NOTES WEEK 14 DAY 2 SCOT ADAMS

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1 NOTES WEEK 14 DAY 2 SCOT ADAMS We igligt tat it s possible to ave two topological spaces and a continuous bijection from te one to te oter wose inverse is not continuous: Let I : r0, 2πq and let C : tpx, yq P R 2 x 2 ` y 2 u Define f : I Ñ C by fptq pcos t, sin tq Ten f : I Ñ C is a continuous bijection, but we argued tat f 1 : C Ñ I is not continuous We ave fp0q p1, 0q and fp099 2πq p , q Note tat fp0q and fp099 2πq are close togeter Ten f takes points far apart, like 0 and 099 2π, and brings tem very close togeter If we tink of I r0, 2πq as a piece of string, ten f attaces te one end to te oter, creating a circle of string Te points 0 and 099 2π are on opposite sides of te piece of string, so it s not surprising tat tey end up close togeter on te circle Gluing is a continuous operation and te mapping f : I Ñ C turns out to be a continuous bijection However, f 1 : C Ñ I corresponds to cutting te circle of string and laying it out onto an interval Tat operation takes points tat are close togeter, like fp0q and fp099 2πq, and moves tem to 0 and 099 2π, wic are far apart Tis would lead us to believe tat f 1 : C Ñ I is not continuous at p1, 0q We elaborate furter on tis: Let p : p1, 0q Ten f 1 ppq 0 We played a little game: I coose ε ą 0 Ten you coose δ ą 0 Ten I coose q P C satisfying d 2 pp, qq ă δ, Ten we ceck weter d R pf 1 ppq, f 1 pqqq ă ε If d R pf 1 ppq, f 1 pqqq ă ε, ten you win Oterwise, I do I cose ε 1 You cose δ 001 I cose q fp099 2πq p , q Note tat d 2 pp, qq? ` ă 001 Ten f 1 ppq 0 and f 1 pqq 099 2π, so d R p f 1 ppq, f 1 pqq q 099 2π ą 1 ε, and I won Coosing ε 1, I ll win every time In fact, once I know your δ, I can coose a point q tat is δ-close to p and just sligtly below p p1, 0q on C Tat point q gets moved close to 2π, and, Date: December 8, 2016 Printout date: December 15,

2 2 SCOT ADAMS ence, far away from f 1 p1, 0q 0 How far away? About 2π away, and 2π 628 is muc greater tan 1 wic is my ε Te fact tat I win indicates tat f 1 : C Ñ I is not continuous at p1, 0q Note tat domrfs I r0, 2πq is not sequentially compact Te fact tat continuity of a bijection doesn t imply continuity of its inverse sometimes presents difficulty Say we ave two topological spaces X and Y and we tink tey may be omeomorpic Say we ve constructed some function f : X Ñ Y tat is cleary a continuous bijection and we re oping tat its a omeomorpism We need to ceck tat f 1 : Y Ñ X is continuous It may be callenging to compute a formula for f 1 : Y Ñ X, so it would be nice if tere were a teorem tat guarantees tat, in some situations, te continuity of f 1 : Y Ñ X follows from tat of f : X Ñ Y We will aim for a result (Corollary 05) tat says tat, if X is sequentially compact and Y is metrizable, ten continuity of f : X Ñ Y implies continuity of f 1 : Y Ñ X DEFINITION 01 Let X, Y be topological spaces Let f : X Ñ Y Ten f : X Ñ Y is open means: for any open subset U of X, te set f puq is open in Y Also, f : X Ñ Y is closed means: for any closed subset C of X, te set f pcq is closed in Y REMARK 02 Let X, Y be topological spaces Let f : X Ñ Y be a bijection Ten ( f : X Ñ Y is open ) iff ( f : X Ñ Y is closed ) Proof For all S Ď X, we ave pf 1 q psq f psq Ten U in X, pf 1 q puq is open in Y ) iff U in X, f puq is open in Y ) iff ( f : X Ñ Y is open ), It remains to sow tat We ave ( f : X Ñ Y is closed ) C in X, pf 1 q pcq is closed in Y ) iff C in X, f pcq is closed in Y ) iff

3 as desired ( f : X Ñ Y is closed ), NOTES WEEK 14 DAY 2 3 COROLLARY 03 Let X, Y be topological spaces Let f : X Ñ Y be a bijection Ten ( f : X Ñ Y is a omeomorpism ) iff ( f : X Ñ Y is continuous and open ) iff ( f : X Ñ Y is continuous and closed ) Proof We ave ( f : X Ñ Y is a omeomorpism ) iff ( f : X Ñ Y is continuous and f 1 : Y Ñ X is continuous ) Ten, by Remark 02, te result follows THEOREM 04 Let X, Y be topological spaces Let f : X Ñ Y be a continuous injection Assume X is sequentially compact and Y is metrizable Let Y 0 : imrfs Ten f : X Ñ Y 0 is a omeomorpism Proof Since f : X Ñ Y is continuous, by HW#43, we conclude tat f : X Ñ Y 0 is continuous By Corollary 03, it suffices to sow tat f : X Ñ Y 0 is closed Tat is, we wis to sow, for any closed subset C of X, tat f pcq is closed in Y 0 Let a closed subset C of X be given We wis to sow tat f pcq is closed in Y 0 Since C is closed in X and X is sequentially compact, we see tat C is sequentially compact Since f : X Ñ Y 0 is continuous, we conclude tat f C : C Ñ Y 0 is continuous So, since C is sequentially compact, we see tat imrf Cs is sequentially compact We ave f pcq imrf Cs, so we see tat f pcq is sequentially compact Since Y is metrizable, coose a metric d on Y suc tat T d is te topology on Y Let d 0 : d py 0 ˆ Y 0 q Ten d 0 is a metric on Y 0 and T d0 is te topology on Y 0 Since f pcq is sequentially compact, by Lemma 08 of Week 12 Day 1, we see tat f pcq is closed in Y 0 and tat f pcq is bounded in py 0, d 0 q In particular, f pcq is closed in Y 0, as desired COROLLARY 05 Let X, Y be topological spaces Let f : X Ñ Y be a continuous bijection Assume X is sequentially compact and Y is metrizable Ten f : X Ñ Y is a omeomorpism Proof Let Y 0 : imrfs By Teorem 04, f : X Ñ Y 0 is a omeomorpism Since f : X Ñ Y is a bijection, it follows tat imrfs Y Ten Y 0 imrfs Y Ten f : X Ñ Y is a omeomorpism

4 4 SCOT ADAMS DEFINITION 06 A vector space is a set V togeter wit (α) a zero element 0 V P V (β) a negation v ÞÑ v : V Ñ V (γ) a vector addition pv, wq ÞÑ v ` w : V ˆ V Ñ V (δ) a scalar multiplication pc, vq ÞÑ cv : R ˆ V Ñ V suc tat w P V, v ` w w ` v w, x P V, pv ` wq ` x v ` pw ` xq P V, 0 V ` v v P V, v ` p vq 0 V P V, 1v v b P P V, pabqv apbvq b P P V, pa ` bqv av ` bv and P w P V, apv ` wq av ` aw and Wat we call a vector space ere is often called a real vector space or R-vector space or vector space over te reals In tis exposition, we will not be using oter scalar fields besides R For any vector space V, te vector addition (γ) and scalar multiplication (δ) are called te linear operations in V, and we will refer to te zero element (α) and te negation (β) as te extras in V It s te linear operations in a vector space tat are most important; te extras are used, but, as we ll see, aren t nearly as important Requirements (1)-(4) are expressed by saying tat pv, 0 V,, `q is an additive group Requirements (5)-(6) are expressed by saying tat te monoid pr, 1q acts on V by scalar multiplication Requirement (7) is expressed by saying tat scalar multiplication distributes over scalar addition (from te rigt) Requiment (8) is expressed by saying tat scalar multiplication distributes over vector addition (from te left) So we ave four group laws, two action laws and two distributive laws Wen specifying a vector space, we often just give V and expect te reader to intuit wat (α), (β), (γ) and (δ) are Occasionally, in case of confusion, we specify te linear operations (γ) and (δ), but almost always leave it to te reader to find te unique zero element (α) and negation (β) So, for example, R is a vector space, were vector addition is just te usual addition of real numbers and scalar multiplication is just te usual multiplication of real numbers Below,

5 NOTES WEEK 14 DAY 2 5 we will explain ow, for every n P N, R n is a vector space Also, we will explain ow, for every set S, R S is a vector space Let V be a vector space For all v, w P V, we define v w : v`p wq For all c P R, for all v P V, we define c v : cv For all v P V, for all P Rzt0u, we define v{ : r1{s v DEFINITION 07 Let V be a vector space, let S Ď V and let x P S We define S ` x : ts ` x s P Su and S x ts x s P Su DEFINITION 08 Let V be a vector space, let S be a set and let x P V Let g : V S We define gpx ` q : pdomrgsq x Ñ S by pgpx ` qqpvq gpx ` vq DEFINITION 09 Let V and W be vector spaces, let f : V W, let x P domrfs and let v P V We define DSS f x,v : t P Rzt0u x ` v P domrfs u, and we define SS f x,v : DSS f x,v Ñ W by SS f x,vpq rfpx ` vqs rfpxqs In Definition 09, te notation DSS stands for domain of secant slope and SS stands for secant slope We call SS f x,v te secant slope function for f at x in te direction v DEFINITION 010 Let V be a vector space, let S Ď V Ten te span of S in V is xsy lin : t c 1 v 1 ` ` c k v k k P N, c 1,, c k P R, v 1,, v k P S u DEFINITION 011 Let V be a vector space and let S Ď V Ten S is linearly independent in V P 1,, c k P 1,, v k P S, r p c 1 v 1 ` ` c k v k 0 V q ñ p c 1 c k 0 q s DEFINITION 012 Let V be a vector space, let B Ď V Ten B is a basis of V means: ( xby lin V ) and ( B is linearly independent in V ) DEFINITION 013 Let V be a vector space Ten BaspV q denotes te set of all bases of V FACT 014 Let V be a vector space Let A and B be bases of V Ten #A #B

6 6 SCOT ADAMS Proof Omitted Let V be a vector space Ten t#b B P BaspV qu Ď N 0 Y t8u and, by Fact 014, t#b B P BaspV qu as exactly one element DEFINITION 015 Let V be a vector space By te dimension of V, we mean: dim V : ELTpt#B B P BaspV quq DEFINITION 016 Let S be a set, A Ď S Ten te caracteristic function of A in S is te function χ S A : S Ñ t0, 1u defined by # χ S 1, if x P A Apxq 0, if x R A Let S be a set Ten te set R S of all functions S Ñ R is a vector space, were te linear operations are defined g P R S, pf ` gqpxq rfpxqs ` P P R S,pcfqpxq c rfpxqs and We leave it as an exercise to te reader to figure out wat te extras 0 R S and : R S Ñ R S are We developed a way of visualizing elements of R S as sections of an R-bundle over S DEFINITION 017 Let S be a finite set Ten te standard basis of R S is: t χ S tsu s P S u We leave it to te reader to ceck, for any finite set S, tat te standard basis of R S is, in fact a basis of R S ; ten dim R S #S Let n P N Recall tat R n : R t1,,nu Te standard basis of R n is t p1, 0, 0,, 0, 0q, p0, 1, 0,, 0q,, p0, 0, 0,, 0, 1q u Ten dim R n n DEFINITION 018 Let V be a finite dimensional vector space and let n : dim V Let B P V t1,,nu Ten B is an ordered basis of V means imrbs is a basis of V DEFINITION 019 Let X be a set, let n P N and let p 1,, p n P X We define te function pp 1,, p n q P X t1,,nu by pp 1,, p n q i p i DEFINITION 020 Let n P N Te standard ordered basis of R n is p p1, 0, 0,, 0, 0q, p0, 1, 0,, 0q,, p0, 0, 0,, 0, 1q q

7 NOTES WEEK 14 DAY 2 7 Let n P N, and let e denote te standard ordered basis of R n Ten e : t1,, nu Ñ R n P t1,, nu, e i p0,, 0, 1, 0,, 0q, were te unique 1 is in te it coordinate DEFINITION 021 Let m P N, W a vector space, f : R m W, x P domrfs, i P t1,, mu Let e be te standard ordered basis of R m Ten we define DSS f x;i : DSSf x,e i and SS f x;i : SSf x,e i Recall tat, for any vector spaces V and W, for any f : V W, for any x P domrfs, for any v P V, for any P R, we ave SSx,vpq f rfpx ` vqs rfpxqs Let m P N and let e denote te standard basis of R m Ten, for any f : R m W, for any x P domrfs, for any i P t1,, mu, for any P R, we ave SS f x;i pq rfpx ` e i qs rfpxqs DEFINITION 022 Let W be a vector space, let f : R m W and let x P domrfs Ten DSSx f : DSS f x,1 and SSx f : SSx,1 f For any vector space W, for any f : R W, for any x P domrfs, for any P R, we ave SSxpq f rfpx ` qs rfpxqs To compute derivatives, we now need to take te limit near 0 of tese various secant slope functions Since tese functions take values in vector spaces, it is important to understand wen tere s a standard topology on a vector space We will address tis next semester