Topological SpacesDefinition 1.2.6. A topological space is a pair (X, T) where X is a set, T is a collectionof subsets of X, and T satisfies:(1) T, X T(2) V1 , . . . , Vn T impliesTn(3) Vi T for i I impliesi 1SVi Ti IVi TT is called the topology of the space, and the elements of T are called the open sets. Wewill refer to X as the space when T is clear.Definition 1.2.7. We say T is generated from U T if T consists of arbitrary unions ofsets from U (note then U must be closed under finite intersection). U is called a basis forT. Elements of U are basic open sets.Example 1.2.8. If (L; ) is a linear order with no endpoints, the open intervals generate atopology, called the order topology.Definition 1.2.9. V is an (open) neighborhood of x if V is open and x V . V is a basicopen neighborhood if in addition V is a basic open set.A basis for the neighborhoods of x is any U consisting of neighborhoods of x such that everyneighborhood of x contains some V U.Proposition 1.2.10. A X is open if and only if for all x A, there exists an openneighborhood V of x such that V A.Proof. ( ) If A is open, then A A is a neighborhood of x.( ) For each x A, there is a neighborhood V of x such that V A. Then A is theunion of all such neighborhoods, and is therefore open.8

Definition 1.2.11. The interior of E X is the union of all open subsets of X containedin E. It is the largest open subset of X contained in E.The exterior of E X is the union of all open subsets of X which have empty intersectionwith E. It is the largest open subset of X contained in X \ E, hence is the interior of X \ E.The boundary of E X is the set of all points of X which are not in Int E or Ext E.Definition 1.2.12. A X is closed if X \ A is open. Note that arbitrary intersections ofclosed sets are closed, as are finite unions.Definition 1.2.13. The closure of A in X, denoted A, is the intersection of all closed setsin X containing A. It is the smallest closed set of X containing A.Definition 1.2.14. D X is dense in X if every nonempty open subset V of X containsa point of D.b X, T a topology on X. Then the relative or induced topolDefinition 1.2.15. Let Xbb {V Xb is defined to be Tb V T}.ogy T on XDefinition1.2.16. An open cover of X is a collection {Vi }i I of open subsetsSS of X suchthat i I Vi X. A subcover of X is a subcollection {Vi }i J , J I such that i J Vi X.Definition 1.2.17. X is compact if every open cover has a finite subcover. Y X iscompact if Y isScompact in the relative topology. Equivalently, whenever{Vi }i I are openSin X such that i I Vi Y , then there exists J I finite such that i J Vi Y .Proposition 1.2.18. If N X is compact, and V is open, N \ V is compact.Proof. Let U be an open cover of N \ V . Since V is open, U {V } is an open cover of N .Since N is compact, there is a finite subcover U0 U of N . If V U0 , replace U0 by U0 \ {V },which yields a finite subcover of N \ V .Definition 1.2.19. (X, T) is locally compact if for every x X, there is a compact Ncontaining a neighborhood of x.Definition 1.2.20. (X, T) is connected if it cannot be partitioned into two nonempty opensets, i.e. there are no open, disjoint A, B such that A B X.Proposition 1.2.21. R (with the order topology) is connected.Proof. Suppose R A B for some nonempty open, disjoint subsets A, B such that A B . Let a A, b B; WLOG, a b. Put E {x A x b}; since a E, E is nonemptyand bounded above. Let z sup(E). Since R is Dedekind complete, we must have z Aor z B. We break into cases:Case 1: Suppose z A. Since A is open, there is an open interval (x, y) such thatz (x, y) A. Since R is dense, we can find ẑ z with z ẑ min{b, y}. Then ẑ (x, y),so ẑ A, but ẑ b, contradicting that z is an upper bound for E.Case 2: Suppose z B. Since B is open, there is an open interval (x, y) such thatz (x, y) B. Then x z, so x is not an upper bound for E (z is the least upper boundfor E). Thus, we can find ẑ E such that ẑ x. Since z is an upper bound for E, hatz 6 z,so ẑ (x, y) B. This is a contradiction, since ẑ A.9

Proposition 1.2.22. R is locally compact.Proof. We show for all a b R, the closed interval [a, b] {x a 6 x 6 b} is compact. Let{Vi }i I be an open cover of [a, b]. Suppose for contradiction that there is no finite subcover.The strategy we will pursue is as follows: we will find the greatest point x of [a, b] such that[a, x] has a finite subcover. Of course, this x must lie in some open set, which allows us topush the finite open cover to cover [a, x ], contradictingthe maximality of x.SLet A {x [a, b] there is finite J I such that i J Vi [a, x]}. A is nonempty andbounded, since every element of A is less than b by hypothesis, and a A since we canjust take J {i} for any Vi containing a (such a Vi must exist since {Vi }i I is an opencover of [a, b]). If b A, we are done. Suppose not, and let c sup(A). Then a 6 c 6 b.Hence, there exists k I such that c Vk . Vk is open, so there is an open interval (u, w)such that c (u, w) Vk . Since c sup(A), we can find x A such that u x 6 c(otherwise, u would be an explictly smaller lower boundSfor A, contradicting the minimalityofS c). Since x A, we can find finite J I such that i J [a, x]. Take z (c, w). Theni (J {k}) [a, z], which contradicts the maximality of c, since z c.Definition 1.2.23. (X, T) is Hausdorff if for all x 6 y X, there exists neighborhoodsVx , Vy T of x and y such that Vx Vy .Proposition 1.2.24. R is Hausdorff.Proof. This essentially boils down to density. Let x, y R be distinct points, and x yWLOG. Then there exists z R such that x z y, so for ε 0, (x ε, z) and (z, y ε)are neighborhoods of x and y respectively with empty intersection.Proposition 1.2.25. Let (X, T) be Hausdorff and locally compact. Then for every open setU and for every x U , there exists a compact Nx U containing a neighborhood of x.Proof. Fix N compact containing a neighborhood U x of x; this is possible by local compactness. If N U , we’re done. If not, we can use the fact that X is Hausdorff to selectively peelaway parts of N not contained in U while retaining a neighborhood of x. For every y N \U ,fix a neighborhoodVy of y, and let Uyx be a neighborhood of x such that Uyx Vy .TNote y N \U Vy N \ U , i.e. {Vy }y N \U form an open cover of N \ U ; but N \ U is compact,since N is compact and U is open. Thus, there exists a finite subcover {Vy1 , . . . , Vyk } suchSb N \Sbbthat i 1,.,k Vyi N \ U . Let Ni 1,.,k Vyi . Then N is compact, and N U .Txxb Further, Ni 1,.,k (Uyi N ), since each Uyi has no points in common with Viy . Finally,Tb.since U x N , i 1,.,k (Uyxi U x ) is a neighborhood of x contained in N1.3Lecture 3 - More on Compactness and the Baire CategoryTheoremToday, we will give a few more results on compactness, and will introduce the Baire CategoryTheorem.Proposition 1.3.1. Let (X, T)T be a compact space.T Let Cii N be a collection of closed subsetsof X. Suppose for all n N, i6n Ci 6 . Then i N Ci 6 .10

Proof. As we will see in this proof, compactness isT often the natural bridge between finitefacts and infinite claims, and vice versa. Suppose i N Ci . Then for every x X, thereis some ix N such that x / Cix . Then since Cix is closed, there is an open neighborhood Uxof x contained in X \ Cix , i.e. Ux Cix . Note {Ux }x X is an open cover of X. Since Xis compact, we have k N and x1 , . . . , xk X such that Ux1 , . . . , Uxk cover the space. SinceCixl Uxl for each l {1, . . . , k}, every y X is outside at least one ofTCix1 , . . . , Cixk ,so Cix1 · · · Cixk . Then for any n larger than max{ix1 , . . . , ixk }, i6n Ci , acontradiction.Proposition 1.3.2. Compact sets in Hausdorff spaces are closed.Proof. Let N be a compact subset of a Hausdorff space X. It suffices to show that eachx X \ N has a neighborhood completely contained in X \ N . For each y Y , let Vy , Uybe neighborhoods of y and x respectively such that Vy Uy . Then {Vy }y N is an opencover of N , so there exists a finite subcover {Vy1 , . . . , Vyk }. Since Uyi Vyi for eachi 1, . . . , k, Uy1 · · · Uyk is a neighborhood of x which is disjoint from N .We now have the tools to present proof of the Baire Category theorem. We will seeanother equivalent formulation of the BCT today, as well as a distinct formulation in Lecture4.Theorem 1.3.3 (Baire Category Theorem). Let (X, T) Tbe Hausdorff and locally compact(e.g., R). Let {Dn }n N be dense open subsets of X. Then n N Dn is dense (and nonempty).TProof. Let U be a nonemptyopen subset of X. We will find a point in U ( n N Dn ). ThisTwill establish that n N Dn is nonemptyT and dense, as we will have shown each nonemptyopen subset of X contains a point of n N Dn . The basic idea is that we will construct asequence of shrinking neighborhoods, making sure we pull in a point from each Dk at everystep. We will then leverage local compactness and Hausdorff-ness as needed to ensure thatour sets shrink to at least one point, using the previous two propositions.Set U0 U . Now by induction on k 1, we construct the following:(i) Take xk Uk 1 Dk 1 ; this is possible since Dk 1 is dense, so its intersection witheach nonempty open subset is nonempty(ii) Find Nk compact containing a neighborhood of xk such that Nk Uk 1 ; this is possiblesince X is locally compact and Hausdorff, using the theorem proved at the end ofLecture 2.(iii) Let Uk be a neighborhood of xk such that Uk Nk 1 (via (ii) above) and Uk 1 Dk 1 ;this is possible since Dk 1 is open, and we can take a neighborhood of xk in Nk 1 andintersect it with Dk 1 (which is nonempty, since (i) guarantees xk Dk 1 )Then we have constructed a chain U0 N1 U1 N2 U2 . . . EachT Nk is compact,hence closed by the previous proposition,and for every k N, weTT have i6k Ni Nk 6 .Thus, by our earlier proposition, i N Ni 6 . Take any y Ti N Ni ; then for every i,y Ni 1T Ui 1 Di , and y N0 U0 U . Thus, y n N Di , and y U , soy U ( n N Di ).11

Definition 1.3.4. A set C is nowhere dense if C has empty interior.Proposition 1.3.5. C is closed nowhere dense if and only if X \ C is open dense.Proof. ( ) If C is closed nowhere dense, then X \ C is open, and C C. Further, letU be a nonempty open subset of X. Since C has empty interior, U has at least one pointu X \ C X \ C.( ) If X \ C is open dense, then C is closed, whence C C. Further, since X \ C isdense, every nonempty open subset U of X contains a point u X \ C X \ C. Thus, Chas empty interior, so C is nowhere dense.Theorem 1.3.6 (Equivalent formulation of the Baire Category Theorem). LetS (X, T) be alocally compact Hausdorff space. Let {Cn }n N be closed nowhere dense. Then n N Cn 6 X.Proposition 1.3.7. Let (X, T) be Hausdorff. Then for every x X, X \ {x} is open.Proof. Since X is Hausdorff,S for each y 6 x X, there is a neighborhood Vy of y such thatx / Vy . Then X \ {x} y6 x X Vy , whence X \ {x} is open.Definition 1.3.8. x X is isolated if {x} is open.Proposition 1.3.9. If x is not isolated, then X \ {x} is dense.Proof. Every nonempty open set clearly contains a point of X \ {x}, except for possibly {x}.Since x is not isolated, {x} is not open, so we’re done.Note: R has no isolated points.Corollary 1.3.10. Let X be Hausdorff and locally compact with no isolated points. ThenX is not countable.Proof. Suppose X were countable, and enumerate X via the sequence (qn )n N . For eachn N, put Dn X \{qn }.TThen Dn is open densetwo propositions,so by theT by the previousTSBaire Category Theorem, n N Dn 6 . But n N Dn n N X \{qn } X \ n N {qn } ,a contradiction.Corollary 1.3.11. R is not countable.Proof. R is Hausdorff and locally compact with no isolated points.Proposition 1.3.12T(F12.5). Let (X, T) be Hausdorff and locally compact. E X is Gδ ifit can be written as n N Gn with Gn open for each n N. Prove that Q is not Gδ .Proof. This is a standard application of the Baire Category Theorem. Let (qn )n N enumerateQ. Note that DTn R \ {qn } is open dense, since R is Hausdorff and has no isolated points.Suppose Q n 1 Gn , where Gn is open for each n N. Notedense for eachT Gn is also Tn N, since Gn Q. Thus, bytheBaireCategoryTheorem,(G) (n N nn N Dn ) 6 .TTBut this is a contradiction, as n N Gn Q and n N Dn R \ Q.12

Metric SpacesDefinition 1.3.13. A metric space is a pair (X, d) where d : (X X) [0, ) satisfies:(1) For all x X, d(x, x) 0(2) For all x 6 y X, d(x, y) 6 0(3) For all x, y X, d(x, y) d(y, x)(4) (Triangle inequality): For all x, y, z X, d(x, z) 6 d(x, y) d(y, z)Intuitively, d(x, y) can be thought of as the distance between x and y.Proposition 1.3.14. Let (X, d) be a metric space. Then the setsB(z, r) {x d(z, x) r}, for z X, r 0generate a topology on X called the metric topology; B(z, r) are the open balls of radiusr centered at z.Proof. Let T be the collection of all unions of open balls. To show T is a topology, it suffiesto check that the intersection of any two open balls is a union of open balls. For this, it isenough to show that for all z1 , z2 X and r1 , r2 (0, ), for all x B(z1 , r1 ) B(z2 , r2 ),there exists s 0 such that B(x, s) B(z1 , r1 ) B(z2 , r2 ).This essentially boils down to the triangle inequality. Let d1 d(z1 , x) r1 ; d2 d(z2 , x) r2 . Let s 0 be small enough such that d(z1 , x) s r1 , and d(z2 , x) s r2 ; this ispossibly because R is dense.Fix y B(x, s). Then d(zi , y) 6 d(zi , x) d(x, y) di s ri for i 1, 2. Hence,B(x, s) B(z1 , r1 ) B(z2 , r2 ).Example 1.3.15. The usual metric on R: d(x, y) x y . Then the metric topologyon R is the usual order topology, generated by open intervals.Example 1.3.16. The discrete metric on any set X:(0, if x yd(x) 1, if x 6 yEvery subset of X is open with respect to this metric.Definition 1.3.17. A metric space is compact if it is compact with the metric topology.Equivalently, any covering of X with open balls has a finite subcover.Y X is compact if (Y, d Y Y ) is compact. Equivalently, any covering of Y with openballs has a finite subcover.Definition 1.3.18. A sequence (xn ) n 1 is Cauchy if for every ε 0, there exists n Nsuch that for all k, l N N, d(xk , xl ) ε. Intuitively, the points of a Cauchy sequencecluster arbitrarily closely together if you go far out enough in the sequence.13

Definition 1.3.19. A sequence (xn ) n 1 converges

Boot Camp: Real Analysis Lecture Notes Lectures by Itay Neeman Notes by Alexander Wertheim August 23, 2016 Introduction Lecture notes from the real analysis class of Summer 2015 Boot Camp, delivered by Professor Itay Neeman. Any errors are my fault, not Professor Neeman's. Corrections are welcome; please send them to [ rstinitial][lastname .