Final Solutions. 1. (25pts) Define the following terms. Be as precise as you can.

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1 Mathematics H104 A. Ogus Fall, 004 Fial Solutios 1. (5ts) Defie the followig terms. Be as recise as you ca. (a) (3ts) A ucoutable set. A ucoutable set is a set which ca ot be ut ito bijectio with a fiite set or with the set of atural umbers. (b) (3ts) A toological sace. A toological sace is a set X together with a set T of subsets of X with the followig roerties: X ad belog to T, the uio of ay family of elemets of T belogs to T, ad the itersectio of ay fiite umber of elemets of T belogs to T. (c) (3ts) A comact toological sace. A toological sace is comact if every oe cover cotais a fiite subcover. (d) (3ts) A coected toological sace. A toological sace is coected if the oly subsets of X which are both oe ad closed are X ad. (e) (3ts) A limit oit of a subset of a metric sace. If E is subset of X ad x X, the x is a limit oit of E if every eighborhood of x meets E \ {x}. (f) (3ts) A cotiuous fuctio from X to Y, where X ad Y are toological saces. A fuctio f: X Y is cotiuous if the iverse image of every oe subset of Y is oe i X. (g) (4ts) Suose that f: E Y, where E X ad X ad Y are toological saces, ad x X. What is the defiitio of lim f(e), e x ad uder what coditios is it uique? A oit y is a limit if for every eighborhood V of y, there exists a eighborhood U of x such that U E \ {x} f 1 (V ). Ofte oe requires that x be a limit oit of E. It is uique if this is the case ad Y is Hausdorff, (h) (3ts) A equicotiuous family of real-valued fuctios o a metric sace. A family F is equicotiuous if for every ɛ > 0, there exists a δ such that f(x) f(x ) < ɛ wheever d(x, s ) < δ ad f F.. (0ts) Give a examle of the followig, or exlai, usig results discussed i class or the book, why o such examle exists. If you give a examle, you eed ot rove that it works.

2 (a) (5ts) A cotiuos real-valued fuctio o a bouded subset of the reals which is ubouded. The fuctio 1/x o (0, 1). (b) (5ts) A uiformly cotiuous fuctio o a bouded subset of the reals which is ubouded. This ca t exist. If f is uiformly cotiuous there exists a δ > 0 such that f(x) f(x ) < 1 wheever x x < δ. The if x x < kδ, f(x) f(x ) < k. Sice the domai of f is bouded, it follows that there exists k such that this is true for ay air of oits, ad it follows that f is bouded. (c) (5ts) A sequece of cotiuous real-valued fuctios o [0, 1] which has o oitwise coverget subsequece. The sequece si(x) has this roerty. So does the sequece f =. (d) A uiformly bouded sequece of real-valued cotiuous fuctios o Q [0, 1] which has o oitwise coverget subsequece. This does t exist, by a rather tricky diagoalizatio argumet. 3. (10ts) Prove that the closed iterval [0, 1] is comact. Let U be a oe cover of [0, 1]. Let S be the set of all x [0, 1] such that [0, x] ca be covered by a fiite subset of U. Clearly 0 S, ad S is bouded, hece has a suremum, s. Evidetly s 1. We claim that s S. If s = 0 this is clear. If s > 0, there is a U U such that s U, ad there is a s [0, s) such that [s, s] U. Sice s < s, there is a s > s with s S, ad hece a fiite subset U of U which covers [0, s ]. The U together with U covers [0, s], ad s S. Furthermore, if s < 1, we see that U also cotais a iterval of the form [s, t] with t (s, 1], so that we also have a fiite cover of [0, t]. Hece s = 1, ad we are doe. 4. (10ts) Let A be a oemty closed subset of a metric sace X. For x X, let d A (x) := if{d(x, a) : a A}. (a) Exlai why the above defiitio makes sese. This is because the set i questio is bouded below ad oemty. Hece it has a ifimum. (b) Prove that d A is cotiuous, ad that x A iff d A (x) = 0. Suose that x ad y are oits of X. The for ay a A, d A (x) d(x, a) d(x, y)+d(y, a). Thus d A (x) d(x, y) is a lower boud for {d(y, a) : a A)} ad hece d A (y) d A (x) d(x, y). I other words, d A (x) d A (y) d(x, y). By symmetry, the same is true with x ad y iterchaged, ad we coclude that d A (x) d A (y) d(x, y). This shows that d A is uiformly cotiuous. Fially, if x A, d A (x) d(x, x) = 0, ad if x A, sice A is closed there is a ɛ > 0 such that B ɛ (x) A =. Hece d(x, a) ɛ for all a ad d A (x) ɛ. (c) Prove that if A ad B are disjoit closed subsets of X, the there exist disjoit oe sets U ad V such that A U ad B V. Hit: cosider {x : d A (x) < d B (x)}.

3 It follows from the above that this set U is oe ad cotais A, sice if a A, d A (x) = 0 ad d B (x) > 0. O the other had, if V := {x : d B (x) < d A (x)}, V is oe ad cotais B. Clearly the two sets are disjoit. 5. (10ts) Prove that if 0 < r < 1, r aroaches 0 as aroaches ifiity. You get more credit for a elemetary roof. You may fid it easier to look at the recirocal. It suffices to rove that if r > 1, r / aroaches ifiity as aroaches ifiity. Write r = 1 + x, where x > 0. The ( ) r / 1 ( 1) 1 + x + x ( 1)x /, which clearly goes to ifiity. 6. (10ts) Let f: R R be a differetiable fuctio. Suose that f (x) 1 for all x R. Prove that there is at most oe umber a such that f(a) = a. If a ad b are fixed oits of f ad a < b, the by the MVT there exists a x (a, b) such that f (x)(b a) = f(b) f(a) = b a. But the f (x) = 1, a cotradictio. 7. (10ts) Suose f(0) = f (0) = f (0) = 0. Prove that if ɛ > 0, there is a δ > 0 such that f(x) ɛ x wheever x < δ. First observe that sice f (0) = 0, 0 = f (0) = lim x 0 f (x) x. Suose ɛ > 0 is give. The there exists a δ > 0 such that f (x) ɛ x wheever x < δ. Furthermore, sice f(x) = 0, for ay such x, the MVT says that there is some x (0, x) such that f (x )x = f(x). Hece f(x) ɛ x x ɛ x. 8. (15ts) Cosider the series z /, where z is a comlex umber. (a) Prove that the series coverges oitwise o [ 1, 1). The first statemet follows from Abel s theorem. For each m, if z 1 ad z [ 1, 1), m A m := z z = 1 z zm+1 1 z 1 z. 1 Thus the sequece of artial sums is bouded ad hece Abel s theorem alies. (b) Prove that the covegece is uiform o [ 1, 1 ɛ) for every ɛ > 0. For the uiformity we have to recall the roof. If ɛ > 0, the there exists a umber M > 0 such that for all z [ 1, ɛ], A m M. Now if q > > 1, q z q = A A 1 3 = q q 1 A 1 A + 1

4 = A q q A 1 A q q q 1 ( 1 + A 1 ) A q A M q + M q 1 + M ( 1 1 ) M + 1. Thus the sequece is uiformly Cauchy ad hece uiformly coverget. Alteratively, we ca use a simler argumet, based o alteratig series. First observe that if r (0, 1), the series coverges absolutely ad uiformly i ay iterval of the form [ r, r], by the ratio test, for examle. So it suffices to rove uiform covergece i the iterval [ 1, 0]. The the usual alteratig series trick shows that if z [ 1, 0], M N z < z N+1 N + 1 < 1 N + 1. This shows that the series is uiformly Caucy, hece uiformly coverget. (c) What ca you deduce about the fuctio f(z) = z / for z [ 1, 1)? Ca you fid ad rove a simle formula for it, valid at all these oits? I articular, what is its value at 1? It follows from the uiform covegece that the limit fuctio f(z) is cotiuous o [ 1, 1). Now we kow that 1 + z + z, coverges to (1 z) 1, ad the covergece is uiform o comact subsets of ( 1, 1). Hece we ca itegrate term by term, ad deduce that z z / + coverges to log(1 z) o the oe iterval ( 1, 1). Hece f(z) = log z if z < 1. But both fuctios are also cotiuous at 1, hece they agree there also. Thus whe z = 1, the series coverges to log. 9. (10ts) Prove that if f: X Y is a cotiuous bijective ma of comact Hausdorff saces, the the iverse maig Y X is also cotiuous. Show that this eed ot be the case if X is ot comact. Let g: Y X be the iverse maig, ad let Z be a closed subset of X. The g 1 (Z) = f(z). But Z is comact, hece f(z) is comact, hece closed. Thus g 1 takes closed sets to closed sets, so g is cotiuous. O the other had, let X := [0, π), let Y := {z C : z = 1}, ad let f: X Y the be ma sedig θ to cos(θ) + i si(θ). The f is cotiuous ad bijective. However its iverse g is ot cotiuous, sice the iverse image of every eighborhood of 1 Y cotais oits ear 0 ad ear π. 10. (15ts) Let C(T ) deote the set of cotiuous fuctios R C which are eriodic with eriod π. Let A C(T ) deote the set of fiite liear combiatios of fuctios of the form e ix with Z. If f ad g belog to C(T ), let (f g)(x) := π f(x t)g(t)dt. 4

5 (a) Show that if g A, the f g A. Let s = x t, ad observe that (f g)(x) = π x π +x f(x t)g(t)dt = f(s)g(x s)ds = f(s)g(x s)ds π+x π x Sice f ad g are eriodic with eriod π, +x π x f(s)g(x s)ds = π f(s)g(x s)ds. Now if g (t) := e it, g(x s) = g(x)g( s), ad hece (f g )(x) = π f(s)g(x)g( s)ds = g (x)(f g)(0). This is a elemet of A. Hece the same is true if g is a liear combiatio of the g s. (b) Let ( ) 1 + cos t ( ) 1 + cos t c := dt = dt π 0 Show that ( ) 1 + cos t c si tdt. 0 Use this last iequality to show that c c/( + 1) for some c. Let ( ) 1 + cos t Q (t) := c 1 ad rove if δ > 0, Q coverges uiformly to 0 o [δ, π]. Sice si(t) 1, c 0 ( ) 1 + cos t si tdt = (1 + cos t)+1 π 0 ( + 1) = Now if δ > 0 ad t [δ, π], ( ) 1+cos t r, where r (0, 1). Hece Q (t) M( + 1)r for some M, which aroaches zero. (c) Deduce that A is uiformly dese i C(T ). You may use the theorem o Dirac families. The above argumets shows that the set of Q form a Dirac family. Hece for ay cotiuous f, f Q coverges uiformly to f. 5