NOTES WEEK 14 DAY 2 SCOT ADAMS

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1 NOTES WEEK 14 DAY 2 SCOT ADAMS For this lecture, we fix the following: Let n P N, let W : R n, let : 2 P N pr n q and let I : id R n : R n Ñ R n be the identity map, defined by Ipxq x. For any h : W W, recall: h is a n.c. means: for all x, y P domrhs, rhpxqs rhpyqs ď r1{2s x y. For any f : W W, recall: f is a n.t. means: f I is a n.c. Recall that n.c. =near constant, and n.t. =near translation. LEMMA 0.1. Let h : W W be n.c. Assume that 0 W P domrhs. Let r : hp0 W q and let D : Bp0 W, 2rq. Assume that D Ď domrhs. For all i P N 0, let x i : h i p0 W q. Let a : 1{2. For all i P N, define s i : 1 ` a ` a 2 ` ` a i 1 and let D i : Bp0 W, s i rq. For all i P N, let P i be the statement r x i P D i s and xi x i 1 ď a i 1 r. Then, for all i P N, P i is true. Proof. We have x 1 h 1 p0 W q hp0 W q. By definition, r hp0 W q. Then x 1 0 W x 1 hp0 W q r. Then x 1 P Bp0 W, rq. We have s 1 1. Then x 1 P Bp0 W, rq Bp0 W, s 1 rq D 1. We have x 0 h 0 p0 W q Ip0 W q 0 W. Then x 1 x 0 x 1 0 W x 1. We have a 0 1. Then x 1 x 0 x 1 r a 0 r. Therefore, r x 1 P D 1 s and x1 x 0 ď a 0 r. That is, P 1 is true. By induction, we wish to P N, P i ñ P i`1. Let i P N be given. We wish to show: P i ñ P i`1. Assume that P i is true. We wish to show that P i`1 is true. Since P i is true, we have r x i P D i s and xi x i 1 ď a i 1 r. We wish to show: r x i`1 P D i`1 s and xi`1 x i ď a i r. Date: April 27, 2017 Printout date: April 28,

2 2 SCOT ADAMS Since h is n.c., we get rhpx i qs rhpx i 1 qs ď r1{2s x i x i 1. Also, we have hpx i q hph i p0 W qq h i`1 p0 W q x i`1. Similarly, we see that hpx i 1 q hph i 1 p0 W qq h i p0 W q x i. Also, we have a 1{2 and x i x i 1 ď a i 1 r. Then x i`1 x i rhpx i qs rhpx i 1 qs ď r1{2s x i x i 1 ď a pa i 1 rq a i r. It remains to show: x i`1 P D i`1 Since x i P D i Bp0 W, s i rq, we see that x i 0 W ď s i r. We have a i ` s i s i 1 ` a i 1 ` a ` a 2 ` ` a i 1 ` a i s i`1. Then x i`1 0 W ď x i`1 x i ` x i 0 W ď a i r ` s i r pa i ` s i qr s i`1 r. Then x i`1 P Bp0 W, s i`1 rq D i`1, as desired. The next result is the Fixpoint Propery for Contractions: THEOREM 0.2. Let h : W W be n.c. Assume that 0 W P domrhs. Let r : hp0 W q and let D : Bp0 W, 2rq. Assume that D Ď domrhs. Then there exists p P D such that hppq p. Proof. Let a : 1{2. For all i P N, define s i : 1 ` a ` a 2 ` ` a i 1 and let D i : Bp0 W, s i rq. For all i P N, we have s i 2 a i 1, so s i ă 2, so Bp0 W, s i rq Ď Bp0 W, 2rq. Thus, for all i P N, we have D i Ď D. By Lemma 0.1, for all i P N, we have h i p0 W q P D i. Define x P W N by x i h i p0 W q. Then, for all i P N, we have x i P D i Ď D. Also, we define x 0 : 0 W. By Lemma 0.1, we P N, x i x i 1 ď a i 1 r. Claim: x is Cauchy in W. Proof of claim: We wish to show: for all ε ą 0, there exists M P N such that, for all i, j P N, we have Let ε ą 0 be given. We wish to show: there exists M P N such that, for all i, j P N, we have Since a 1{2 P p0, 1q, choose M P N such that a M r ă ε. We wish to show: for all i, j P N, we have Let i, j P N be given. We wish to show:

3 NOTES WEEK 14 DAY 2 3 Assume that i, j ě M. We wish to show that x i x j ă ε. Let k : minti, ju and l : maxti, ju. Then x l x k x i x j. We wish to show that x l x k ă ε. We have k P ti, ju ě M, so, because a 1{2 P p0, 1q, we get a k 1 ď a M 1. So, because r hp0 W q ě 0, we get a k 1 r ď a M 1 r. As s l k`1 2 a l k ă 2 a 1 and a ą 0, we get a k s l k`1 ă a k a 1. Then Then a k ` a k`1 ` ` a l a k p1 ` a ` a 2 ` ` a l k q a k s l k`1 ă a k a 1 a k 1. x l x k ď x l x l 1 ` ` x k`1 x k ď a l 1 r ` ` a k r pa k ` ` a l 1 qr ď a k 1 r ă a M 1 r ă ε, as desired. End of proof of claim. By the claim, because Cauchy=convergent, we get: x is converent in W. Let p : lim x. We wish to show: [ ( p P D ) & ( hppq p ) ]. We have lim x p, i.e., x Ñ p. So, since D is closed and P N, x i P D, we see that p P D. It remains to show: hppq p. Let y P W N be the first tail of x, defined by y i x i`1. Since x Ñ p and since y is a subsequence of x, we see that y Ñ p, i.e., that lim y p. As h : W W is n.t., we know, from Lemma 0.9 of the last class, that h : W W is uniformly continuous. In particular, h is continuous at p. So, P N, x i P D Ď domrhs and since p P D Ď domrhs and since x Ñ p, we conclude that hpx q Ñ hppq. For all i P N, hpx i q hph i p0 W qq h i`1 p0 W q x i`1 y i. Then hpx q y. Then y hpx q Ñ hppq, so lim y hppq. Then hppq lim y p. LEMMA 0.3. Let f : W W be a n.t. Assume that 0 W P domrfs and that fp0 W q 0 W. Let r ą 0. Let D : Bp0 W, 2rq, and assume that D Ď domrfs. Then Bp0 W, rq Ď f pdq. Proof. Let q P Bp0 W, rq be given. We wish to show that q P f pdq, i.e., that there exist p P D such that fppq q. Let h : f I. Then domrhs domrfs. Also, h : W W is a n.c. Also, hp0 W q rfp0 W qs rip0 W qs 0 W 0 W 0 W. Also, f I ` h, so, for all p P W, we have fppq p ` rhppqs. We wish to show: there exists p P D such that p ` rhppqs q. That is, we wish to show: there exists p P D such that p q rhppqs.

4 4 SCOT ADAMS Let η : domrhs Ñ W be defined by ηpxq q rhpxqs. Note that domrηs domrhs. We leave it an unassigned exercise to show that, since h : W W as a n.c., η : W W is also a n.c. We have ηp0 W q q rhp0 W qs q 0 W q P Bp0 W, rq, so rηp0 W qs 0 W ď r. Let ρ : ηp0 W q. Then ρ ď r. We define : Bp0 W, 2ρq. Then Bp0 W, 2ρq Ď Bp0 W, 2rq D, and so, because D Ď domrhs domrηs, we see that Ď domrηs. Also, 0 W P domrfs domrhs domrηs. Therefore, by Theorem 0.2 (with h replaced by η, r by ρ and D by ), choose p P such that ηppq p. We have p P Ď D, and we wish to show that p q rhppqs. We have p ηppq q rhppqs, as desired. THEOREM 0.4. Let f : W W be a n.t. Let U be an open subset of W. Assume that U Ď domrfs. Then f puq is an open subset of W. Proof. We wish to show, for all q P f puq, that there exists r ą 0 such that Bpq, rq Ď f puq. Let q P f puq be given. We wish to show that there exists r ą 0 such that Bpq, rq Ď f puq. Since q P f puq, choose p P U such that q fppq. Since U is open in W and p P U, choose s ą 0 such that Bpp, sq Ď U. Let r : s{3. We wish to show: Bpq, rq Ď f puq. We wish to show: for all y P Bpq, rq, y P f puq. Let y P Bpq, rq be given. We wish to show: y P f puq. That is, we wish to show: there exists x P U such that y fpxq. Let φ : f T p. Then domrφs pdomrfsq p. For all x P W, we have φpxq rfpp ` xqs rfppqs. We leave it an unassigned exercise to show that, since f : W W is a n.t., φ : W W is also a n.t. We have Bpp, 2rq Ď Bpp, 3rq Bpp, sq Ď U Ď domrf s. We define D : Bp0 W, 2rq. Then D r Bpp, 2rqs p Ď pdomrfsq p domrφs. Since p P U Ď domrfs, we get 0 W p p P pdomrfsq p domrφs. Also, φp0 W q rfpp ` 0 W qs rfppqs 0 W. By Lemma 0.3 (with f replaced by φ), we see that Bp0 W, rq Ď φ pdq. Since y P Bpq, rq, y q P Bp0 W, rq. Then y q P Bp0 W, rq Ď Bp0 W, rq Ď φ pdq. Choose ξ P D s.t. y q φpξq. Let x : p ` ξ. We wish to show: y fpxq. We have y q φpξq rfpp ` ξqs rfppqs rfpxqs q, so, adding q to both sides, we get y fpxq, as desired.

5 NOTES WEEK 14 DAY 2 5 LEMMA 0.5. Let f : W W be a n.t. Then f is injective. Proof. We wish to y P domrfs, prfpxq fpyqs ñ rx ysq. Let x, y P domrfs be given. We wish to show: rfpxq fpyqs ñ rx ys. Assume that fpxq fpyq. We wish to show: x y. Let h : f I. Then h : W W is a n.c. Because h ` I f, it follows, for all z P W, that rhpzqs ` z fpzq. Then x ` rhpxqs fpxq fpyq y ` rhpyqs. Subtracting rhpxqs ` y from both sides, x y rhpyqs rhpxqs. Let a : x y. Then a rhpyqs rhpxqs. Because h is a n.c., we have rhpxqs rhpyqs ď r1{2s x y. Then a rhpyqs rhpxqs rhpxqs rhpyqs ď r1{2s x y a{2. Subtracting a{2 from both sides, we get a{2 ď 0, and so a ď 0. Then x y a ď 0. By definition of a norm, x y ě 0. Then x y 0. So, by definition of a norm, x y 0 W. Then x y, as desired. THEOREM 0.6. Let f : W W be a n.t. Let U : domrfs and let V : imrfs. Assume U is open in W. Then V is open in W and f : U Ñ V is a homeomorphism. Proof. By Lemma 0.5, f : U Ñ V is injective. By definition of V, f : U Ñ V is surjective. By Lemma 0.10 of Week 14 Day 1, we see that f : U Ñ V is uniformly continuous, and so f : U Ñ V is continuous. It remains to show that f 1 : V Ñ U is continuous. We wish to show: for any open subset U 0 of U, that pf 1 q pu 0 q is open in V. Let an open subset U 0 of U be given. We wish to show that pf 1 q pu 0 q is open in V. Because f : U Ñ V is bijective, we have pf 1 q pu 0 q f pu 0 q. We wish to show that f pu 0 q is open in V. By assumption U is open in W. So, since U 0 is open in U, it follows that U 0 is open in W. Then, by Theorem 0.4 (with U replaced by U 0 ), f pu 0 q is open in W. So, since f pu 0 q Ď imrfs V, we conclude that f pu 0 q is open in V, as desired.