Topologie. Musterlösungen

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1 Fakultät für Mathematik Sommersemester 2018 Marius Hiemisch Topologie Musterlösuge Aufgabe (Beispiel 1.2.h aus Vorlesug). Es sei X eie Mege ud R Abb(X, R) eie Uteralgebra, d.h. {kostate Abbilduge} R ud f, g P, S, O, OR f g, fg R. Die Zariski-Topologie auf X ist diejeige Topologie τ = τ Zar,R, dere abgeschlossee Mege vo der Form V (I) = {x X : f(x) = 0 f I}, für I R Teilmege, sid. Zeige, dass dies eie Topologie auf X defiiert. Lösug: Wir zeige: i), X sid abgeschlosse. ii) A 1,..., A abgeschlosse A i abgeschlosse iii) A j für j J abgeschlosse A j abgeschlosse i=1 Die kostate Abbilduge f 0 ud g 1 sid i R. Die Mege V ({f}) = {x X : f(x) = 0} = X ud V ({g}) = {x X : g(x) = 0} = sid abgeschlosse, also gilt i). Es seie A 1,..., A abgeschlosse, d.h. es existiere I 1,..., I R mit V (I i ) = A i für alle i. Da ist V (I i ) = {x X : f 1 (x)... f (x) = 0 für alle (f 1,..., f ) I 1... I } = V (Ĩ) i=1 für Ĩ = I 1... I = {f 1... f : f i I i } R. Es seie A j für j J abgeschlosse. Da existiere I j, j J mit A j = V (I j ) ud die Mege { V (I j ) = x X : f(x) = 0 f } ( ) I j = V I j ist damit abgeschlosse.

2 Aufgabe (Lemma 1.4 i the lecture). Let (X 1, d 1 ), (X 2, d 2 ) be metric spaces ad f : X 1 X 2 a map. The f is cotiuous via -ε-defiitio preimages of ope sets are ope. Proof. (1) First we show that f is cotiuous i x X 1 iff the preimage U = f 1 (V ) of ay eighbourhood V of f(x) is a eighbourhood of x. : Let V be a eighbourhood of f(x). The there exists ε > 0 such that B ε (f(x)) V. Because f is cotiuous there exists > 0 with f(b (x)) B ε (f(x)) V. Therefore B (x) f 1 (V ) ad we get that f 1 (V ) is a eighbourhood of x. : Let ε > 0 ad y = f(x). The B ε (y) is a eighbourhood of y. From it follows that U := f 1 (B ε (y)) is a eighbourhood of x. There is a > 0 such that B (x) U ad we get f(b (x)) B ε (y) ad therefore the implicatio holds. x : d 1 (x, x) < = d 2 (f(x), f( x) < ε (2) Usig (1) we ow show the claim. : Let f be cotiuous ad V X 2 be ope. Take ay x U = f 1 (V ), the f(x) V ad V is a eighbourhood of f(x) because it is ope. From (1) it follows that U is a eighbourhood of x ad because x is arbitrary, U is a eighbourhood of all of its poits ad therefore ope. : We ow assume that preimages of ope sets are ope. Take x X 1 ad V a eighbourhood of f(x), we show that U = f 1 (V ) is a eighbourhood of x (that suffices to show the cotiuity of f because of (1)). There is a ε > 0 with B ε (f(x)) V. Sice B ε (f(x)) is ope, Ũ = f 1 (B ε (f(x))) is ope. Therefore U Ũ is a eighbourhood of x which shows that f is cotiuous.

3 Übugsblatt 1 Aufgabe 1. We have to fid a map f V \V, i.e. a liear map V R that is ot cotiuous. Because dim V =, we ca fid coutably may liearly idepedet elemets {a 1, a 2,... } = A V. Without loss of geerality we will assume a i = 1 i N (otherwise scale them properly). We ca exted A to a basis B of V. A liear map f : V R is fully determied by its values o the basis elemets. We set { if b = a A for some N f(b) = 0 else for b B ad obtai a liear map f V. The followig calculatio shows that f is ot cotiuous: f = f(x) sup x V \{0} x = sup f(x) sup f(a ) = lim =. N Aufgabe 2. (1) τ V τ : Assume there exists U τ V \τ, so U V is ope i the weak-*-topology but ot i the orm topology. The there exists x U such that x is a boudary poit (i orm topology), so ε > 0 : U ε (x) U (where U ε (x) is the ope ε-ball aroud x i the orm topology). The poit x is a ier poit of U i the weak-*-topology, so there exist > 0 ad x 1,..., x V such that T := {y V : (y x)(x i ) < i} U (see Defiitio (2.9) ad Example (1.2.g)). We set We the see that w := max i=1... x. U ε (x) T U, where ε =, because let y U 2w ε (x), the (y x)(x i ) ε x i = x 2w i i. This 2 cotradicts x beig a boudary poit ad therefore τ V τ, hece ι 2 is cotiuous. (2) τ τ V if dim V = for N: The set A = {U (f) : > 0, f V } is a basis of the orm topology o V ad it suffices to show A τ V. Take f V ad > 0 ad choose a ormalised basis x 1,..., x V, i.e. x i = 1 i. We show that U (f) is ope i the weak-*-topology, i.e.: ( ) g U (f) ε > 0 : h V : h(x i ) g(x i ) < ε h U (f). (This is Example 1.2.g) where we require that i 1,..., i m are the basis elemets x 1,..., x.)

4 We kow We write x = a i x i V ad get i=1 h f = sup h(x) f(x) f g = sup f(x) g(x) <. sup h(x) g(x) + sup g(x) f(x) = sup h(x) g(x) + f g ( ) ) sup h ai x i g( ai x i + f g = sup ai (h(x i ) g(x i )) + f g sup ai h(x i ) g(x i ) + f g (ow use h(x i ) g(x i ) < ε) < ε sup ai + f g. We wat this to be less tha, so we eed ε sup ai < f g. We kow that sup ai because a i 1 i. We ow set f g ε = > 0 ad realise that our claim is prove. Therefore for all f V ad > 0 : U (f) τ V ad ι 1 is cotiuous. (3) τ V τ if dim V = : If dim V =, the U 1 (0) = {ϕ : V R: ϕ = sup ϕ(x) < 1} τ V. The zero fuctio 0 is i U 1 (0) but for ay set of poits A = {x 1,..., x } V ad ε > 0 there is a ψ V with (ψ 0)(x i ) < ε for all i but ψ > 1. Exted A to a basis B of V, choose oe elemet b B\A ad defie ψ via { 2 b, if b = ψ(b) = b 0, else o the basis elemets b B. The ϕ = 2. Ad 2 > 1 (check this!). Therefore ι 1 is ot cotiuous iff dim V =.

5 Aufgabe 3. V is the space of bouded sequeces i R, hece for every sequece x = (x ) N V a umber M R exists with x < M ( N : x < M). We have φ m = sup φ m (x) = sup x m = 1, hece φ m K V V for all m N. The previous exercise tells us that it suffices to show that o subsequece of φ m coverges i the orm topology of V. We have φ m φ = sup φ m (x) φ (x) = 2 m, N (take x = e m e ). This shows that (φ ) N is ot a Cauchy-sequece ad because the distace betwee ay two sequece elemets is costat, o subsequece ca coverge. This proves that K V is ot sequetially compact but by the Theorem of Baach-Alaoglu (Thm. (2.11) i the lecture) it is compact (both with respect to the weak-*-topology). Übugsblatt 2 Aufgabe 1. Aufgabe 2. I this exercise we use the otatio (a, b) Q := (a, b) Q for itervals with a, b R. We will show that o ope eighborhood of 0 (this works for every umber) has a compact eighborhood. This tells us that Q is ot locally compact. Take ay ope eighbourhood U of 0 Q, the there exists N such that ( 1, 1 ) Q U. Choose your favourite irratioal umber r ( 1, 1 ). The we ca fid a strictly icreasig sequece (x ) N i Q with lim x = r. We ca costruct the sequece like this: Take ay x 0 ( 1, r) Q ad the iductively x (r 2 (r x 1 ), r) Q (these sets are ot empty because Q R is dese). The factor 2 guaratees covergece i R. Sice r / Q, the sequece does ot coverge i Q ad because it is mootoe, o subsequece coverges i Q. Hece o sequetially compact eighbourhoods of ( 1, 1 ) Q or U exist ad because i metric spaces compactess ad sequetial compactess coicide, Q is ot locally compact. Aufgabe 3. Übugsblatt 3 Aufgabe 1. Aufgabe 2.

6 Aufgabe Let X be a compact metric space. For every N the set U := {U 1 (x) : x X} is a ope coverig of X. Because X is compact, there exists a fiite subset Q X with x Q U 1 (x). We show that U = N x Q U 1 is a basis for the topology o X. To prove this, take ay ope set U X ad show that there exist N, x Q with U 1 (x) U. It suffices to cosider U = U ε (y) for ay ε > 0, y X. Choose N such that 2 < ε. By costructio, there has to be a 1 -Ball i U cotaiig y. The same ball is a subset of U (draw a sketch). Alteratively oe ca use 2. below ad show that a compact metric space is separable. 2. Let (X, d) be a metric space. We show X is separable X satisfies the Secod Axiom Of Coutability : Because X is separable, there is a at most coutable subset A X with A = X. For every x X ad every ope set V x, there exists N with U 1 (x) V. Because A is dese i X, there is a sequece (a k ) k N with a k A for all k covergig toward x. There is a elemet a := a i for some i i that sequece with d(x, a) < 1. That is equivalet to a U 1 (x). The x U 1 (a) U 1 (x) V. We set U = U 1 (a). N a A The U is coutable because A ad N are (at most) coutable. Therefore X satisfies the Secod Axiom Of Coutability. : Let U = {U 1, U 2,... } be a at most coutable basis. There is a fuctio α : U X with α(u i ) U i i N (this uses the axiom of choice). We defie A := {α(u i ) : i N}. We claim that A is dese i X. For every k N there is a U i U with x U i U 1 (x). Hece we defie a sequece k (a k ) k N via a k = α(u i ) for that specific U i. The (a k ) coverges towards x ad therefore A = X, showig that X is separable. 3. We show that R with the cofiite topology (see Example (1.2) f)) is separable but the First Axiom Of Coutability does ot hold. The Secod Axiom Of Coutability implies the first oe, because the coutable basis of a topological space cotais a eighbourhood basis for each poit. The topological space R with the cofiite topology is separable because Q R is coutable ad dese: Q = V = R Q V closed (x)

7 because i cofiite topology o R a set V R is closed if ad oly if V = R or V is fiite. The poit 0 R has o coutable eighbourhood basis: Assume U = {U i : i I}, where I is at most coutable, is such a basis. We ca write U i = R\{x i,1,..., x i,i } ad we set A = i I {x i,1,..., x i,i }. The A is at most coutable ad it exists x R\A. The set R\{x} is ope ad cotais o U i, i I.