SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 4. Function spaces
|
|
- Domenic Baker
- 5 years ago
- Views:
Transcription
1 SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 4 Fall 2008 IV. Functin spaces IV.1 : General prperties (Munkres, 45 47) Additinal exercises 1. Suppse that X and Y are metric spaces such that X is cmpact. Let Y X dente the cartesian prduct f the spaces Y {x} = Y with the prduct tplgy, and fr each x X let p x : Y X Y dente prjectin nt the factr crrespnding t x. Let q : C(X, Y ) Y X be the map such that fr each x the cmpsite p x q sends f t f(x). Prve that q is a cntinuus 1 1 mapping. The mapping q is 1 1 because q(f) = q(g) implies that fr all x we have f(x) = p x q(f) = p x q(g) = g(x), which means that f = g. T prve cntinuity, we need t shw that the inverse images f subbasic pen sets in Y X are pen in C(X, Y ). The standard subbasic pen subsets have the frm W({x}, U) = p 1 x (U) where x X and U is pen in Y. In fact, there is a smaller subbasis cnsisting f all such sets W({x}, U) such that U = N ε (y) fr sme y Y and ε > 0. Suppse that f is a cntinuus functin such that q(f) lies in W({x}, U). By definitin the later cnditin means that f(x) U. The latter in turn implies that δ = ε d( f(x), y) > 0, and if d(f, g) < δ then the Triangle Inequality implies that d( g(x), y) < ε, which in turn means that g(x) U. Therefre q is cntinuus at f, and since f is arbitrary this shws q is a cntinuus mapping. 2. Suppse that X, Y and Z are metric spaces such that X is cmpact, and let p Y, p Z dente prjectins frm Y Z t Y and Z respectively. Assume that ne takes the d r maximum metric n the prduct (i.e., the distance between (x 1, y 1 ) and (x 2, y 2 ) is the larger f d(x 1, y 1 ) and d(x 2, y 2 ) ). Prve that the map frm C(X, Y Z) t C(X, Y ) C(X, Z) sending f t (p y f, p Z f) is an ismetry (where the cdmain als has the crrespnding d metric). It suffices t shw that the map in questin is nt and distance-preserving. The map is nt because if u and v are cntinuus functins int Y and Z respectively, then we can retrieve f by the frmula f(x) = ( u(x), v(x) ). Suppse nw that f and g are cntinuus functins frm X t Y Z. Then the distance frm f t g is the maximum f d ( f(x), g(x) ). The latter is less than r equal t the greater f d ( p Y f(s), p Y g(s) ) and d ( p Y f(t), p Y g(t) ). Thus if Φ : C(X, Y Z) C(X, Y ) C(X, Z) 1
2 then the distance frm Φ(f) t Φ(g) is greater than r equal t the distance frm f t g. This means that the map Φ 1 is unifrmly cntinuus. Cnversely, we claim that the distance frm f t g is greater than r equal t the distance between Φ(f) and Φ(g). The latter is equal t the larger f the maximum values f d ( p Y f(s), p Y g(s) ) and d ( p Z f(t), p Z g(t) ). If w X is where d(f, g) takes its maximum, it fllws that d ( f(w), g(t) ) = { max d ( p Y f(w), p Y g(w) ), d ( p Z f(w), p Z g(w) ) } which is less than r equal t the distance between Φ(f) and Φ(g) as described abve. 3. Suppse that X and Y are metric spaces such that X is cmpact, and assume further that X is a unin f tw disjint pen and clsed subsets A and B. Prve that the map frm C(X, Y ) t C(A, Y ) C(B, Y ) sending f t (f A, f B) is als an ismetry, where as befre the prduct has the d metric. The distance between f and g is the maximum value f the distance between f(x) and g(x) as x runs thrugh the elements f x, which is the greater f the maximum distances between f(x) and g(x) as x runs thrugh the elements f C, where C runs thrugh the set {A, B}. But the secnd expressin is equal t the larger f the distances between d(f A, g A) and d(f B, g B). Therefre the map described in the prblem is distance preserving. As in the previus exercise, t cmplete the prf it will suffice t verify that the map is nt. The surjectivity is equivalent t saying that a functin is cntinuus if its restrictins t the clsed subsets A and B are cntinuus. But we knw the latter is true. 4. Suppse that X, Y, Z, W are metric spaces such that X and Z are cmpact. Let P : C(X, Y ) C(Z, W ) C(X Z, Y W ) be the map sending (f, g) t the prduct map f g (recall that f g(x, y) = (f(x), g(y)) ). Prve that P is cntinuus, where again ne uses the assciated d metrics n all prducts. Let ε > 0 be given. We claim that the distance between f g and f g is less than ε if the distance between f and f is less than ε and the distance between g and g is less than ε. Chse u 0 X and v 0 Z s that d ( f g (u 0, v 0 ), f g(u 0, v 0 ), ) is maximal and hence equal t the distance between f g and f g. The displayed quantity is equal t the greater f d ( f (u 0 ), f(u 0 ) ) and d ( g (v 0 ), g(v 0 ) ). These quantities in turn are less than r equal t d(f, f) and d(g, g) respectively. Therefre if bth f the latter are less than ε it fllws that the distance between f g and f g is less than ε. 5. Suppse that X and Y are metric spaces and f : X Y is a hmemrphism. (i) If A is a cmpact metric space, shw that the map V (f) defines a hmemrphism frm C(A, X) t C(A, Y ). [Hint: Cnsider V (h), where h = f 1.] Fllw the hint. We then have V (h) V (f) = V (h f) = V (id), which is the identity. Likewise, we als have V (f) V (h) = V (f h) = V (id), which is the identity. 2
3 (ii) If X and Y are cmpact and B is a metric space, shw that the map U(f) defines a hmemrphism frm C(Y, B) t C(X, B). [Hint: Cnsider U(h), where h = f 1.] Again fllw the hint. We then have U(f) U(h) = U(h f) = V (id), which is the identity. Likewise, we als have U(h) V (f) = V (f h) = V (id), which is the identity. (iii) Suppse that A and A are hmemrphic cmpact metric spaces and B and B are hmemrphic metric spaces. Prve that C(A, B) is hmemrphic t C(A, B ). Let f : A A and g : B B be the hmemrphisms. Let U(f 1 ) V (g) = V (g) U(f 1 ) equality hlds by assciativity f cmpsitin. By the first tw parts f the exercise this map is the required hmemrphism frm C(A, B) t C(A, B ). 6. Given a metric space X and a, b X, let P(X; a, b) be the space f all cntinuus functins r curves γ frm [0, 1] t X such that γ(0) = a and γ(1) = b. Given a, b, c X define a cncatenatin map α : P(X; a, b) P(X; b, c) P(X; a, c) such that α(γ, γ )(t) = γ(2t) if t 1 2 and γ (2t 1) if t 1. Infrmally, this is a reparametrizatin f 2 the curve frmed by first ging frm a t b by γ, and then ging frm b t c by γ. Prve that this cncatenatin map is unifrmly cntinuus (again take the d metric n the prduct). It will be cnvenient t dente α(γ, γ ) generically by γ+γ. The cnstructin then implies that the distance between ξ +ξ and η+η is the larger d(ξ, η) and d(ξ, η ). Therefre the cncatenatin map is distance preserving. 7. Let X and Y be tplgical spaces, take the cmpact-pen tplgy n C(X, Y ), and let k : Y C(X, Y ) be the map sending a pint y Y t the cnstant functin k(y) whse value at every pint is equal t y. Prve that k maps Y hmemrphically t k(y ). [Hints: First verify that k is 1 1. If W(K, U) is a basic pen set fr the cmpact-pen tplgy with K X cmpact and U Y pen, prve that k 1 [ W(K, U) ] = U and k[u] = W(K, U) Image(k). Why d the latter identities imply the cnclusin f the exercise?] Fllw the hint. If y y, then the values f k(y) at every pint is X, and hence it is nt equal t y, the value f k(y ) at every pint f X. Therefre k is 1 1. Next, we shall verify the set-theretic identities described abve. If y U then since k(y) is the functin whse value is y at every pint we clearly have k(y) W(K, U), and hence k(u) W(K, U) Image(k). Cnversely, any cnstant functin in the image f the latter is equal t k(y) fr sme y U. The identity k 1 ( W(K, U) ) = U fllws similarly. IV.2 : Adjint equivalences (Munkres, 45 46) Additinal exercises 1. State and prve an analg f the main result f this sectin fr in which X, Y, Z are untplgized sets (with n cardinality restrictins!) and spaces f cntinuus functins are replaced by sets f set-theretic functins. 3
4 The map A : F(X Y, Z) F(X, F(Y, Z) ) sends h : X Y Z t the functin h such that [h (x)](y) = h(x, y). The argument prving the adjint frmula fr spaces f cntinuus functins mdifies easily t cver these examples, and in fact in this case the prf is a bit easier because it is nt necessary t cnsider metrics r tplgies. 2. Suppse that X is a cmpact metric space and Y is a cnvex subset f R n. Prve that C(X, Y ) is arcwise cnnected. [Hint: Let f : X Y be a cntinuus functin. Pick sme y 0 Y and cnsider the map H : X [0, 1] Y sending (x, t) t the pint (1 t)f(x) + ty 0 n the line segment jining f(x) t y 0.] Let y 0 Y, and let L : Y [0, 1] Y be the map sending (y, t) t the pint (1 t)y + ty 0 n the line segment jining y t y 0. If we set H(x, t) = L( f(x), t), then H satisfies the cnditins in the hint and defines a cntinuus map in C(X, Y ) jining f t the cnstant functin whse values is always y 0. Thus fr each f we knw that f and the cnstant functin with value y 0 lie in the same arc cmpnent f C(X, Y ). Therefre there must be nly ne arc cmpnent. 3. Suppse that X, Y, Z are metric spaces such that X and Y are cmpact. Prve that there is a 1 1 crrespndence between cntinuus functins frm X t C(Y, Z) and cntinuus functins frm Y t C(X, Z). As usual, assume the functin spaces have the tplgies determined by the unifrm metric. By the adjint frmula there are hmemrphisms C ( X, C(Y, Z) ) = C(X Y, Z) = C(Y X, Z) = C ( Y, C(X, Z) ) and this yields the desired 1 1 crrespndence f sets. 4
5 V. Cnstructins n spaces V.1 : Qutient spaces Prblem frm Munkres, 22, pp (a) Define an equivalence relatin n the Euclidean crdinate plane by defining (x 0, y 0 ) (x 1, y 1 ) if and nly if x 0 + y0 2 = x 1 + y1 2. It is hmemrphic t a familiar space. What is it? [Hint: Set g(x, y) = x + y 2.] The hint describes a well-defined cntinuus map frm the qutient space W t the real numbers. The equivalence classes are simply the curves g(x, y) = C fr varius values f C, and they are parablas that pen t the left and whse axes f symmetry are the x-axis. It fllws that there is a 1 1 nt cntinuus map frm W t R. Hw d we shw it has a cntinuus inverse? The trick is t find a cntinuus map in the ther directin. Specifically, this can be dne by cmpsing the inclusin f R in R 2 as the x-axis with the qutient prjectin frm R 2 t W. This gives the set-theretic inverse t R 2 W and by cnstructin it is cntinuus. Therefre the qutient space is hmemrphic t R with the usual tplgy. (b) Answer the same questin fr the equivalence relatin (x 0, y 0 ) (x 1, y 1 ) if and nly if x y0 2 = x y1. 2 Here we define g(x, y) = x 2 + y 2 and the equivalence classes are the circles g(x, y) = C fr C > 0 alng with the rigin. In this case we have a cntinuus 1 1 nt map frm the qutient space V t the nnnegative real numbers, which we dente by [0, ) as usual. T verify that this map is a hmemrphism, cnsider the map frm [0, ) t V given by cmpsing the standard inclusin f the frmer as part f the x-axis with the qutient map R 2 V. This is a set-theretic inverse t the map frm V t [0, ) and by cnstructin it is cntinuus. Additinal exercises 0. Suppse that X is a space with the discrete tplgy and R is an equivalence relatin n X. Prve that the qutient tplgy n X/R is discrete. We claim that every subset f X/R is bth pen and clsed. But a subset f the qutient is pen and clsed if and nly if the inverse image has these prperties, and every subset f a discrete space has these prperties. 1. If A is a subspace f X, a cntinuus map r : X A is called a retractin if the restrictin f r t A is the identity. Shw that a retractin is a qutient map. We need t shw that U A is pen if and nly if r 1 [U] is pen in X. The ( = ) implicatin is immediate frm the cntinuity f r. T prve the ther directin, nte that r i = id A implies that U = i 1 [ r 1 [U] ] = A r 1 [U] 5
6 and thus if the inverse image f U is pen in X then U must be pen in A. 2. Let R be an equivalence relatin n a space X, and assume that A X cntains pints frm every equivalence class f R. Let R 0 be the induced equivalence relatin n A, and let j : A/R 0 X/R be the assciated 1 1 crrespndence f equivalence classes. Prve that j is a hmemrphism if there is a retractin r : X A sending R-equivalent pints f X t R 0 -equivalent pints f A. such that each set r 1 [ {a} ] is cntained in an R-equivalence class. Let p X and p A dente the qutient space prjectins fr X and A respectively. By cnstructin, j is the unique functin such that j p A = p X A and therefre j is cntinuus. We shall define an explicit cntinuus inverse k : X/R A/R 0. T define the latter, cnsider the cntinuus map p A r : X A/R 0. If yrz hlds then r(y)r 0 r(z), and therefre the images f y and z in A/R 0 are equal. Therefre there is a unique cntinuus map k f qutient spaces such that k p X = p A r. This map is a set-theretic inverse t j and therefre j is a hmemrphism. 3. (a) Let 0 dente the rigin in R 3. In R 3 {0} define xry if y is a nnzer multiple f x (gemetrically, if x and y lie n a line thrugh the rigin). Shw that R is an equivalence relatin; the qutient space is called the real prjective plane and dented by RP 2. The relatin is reflexive because x = 1 x, and it is reflexive because y = αx fr sme α 0 implies x = α 1 y. The relatin is transitive because y = αx fr α 0 and z = βy fr y 0 implies z = βαx, and βα 0 because the prduct f nnzer real numbers is nnzer. (b) Using the previus exercise shw that RP 2 can als be viewed as the qutient f S 2 mdul the equivalence relatin x y y = ± x. In particular, this shws that RP 2 is cmpact. [Hint: Let r be the radial cmpressin map that sends v t v 1 v.] Use the hint t define r; we may apply the preceding exercise if we can shw that fr each a S 2 the set r 1 ({a}) is cntained in an R-equivalence class. By cnstructin r(v) = v 1 v, s r(x) = a if and nly if x is a psitive multiple f a (if x = ρ a then x = ρ and r(x) = a, while if a = r(x) then by definitin a and x are psitive multiples f each ther). Therefre if xry then r(x) = ± r(y), s that r(x)r 0 r(y) and the map is a hmemrphism. S 2 /[x ±x] RP 2 4. In D 2 = {x R 2 x 1}, cnsider the equivalence relatin generated by the cnditin xr y if x = y = 1 and y = x. Shw that this qutient space is hmemrphic t RP 2. [Hints: Use the descriptin f RP 2 as a qutient space f S 2 frm the previus exercise, and let h : D 2 S 2 be defined by h(x, y) = (x, y, 1 x 2 y 2 ). Verify that h preserves equivalence classes and therefre induces a cntinuus map h n qutient spaces. Why is h a 1 1 and nt mapping? Finally, prve that RP 2 is Hausdrff and h is a clsed mapping.] 6
7 Needless t say we shall fllw the hints in a step by step manner. Let h : D 2 S 2 be defined by h(x, y) = (x, y, 1 x 2 y 2 ). Verify that h preserves equivalence classes and therefre induces a cntinuus map h n qutient spaces. T shw that h is well-defined it is nly necessary t shw that its values n the R -equivalence classes with tw elements are the same fr bth representatives. If π : S 2 RP 2 is the qutient prjectin, this means that we need π h(u) = π h(v) if u = v = 1 and u = v. This is immediate frm the definitin f the equivalence relatin n S 2 and the fact that h(w) = w if w = 1. Why is h a 1 1 and nt mapping? By cnstructin h maps the equivalence classes f pints n the unit circle nt the pints f S 2 with z = 0 in a 1 1 nt fashin. On the ther hand, if u and v are distinct pints that are nt n the unit circle, then h(u) cannt be equal t ± h(v). The inequality h(u) h(v) fllws because the first pint has a psitive z-crdinate while the secnd has a negative z-crdinate. The ther inequality h(u) h(v) fllws because the prjectins f these pints nt the first tw crdinates are u and v respectively. This shws that h is 1 1. T see that it is nt, recall that we already knw this if the third crdinate is zer. But every pint n S 2 with nnzer third crdinate is equivalent t ne with psitive third crdinate, and if (x, y, z) S 2 with z > 0 then simple algebra shws that the pint is equal t h(x, y). Finally, prve that RP 2 is Hausdrff and h is a clsed mapping. If the first statement is true, then the secnd ne fllws because the dmain f h is a qutient space f a cmpact space and cntinuus maps frm cmpact spaces t Hausdrff spaces are always clsed. Since h is already knwn t be cntinuus, 1 1 and nt, this will prve that it is a hmemrphism. S hw d we prve that RP 2 is Hausdrff? Let v and w be pints f S 2 whse images in RP 2 are distinct, and let P v and P w be their rthgnal cmplements in R 3 (hence each is a 2- dimensinal vectr subspace and a clsed subset). Since Euclidean spaces are Hausdrff, we can find an ε > 0 such that N ε (v) P v =, N ε (w) P w =, N ε (v) N ε (w) =, and N ε ( v) N ε (w) =. If T dentes multiplicatin by 1 n R 3, then these cnditins imply that the fur pen sets N ε (v), N ε (w), N ε ( v) = T (N ε (v)), N ε ( w) = T (N ε (w)) are pairwise disjint. This implies that the images f the distinct pints π(v) and π(w) in RP 2 lie in the disjint subsets π [ N ε (v) ] and π [ N ε (w) ] respectively. These are pen subsets in RP 2 because their inverse images are given by the pen sets N ε (v) N ε ( v) and N ε (w) N ε ( w) respectively. 5. Suppse that X is a tplgical space with tplgy T, and suppse als that Y and Z are sets with set-theretic maps f : X Y and g : Y Z. Prve that the qutient tplgies satisfy the cnditin (g f) T = g (f T). (Infrmally, a qutient f a qutient is a qutient.) 7
8 A set W belngs t (g f) T if and nly if (g f) 1 [W ] is pen in X. But (g f) 1 [W ] = f 1 [ g 1 [W ] ] s the cnditin n W hlds if and nly if g 1 [W ] belngs t f T. The latter in turn hlds if and nly if w belngs t g (f T). 6. If Y is a tplgical space with a tplgy T and f; X Y is a set-theretic map, then the induced tplgy f T n X is defined t be the set f all subsets W X having the frm f 1 [U] fr sme pen set U T. Prve that f T defines a tplgy n X, that it is the unique smallest tplgy n X fr which f is cntinuus, and that if h : Z X is anther set-theretic map then (f h) T = h (f T). The bject n the left hand side is the family f all sets having the frm (f h) 1 [V ] where V belngs t T. As in the preceding exercise we have s the family in questin is just h (f T). (f h) 1 [V ] = h 1 [ f 1 [V ] ] 7. Let X and Y be tplgical spaces, and define an equivalence relatin R n X Y by (x, y) (x, y ) if and nly if x = x. Shw that X Y/R is hmemrphic t X. Let p : X Y X be prjectin nt the first crdinate. Then urv implies p(u) = p(v) and therefre there is a unique cntinuus map X Y/R X sending the equivalence class f (x, y) t x. Set-theretic cnsideratins imply this map is 1 1 and nt, and it is a hmemrphism because p is an pen mapping. 8. Let R be an equivalence relatin n a tplgical space X, let Γ R be the graph f R, and let π : X X/R be the qutient prjectin. Prve the fllwing statements: (a) If X/R satisfies the Hausdrff Separatin Prperty then Γ R is clsed in X X. If X/R is Hausdrff then the diagnal (X/R) is a clsed subset f (X/R) (X/R). But π π is cntinuus, and therefre the inverse image f (X/R) must be a clsed subset f X X. But this set is simply the graph f R. (b) If Γ R is clsed and π is pen, then X/R is Hausdrff. If π is pen then s is pi π, fr the penness f π implies that π π takes basic pen subsets f X X int pen subsets f (X/R) (X/R). By hypthesis the cmplementary set X X Γ R is pen in X X, and therefre its image, which is (X/R) (X/R) (X/R) 8
9 must be pen in (X/R) (X/R). But this means that the diagnal (X/R) must be a clsed subset f (X/R) (X/R) and therefre that X/R must satisfy the Hausdrff Separatin Prperty. (c) If Γ R is pen then X/R is discrete. The cnditin n Γ R implies that each equivalence class is pen. But this means that each pint in X/R must be pen and hence the latter must be discrete. V.2 : Sums and cutting and pasting Additinal exercises 1. Let {A α α A} be a family f tplgical spaces, and let X = α A α. Prve that X is lcally cnnected if and nly if each A α is lcally cnnected. ( = ) If X is lcally cnnected then s is every pen subset. But each A α is an pen subset, s each is lcally cnnected. ( = ) We need t shw that fr each x X and each pen set U cntaining x there is an pen subset V U such that x V and V is cnnected. There is a unique α such that x = i α (a) fr sme a A α. Let U 0 = i 1 α (U). Then by the lcal cnnectedness f A α and the penness f U 0 there is an pen cnnected set V 0 such that x V 0 U 0. If V = i α (V 0 ), then V has the required prperties. 2. In the preceding exercise, frmulate and prve necessary and sufficient cnditins n A and the sets A α fr the space X t be cmpact. X is cmpact if and nly if each A α is cmpact and there are nly finitely many (nnempty) subsets in the cllectin. The ( = ) implicatin fllws because each A α is an pen and clsed subspace f the cmpact space X and hence cmpact, and the nly way that the pen cvering { A α } f X, which cnsists f pairwise disjint subsets, can have a finite subcvering is if it cntains nly finitely many subsets. T prve the reverse implicatin, ne need nly use a previus exercise which shws that a finite unin f cmpact subspaces is cmpact. 3. Prve that RP 2 can be cnstructed by identifying the edge f a Möbius strip with the edge circle n a clsed 2-dimensinal disk by filling in the details f the fllwing argument: Let A S 2 be the set f all pints (x, y, z) S 2 such that z 1 2, and let B be the set f all pints where z 1 2. If T (x) = x, then T [A] = A and T [B] = B s that each f A and B (as well as their intersectin) can be viewed as a unin f equivalence classes fr the equivalence relatin that prduces RP 2. By cnstructin B is a disjint unin f tw pieces B ± cnsisting f all pints where sign(z) = ± 1, and thus it fllws that the image f B in the qutient space is hmemrphic t B + = D 2. Nw cnsider A. There is a hmemrphism h frm S 1 [ 1, 1] t A sending (x, y, t) t (α(t)x, α(t)y, 1 2 t) where α(t) = 9 1 t2 4
10 and by cnstructin h( v) = h(v). The image f A in the qutient space is thus the qutient f S 1 [ 1, 1] mdul the equivalence relatin u v u = ±v. This qutient space is in turn hmemrphic t the qutient space f the upper semicircular arc S+ 1 (all pints with nnnegative y-crdinate) mdul the equivalence relatin generated by setting ( 1, 0, t) equivalent t (1, 0, t), which yields the Möbius strip. The intersectin f this subset in the qutient with the image f B is just the image f the clsed curve n the edge f B +, which als represents the edge curve n the Möbius strip. FURTHER DETAILS. We shall fill in sme f the reasns that were left unstated in the sketch given abve. Let A S 2 be the set f all pints (x, y, z) S 2 such that z 1, and let B be the set f all 2 pints where z 1. If T (x) = x, then T [A] = A and T (B) = B 2 [etc.] This is true because if T (v) = w, then the third crdinates f bth pints have the same abslute values and f curse they satisfy the same inequality relatin with respect t 1 2. By cnstructin B is a disjint unin f tw pieces B ± cnsisting f all pints where sign(z) = ± 1, This is true the third crdinates f all pints in B are nnzer. There is a hmemrphism h frm S 1 [ 1, 1] t A sending (x, y, t) t (α(t)x, α(t)y, 1 2t) where α(t)s = 1 t2 4 One needs t verify that h is 1 1 nt; this is essentially an exercise in algebra. Since we are dealing with cmpact Hausdrff spaces, cntinuus mappings that are 1 1 nt are autmatically hmemrphisms. This qutient space [S 1 [ 1, 1] mdul the equivalence relatin u v u = ±v] is in turn hmemrphic t the qutient space f the upper semicircular arc S+ 1 (all pints with nnnegative y-crdinate) mdul the equivalence relatin generated by setting ( 1, 0, t) equivalent t (1, 0, t), which yields the Möbius strip. Let A and B be the respective equivalence relatins n S+ 1 [ 1, 1] and S 1 [ 1, 1], and let A and B be the respective qutient spaces. By cnstructin the inclusin S+ 1 [ 1, 1] S 1 [ 1, 1] passes t a cntinuus map f qutients, and it is necessary and sufficient t check that this map is 1 1 and nt. This is similar t a previus exercise. Pints in S 1 S+ 1 all have negative secnd crdinates and are equivalent t unique pints with psitive secnd crdinates. This implies that the mapping frm A t B is 1 1 and nt at all pints except perhaps thse whse secnd crdinates are zer. Fr such pints the equivalence relatins given by A and B are identical, and therefre the mapping frm A t B is als 1 1 and nt at all remaining pints. 4. Suppse that the tplgical space X is a unin f tw clsed subspaces A and B, let C = A B, let h : C C be a hmemrphism, and let A h B be the space frmed frm A B by identifying x C A with h(x) C B. Prve that A h B is hmemrphic t X if h extends t a hmemrphism H : A A, and give an example fr which X is nt hmemrphic t A h B. [Hint: Cnstruct the hmemrphism using H in the first case, and cnsider als the case where X = S 1 S 1, with A ± == S 1 ± S 1 ±; then C = {± 1} {1, 2}, and there is a hmemrphism frm h t itself such that A + h A is cnnected.] 10
11 We can and shall view X as A id B. Cnsider the map F 0 : A B A B defined by H 1 n A and the identity n B. We claim that this passes t a unique cntinuus map f qutients frm X t A h B; i.e., the map F 0 sends each nnatmic equivalence classes { (c, 1), (c, 2) } fr X = A id B t a nnatmic equivalence class f the frm { (u, 1), (h(u), 2) } fr A h B. Since F 0 sends (c, 1) t (h 1 (c), 1) and (c, 2) t itself, we can verify the cmpatibility f F 0 with the equivalence relatins by taking u = h 1 (c). Passage t the qutients then yields the desired map F : X A h B. T shw this map is a hmemrphism, it suffices t define Specifically, start with G 0 = F 1 0, s that G 0 = H n A and the identity n B. In this case it is necessary t shw that a nnatmic equivalence class f the frm { (u, 1), (h(u), 2) } fr A h B gets sent t a nnatmic equivalence class f the frm { (c, 1), (c, 2) } fr X = A id B. Since G 0 maps the first set t { (h(u), 1), (h(u), 2) } this is indeed the case, and therefre G 0 als passes t a map f qutients which we shall call G. Finally we need t verify that F and G are inverses t each ther. By cnstructin the maps F 0 and G 0 satisfy F ([y]) = [F 0 (y)] and G([z]) = [G 0 (z)], where square brackets dente equivalence classes. Therefre we have G F ([y]) = G ([F 0 (y)]) = [G 0 (F 0 (y))] which is equal t [y] because F 0 and G 0 are inverse t each ther. Therefre G F is the identity n X. A similar argument shws that F G is the identity n A h B. T cnstruct the example where X is nt hmemrphic t A h B, we fllw the hint and try t find a hmemrphism f the fur pint space {± 1} {1, 2} t itself such that X is nt hmemrphic t A h B is cnnected; this suffices because we knw that X is nt cnnected. Sketches n paper r physical experimentatin with wires r string are helpful in finding the right frmula. Specifically, the hmemrphism we want is given as fllws: ( 1, 1) A + (1, 2) A (1, 1) A + (1, 1) A (1, 2) A + ( 1, 1) A ( 1, 2) A + ( 1, 2) A The first f these implies that the images f S+ 1 {2} and S1 {1} lie in the same cmpnent f the qutient space, the secnd f these implies that the images f S 1 {1} and S1 + {1} bth lie in the same cmpnent, and the third f these implies that the images f S+ 1 {2} and S1 {2} als lie in the same cmpnent. Since the entire space is the unin f the images f the cnnected subsets S± 1 {1} and S± 1 {2} it fllws that A h B is cnnected. FOOTNOTE. The argument in the first part f the exercise remains valid if A and B are pen rather than clsed subsets. 5. One-pint unins. One cnceptual prblem with the disjint unin f tplgical spaces is that it is never cnnected except fr the trivial case f ne summand. In many gemetrical and tplgical cntexts it is extremely useful t cnstruct a mdified versin f disjint unins that is cnnected if all the pieces are. Usually sme additinal structure is needed in rder t make such cnstructins. In this exercise we shall describe such a cnstructin fr bjects knwn as pinted spaces that are indispensable fr many purpses (e.g., the definitin f fundamental grups as in Munkres). A pinted 11
12 space is a pair (X, x) cnsisting f a tplgical space X and a pint x X; we ften call x the base pint, and unless stated therwise the ne pint subset cnsisting f the base pint is assumed t be clsed. If (Y, y) is anther pinted space and f : X Y is cntinuus, we shall say that f is a base pint preserving cntinuus map frm (X, x) t (Y, y) if f(x) = y, In this case we shall ften write f : (X, x) (Y, y). Identity maps are base pint preserving, and cmpsites f base pint preserving maps are als base pint preserving. Given a finite cllectin f pinted spaces (X i, x i ), define an equivalence relatin n i X i whse equivalence classes cnsist f j {x j} and all ne pint sets y such that y j {x j}. Define the ne pint unin r wedge n (X j, x j ) = (X 1, x 1 ) (X n, x n ) i=1 t be the qutient space f this equivalence relatin with the qutient tplgy. The base pint f this space is taken t be the class f j {x j}. (a) Prve that the wedge is a unin f clsed subspaces Y j such that each Y j is hmemrphic t X j and if j k then Y j Y k is the base pint. Explain why k (X k, x k ) is Hausdrff if and nly if each X j is Hausdrff, why k (X k, x k ) is cmpact if and nly if each X j is cmpact, and why k (X k, x k ) is cnnected if and nly if each X j is cnnected (and the same hlds fr arcwise cnnectedness). Fr each j let in j : X j k X k be the standard injectin int the disjint unin, and let P : k X k k (X k, x k ) be the qutient map defining the wedge. Define Y j t be P in j [ X j ]. By cnstructin the map P in j is cntinuus and 1 1; we claim it als sends clsed subsets f X j t clsed subsets f the wedge. Suppse that F X j is clsed; then P in j [F ] is clsed in the wedge if and nly if its inverse image under P is clsed. But this inverse image is the unin f the clsed subsets in j [F ] and k {x k} (which is a finite unin f ne pint subsets that are assumed t be clsed). It fllws that Y j is hmemrphic t X + j. The cnditin n Y k Y l fr k l is an immediate cnsequence f the cnstructin. The assertin that the wedge is Hausdrff if and nly if each summand is fllws because a subspace f a Hausdrff space is Hausdrff, and a finite unin f clsed Hausdrff subspaces is always Hausdrff (by a previus exercise). T verify the assertins abut cmpactness, nte first that fr each j there is a cntinuus cllapsing map q j frm k (X k, x k ) t (X j, x j ), defined by the identity n the image f (X j, x j ) and by sending everything t the base pint n every ther summand. If the whle wedge is cmpact, then its cntinuus under q j, which is the image f X j, must als be cmpact. Cnversely if the sets X j are cmpact fr all j, then the (finite!) unin f their images, which is the entire wedge, must be cmpact. T verify the assertins abut cnnectedness, nte first that fr each j there is a cntinuus cllapsing map q j frm k (X k, x k ) t (X j, x j ), defined by the identity n the image f (X j, x j ) and by sending everything t the base pint n every ther summand. If the whle wedge is cnnected, then its cntinuus under q j, which is the image f X j, must als be cnnected. Cnversely if the sets X j are cnnected fr all j, then the unin f their images, which is the entire wedge, must be cnnected because all these images cntain the base pint. Similar statements hld fr arcwise cnnectedness and fllw by inserting arcwise in frnt f cnnected at every step f the argument. 12
13 (b) Let ϕ j : (X j, x j ) k (X k, x k ) be the cmpsite f the injectin X j k X k with the qutient prjectin; by cnstructin ϕ j is base pint preserving. Suppse that (Y, y) is sme arbitrary pinted space and we are given a sequence f base pint preserving cntinuus maps F j : (X j, x j ) (Y, y). Prve that there is a unique base pint preserving cntinuus mapping such that F ϕ j = F j fr all j. F : k (X k, x k ) (Y, y) T prve existence, first bserve that there is a unique cntinuus map F : k X k Y such that in j F = Fj fr all j. This passes t a unique cntinuus map F n the qutient space k (X k, x k ) because F is cnstant n the equivalence classes assciated t the qutient prjectin P. This cnstructs the map we want; uniqueness fllws because the cnditins prescribe the definitin at every pint f the wedge. (c) In the infinite case ne can carry ut the set-theretic cnstructin as abve but sme care is needed in defining the tplgy. Shw that if each X j is Hausdrff and ne takes the s-called weak tplgy whse clsed subsets are generated by the family f subsets ϕ j [F ] where F is clsed in X j fr sme j, then [1] a functin h frm the wedge int sme ther space Y is cntinuus if and nly if each cmpsite h ϕ j is cntinuus, [2] the existence and uniqueness therem fr mappings frm the wedge (in the previus prtin f the exercise) generalizes t infinite wedges with the s-called weak tplgies. Strictly speaking, ne shuld verify that the s-called weak tplgy is indeed a tplgy n the wedge. We shall leave this t the reader. T prve [1], nte that ( = ) is trivial. Fr the reverse directin, we need t shw that if E is clsed in Y then h 1 [E] is clsed with respect t the s-called weak tplgy we have defined. The subset in questin is clsed with respect t this tplgy if and nly if h 1 [E] ϕ[x j ] is clsed in ϕ[x j ] fr all j, and since ϕ j maps its dmain hmemrphically nt its image, the latter is true if and nly if ϕ 1 h 1 [E] is clsed in X j fr all j. But these cnditins hld because each f the maps ϕ j h is cntinuus. T prve [2], nte first that there is a unique set-theretic map, and then use [1] t cnclude that it is cntinuus. (d) Suppse that we are given an infinite wedge such that each summand is Hausdrff and cntains at least tw pints. Prve that the wedge with the s-called weak tplgy is nt cmpact. Fr each j let y j X j be a pint ther than x j, and cnsider the set E f all pints y j. This is a clsed subset f the wedge because its intersectin with each set ϕ[x j ] is a ne pint subset and hence clsed. In fact, every subset f E is als clsed by a similar argument (the intersectins with the summands are either empty r cntain nly ne pint), s E is a discrete clsed subset f the wedge. Cmpact spaces d nt have infinite discrete clsed subspaces, and therefre it fllws that the infinite wedge with the weak tplgy is nt cmpact. Remark. If each f the summands in (d) is cmpact Hausdrff, then there is a natural candidate fr a strng tplgy n a cuntably infinite wedge which makes the latter int a cmpact Hausdrff space. In sme cases this tplgy can be viewed mre gemetrically; fr example, if each (X j, x j ) 13
14 is equal t (S 1, 1) and there are cuntably infinitely many f them, then the space ne btains is the Hawaiian earring in R 2 given by the unin f the circles defined by the equatins ( x 1 ) 2 2 k + + y 2 = 1 2 2k. As usual, drawing a picture may be helpful. The k th circle has center (1/2 k, 0) and passes thrugh the rigin; the y-axis is the tangent line t each circle at the rigin. SKETCHES OF VERIFICATIONS OF ASSERTIONS. If we are given an infinite sequence f cmpact Hausdrff pinted spaces { (X n, x n ) } we can put a cmpact Hausdrff tplgy n their wedge as fllws. Let W k be the wedge f the first k spaces; then fr each k there is a cntinuus map q k : n (X n, x n ) W k (with the s-called weak tplgy n the wedge) that is the identity n the first k summands and cllapses the remaining nes t the base pint. These maps are in turn define a cntinuus functin q : n (X n, x n ) k W k whse prjectin nt W k is q k. This mapping is cntinuus and 1 1; if its image is clsed in the (cmpact!) prduct tplgy, then this defines a cmpact Hausdrff tplgy n the infinite wedge n (X n, x n ). Here is ne way f verifying that the image is clsed. Fr each k let c k : W k W k 1 be the map that is the identity n the first (k 1) summands and cllapses the last ne t a pint. Then we may define a cntinuus map C n k 1 W k by first prjecting nt the prduct k 2 W k (frget the first factr) and then frming the map k 2 W k. The image f q turns ut t be the set f all pints x in the prduct such that C(x) = x. Since the prduct is Hausdrff the image set is clsed in the prduct and thus cmpact. A cmment abut the cmpactness f the Hawaiian earring E might be useful. Let F k be the unin f the circles f radius 2 j that are cntained in E, where j k, tgether with the clsed disk bunded by the circle f radius 2 (k+1) in E. Then F k is certainly clsed and cmpact. Since E is the intersectin f all the sets F k it fllws that E is als clsed and cmpact. 14
SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 4. Function spaces
SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 4 Fall 2014 IV. Functin spaces IV.1 : General prperties Additinal exercises 1. The mapping q is 1 1 because q(f) = q(g) implies that fr all x we have f(x)
More informationHomology groups of disks with holes
Hmlgy grups f disks with hles THEOREM. Let p 1,, p k } be a sequence f distinct pints in the interir unit disk D n where n 2, and suppse that fr all j the sets E j Int D n are clsed, pairwise disjint subdisks.
More informationBuilding to Transformations on Coordinate Axis Grade 5: Geometry Graph points on the coordinate plane to solve real-world and mathematical problems.
Building t Transfrmatins n Crdinate Axis Grade 5: Gemetry Graph pints n the crdinate plane t slve real-wrld and mathematical prblems. 5.G.1. Use a pair f perpendicular number lines, called axes, t define
More informationMATHEMATICS SYLLABUS SECONDARY 5th YEAR
Eurpean Schls Office f the Secretary-General Pedaggical Develpment Unit Ref. : 011-01-D-8-en- Orig. : EN MATHEMATICS SYLLABUS SECONDARY 5th YEAR 6 perid/week curse APPROVED BY THE JOINT TEACHING COMMITTEE
More informationOn Topological Structures and. Fuzzy Sets
L - ZHR UNIVERSIT - GZ DENSHIP OF GRDUTE STUDIES & SCIENTIFIC RESERCH On Tplgical Structures and Fuzzy Sets y Nashaat hmed Saleem Raab Supervised by Dr. Mhammed Jamal Iqelan Thesis Submitted in Partial
More informationLyapunov Stability Stability of Equilibrium Points
Lyapunv Stability Stability f Equilibrium Pints 1. Stability f Equilibrium Pints - Definitins In this sectin we cnsider n-th rder nnlinear time varying cntinuus time (C) systems f the frm x = f ( t, x),
More informationChapter Summary. Mathematical Induction Strong Induction Recursive Definitions Structural Induction Recursive Algorithms
Chapter 5 1 Chapter Summary Mathematical Inductin Strng Inductin Recursive Definitins Structural Inductin Recursive Algrithms Sectin 5.1 3 Sectin Summary Mathematical Inductin Examples f Prf by Mathematical
More informationMODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b
. REVIEW OF SOME BASIC ALGEBRA MODULE () Slving Equatins Yu shuld be able t slve fr x: a + b = c a d + e x + c and get x = e(ba +) b(c a) d(ba +) c Cmmn mistakes and strategies:. a b + c a b + a c, but
More informationChapter 9 Vector Differential Calculus, Grad, Div, Curl
Chapter 9 Vectr Differential Calculus, Grad, Div, Curl 9.1 Vectrs in 2-Space and 3-Space 9.2 Inner Prduct (Dt Prduct) 9.3 Vectr Prduct (Crss Prduct, Outer Prduct) 9.4 Vectr and Scalar Functins and Fields
More informationENGI 4430 Parametric Vector Functions Page 2-01
ENGI 4430 Parametric Vectr Functins Page -01. Parametric Vectr Functins (cntinued) Any nn-zer vectr r can be decmpsed int its magnitude r and its directin: r rrˆ, where r r 0 Tangent Vectr: dx dy dz dr
More informationChapter 2 GAUSS LAW Recommended Problems:
Chapter GAUSS LAW Recmmended Prblems: 1,4,5,6,7,9,11,13,15,18,19,1,7,9,31,35,37,39,41,43,45,47,49,51,55,57,61,6,69. LCTRIC FLUX lectric flux is a measure f the number f electric filed lines penetrating
More informationTHE FINITENESS OF THE MAPPING CLASS GROUP FOR ATOROIDAL 3-MANIFOLDS WITH GENUINE LAMINATIONS
j. differential gemetry 50 (1998) 123-127 THE FINITENESS OF THE MAPPING CLASS GROUP FOR ATOROIDAL 3-MANIFOLDS WITH GENUINE LAMINATIONS DAVID GABAI & WILLIAM H. KAZEZ Essential laminatins were intrduced
More informationFINITE BOOLEAN ALGEBRA. 1. Deconstructing Boolean algebras with atoms. Let B = <B,,,,,0,1> be a Boolean algebra and c B.
FINITE BOOLEAN ALGEBRA 1. Decnstructing Blean algebras with atms. Let B = be a Blean algebra and c B. The ideal generated by c, (c], is: (c] = {b B: b c} The filter generated by c, [c), is:
More informationDifferentiation Applications 1: Related Rates
Differentiatin Applicatins 1: Related Rates 151 Differentiatin Applicatins 1: Related Rates Mdel 1: Sliding Ladder 10 ladder y 10 ladder 10 ladder A 10 ft ladder is leaning against a wall when the bttm
More informationAdmissibility Conditions and Asymptotic Behavior of Strongly Regular Graphs
Admissibility Cnditins and Asympttic Behavir f Strngly Regular Graphs VASCO MOÇO MANO Department f Mathematics University f Prt Oprt PORTUGAL vascmcman@gmailcm LUÍS ANTÓNIO DE ALMEIDA VIEIRA Department
More informationSupport-Vector Machines
Supprt-Vectr Machines Intrductin Supprt vectr machine is a linear machine with sme very nice prperties. Haykin chapter 6. See Alpaydin chapter 13 fr similar cntent. Nte: Part f this lecture drew material
More informationA crash course in Galois theory
A crash curse in Galis thery First versin 0.1 14. september 2013 klkken 14:50 In these ntes K dentes a field. Embeddings Assume that is a field and that : K! and embedding. If K L is an extensin, we say
More informationTHE DEVELOPMENT OF AND GAPS IN THE THEORY OF PRODUCTS OF INITIALLY M-COMPACT SPACES
Vlume 6, 1981 Pages 99 113 http://tplgy.auburn.edu/tp/ THE DEVELOPMENT OF AND GAPS IN THE THEORY OF PRODUCTS OF INITIALLY M-COMPACT SPACES by R. M. Stephensn, Jr. Tplgy Prceedings Web: http://tplgy.auburn.edu/tp/
More informationEquilibrium of Stress
Equilibrium f Stress Cnsider tw perpendicular planes passing thrugh a pint p. The stress cmpnents acting n these planes are as shwn in ig. 3.4.1a. These stresses are usuall shwn tgether acting n a small
More information5 th grade Common Core Standards
5 th grade Cmmn Cre Standards In Grade 5, instructinal time shuld fcus n three critical areas: (1) develping fluency with additin and subtractin f fractins, and develping understanding f the multiplicatin
More informationSection 6-2: Simplex Method: Maximization with Problem Constraints of the Form ~
Sectin 6-2: Simplex Methd: Maximizatin with Prblem Cnstraints f the Frm ~ Nte: This methd was develped by Gerge B. Dantzig in 1947 while n assignment t the U.S. Department f the Air Frce. Definitin: Standard
More informationA proposition is a statement that can be either true (T) or false (F), (but not both).
400 lecture nte #1 [Ch 2, 3] Lgic and Prfs 1.1 Prpsitins (Prpsitinal Lgic) A prpsitin is a statement that can be either true (T) r false (F), (but nt bth). "The earth is flat." -- F "March has 31 days."
More informationSEMILATTICE STRUCTURES ON DENDRITIC SPACES
Vlume 2, 1977 Pages 243 260 http://tplgy.auburn.edu/tp/ SEMILATTICE STRUCTURES ON DENDRITIC SPACES by T. B. Muenzenberger and R. E. Smithsn Tplgy Prceedings Web: http://tplgy.auburn.edu/tp/ Mail: Tplgy
More information[COLLEGE ALGEBRA EXAM I REVIEW TOPICS] ( u s e t h i s t o m a k e s u r e y o u a r e r e a d y )
(Abut the final) [COLLEGE ALGEBRA EXAM I REVIEW TOPICS] ( u s e t h i s t m a k e s u r e y u a r e r e a d y ) The department writes the final exam s I dn't really knw what's n it and I can't very well
More informationChapter 3 Kinematics in Two Dimensions; Vectors
Chapter 3 Kinematics in Tw Dimensins; Vectrs Vectrs and Scalars Additin f Vectrs Graphical Methds (One and Tw- Dimensin) Multiplicatin f a Vectr b a Scalar Subtractin f Vectrs Graphical Methds Adding Vectrs
More informationDepartment of Economics, University of California, Davis Ecn 200C Micro Theory Professor Giacomo Bonanno. Insurance Markets
Department f Ecnmics, University f alifrnia, Davis Ecn 200 Micr Thery Prfessr Giacm Bnann Insurance Markets nsider an individual wh has an initial wealth f. ith sme prbability p he faces a lss f x (0
More informationRevisiting the Socrates Example
Sectin 1.6 Sectin Summary Valid Arguments Inference Rules fr Prpsitinal Lgic Using Rules f Inference t Build Arguments Rules f Inference fr Quantified Statements Building Arguments fr Quantified Statements
More informationA new Type of Fuzzy Functions in Fuzzy Topological Spaces
IOSR Jurnal f Mathematics (IOSR-JM e-issn: 78-578, p-issn: 39-765X Vlume, Issue 5 Ver I (Sep - Oct06, PP 8-4 wwwisrjurnalsrg A new Type f Fuzzy Functins in Fuzzy Tplgical Spaces Assist Prf Dr Munir Abdul
More informationPattern Recognition 2014 Support Vector Machines
Pattern Recgnitin 2014 Supprt Vectr Machines Ad Feelders Universiteit Utrecht Ad Feelders ( Universiteit Utrecht ) Pattern Recgnitin 1 / 55 Overview 1 Separable Case 2 Kernel Functins 3 Allwing Errrs (Sft
More informationREPRESENTATIONS OF sp(2n; C ) SVATOPLUK KR YSL. Abstract. In this paper we have shown how a tensor product of an innite dimensional
ON A DISTINGUISHED CLASS OF INFINITE DIMENSIONAL REPRESENTATIONS OF sp(2n; C ) SVATOPLUK KR YSL Abstract. In this paper we have shwn hw a tensr prduct f an innite dimensinal representatin within a certain
More information37 Maxwell s Equations
37 Maxwell s quatins In this chapter, the plan is t summarize much f what we knw abut electricity and magnetism in a manner similar t the way in which James Clerk Maxwell summarized what was knwn abut
More informationWe can see from the graph above that the intersection is, i.e., [ ).
MTH 111 Cllege Algebra Lecture Ntes July 2, 2014 Functin Arithmetic: With nt t much difficulty, we ntice that inputs f functins are numbers, and utputs f functins are numbers. S whatever we can d with
More informationA solution of certain Diophantine problems
A slutin f certain Diphantine prblems Authr L. Euler* E7 Nvi Cmmentarii academiae scientiarum Petrplitanae 0, 1776, pp. 8-58 Opera Omnia: Series 1, Vlume 3, pp. 05-17 Reprinted in Cmmentat. arithm. 1,
More information, which yields. where z1. and z2
The Gaussian r Nrmal PDF, Page 1 The Gaussian r Nrmal Prbability Density Functin Authr: Jhn M Cimbala, Penn State University Latest revisin: 11 September 13 The Gaussian r Nrmal Prbability Density Functin
More informationLim f (x) e. Find the largest possible domain and its discontinuity points. Why is it discontinuous at those points (if any)?
THESE ARE SAMPLE QUESTIONS FOR EACH OF THE STUDENT LEARNING OUTCOMES (SLO) SET FOR THIS COURSE. SLO 1: Understand and use the cncept f the limit f a functin i. Use prperties f limits and ther techniques,
More informationCOVERS OF DEHN FILLINGS ON ONCE-PUNCTURED TORUS BUNDLES
prceedings f the american mathematical sciety Vlume 105, Number 3, March 1989 COVERS OF DEHN FILLINGS ON ONCE-PUNCTURED TORUS BUNDLES MARK D. BAKER (Cmmunicated by Frederick R. Chen) Abstract. Let M be
More informationMath 105: Review for Exam I - Solutions
1. Let f(x) = 3 + x + 5. Math 105: Review fr Exam I - Slutins (a) What is the natural dmain f f? [ 5, ), which means all reals greater than r equal t 5 (b) What is the range f f? [3, ), which means all
More informationMath 302 Learning Objectives
Multivariable Calculus (Part I) 13.1 Vectrs in Three-Dimensinal Space Math 302 Learning Objectives Plt pints in three-dimensinal space. Find the distance between tw pints in three-dimensinal space. Write
More informationThermodynamics Partial Outline of Topics
Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)
More informationSections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic.
Tpic : AC Fundamentals, Sinusidal Wavefrm, and Phasrs Sectins 5. t 5., 6. and 6. f the textbk (Rbbins-Miller) cver the materials required fr this tpic.. Wavefrms in electrical systems are current r vltage
More informationAn Introduction to Complex Numbers - A Complex Solution to a Simple Problem ( If i didn t exist, it would be necessary invent me.
An Intrductin t Cmple Numbers - A Cmple Slutin t a Simple Prblem ( If i didn t eist, it wuld be necessary invent me. ) Our Prblem. The rules fr multiplying real numbers tell us that the prduct f tw negative
More information1996 Engineering Systems Design and Analysis Conference, Montpellier, France, July 1-4, 1996, Vol. 7, pp
THE POWER AND LIMIT OF NEURAL NETWORKS T. Y. Lin Department f Mathematics and Cmputer Science San Jse State University San Jse, Califrnia 959-003 tylin@cs.ssu.edu and Bereley Initiative in Sft Cmputing*
More information45 K. M. Dyaknv Garsia nrm kfk G = sup zd jfj d z ;jf(z)j = dened riginally fr f L (T m), is in fact an equivalent nrm n BMO. We shall als be cncerned
Revista Matematica Iberamericana Vl. 5, N. 3, 999 Abslute values f BMOA functins Knstantin M. Dyaknv Abstract. The paper cntains a cmplete characterizatin f the mduli f BMOA functins. These are described
More informationPhysics 212. Lecture 12. Today's Concept: Magnetic Force on moving charges. Physics 212 Lecture 12, Slide 1
Physics 1 Lecture 1 Tday's Cncept: Magnetic Frce n mving charges F qv Physics 1 Lecture 1, Slide 1 Music Wh is the Artist? A) The Meters ) The Neville rthers C) Trmbne Shrty D) Michael Franti E) Radiatrs
More informationCambridge Assessment International Education Cambridge Ordinary Level. Published
Cambridge Assessment Internatinal Educatin Cambridge Ordinary Level ADDITIONAL MATHEMATICS 4037/1 Paper 1 Octber/Nvember 017 MARK SCHEME Maximum Mark: 80 Published This mark scheme is published as an aid
More informationCONSTRUCTING STATECHART DIAGRAMS
CONSTRUCTING STATECHART DIAGRAMS The fllwing checklist shws the necessary steps fr cnstructing the statechart diagrams f a class. Subsequently, we will explain the individual steps further. Checklist 4.6
More informationIntroduction to Spacetime Geometry
Intrductin t Spacetime Gemetry Let s start with a review f a basic feature f Euclidean gemetry, the Pythagrean therem. In a twdimensinal crdinate system we can relate the length f a line segment t the
More informationmaking triangle (ie same reference angle) ). This is a standard form that will allow us all to have the X= y=
Intrductin t Vectrs I 21 Intrductin t Vectrs I 22 I. Determine the hrizntal and vertical cmpnents f the resultant vectr by cunting n the grid. X= y= J. Draw a mangle with hrizntal and vertical cmpnents
More informationInterference is when two (or more) sets of waves meet and combine to produce a new pattern.
Interference Interference is when tw (r mre) sets f waves meet and cmbine t prduce a new pattern. This pattern can vary depending n the riginal wave directin, wavelength, amplitude, etc. The tw mst extreme
More information22.54 Neutron Interactions and Applications (Spring 2004) Chapter 11 (3/11/04) Neutron Diffusion
.54 Neutrn Interactins and Applicatins (Spring 004) Chapter (3//04) Neutrn Diffusin References -- J. R. Lamarsh, Intrductin t Nuclear Reactr Thery (Addisn-Wesley, Reading, 966) T study neutrn diffusin
More informationSource Coding and Compression
Surce Cding and Cmpressin Heik Schwarz Cntact: Dr.-Ing. Heik Schwarz heik.schwarz@hhi.fraunhfer.de Heik Schwarz Surce Cding and Cmpressin September 22, 2013 1 / 60 PartI: Surce Cding Fundamentals Heik
More informationModule 4: General Formulation of Electric Circuit Theory
Mdule 4: General Frmulatin f Electric Circuit Thery 4. General Frmulatin f Electric Circuit Thery All electrmagnetic phenmena are described at a fundamental level by Maxwell's equatins and the assciated
More informationCOMP 551 Applied Machine Learning Lecture 11: Support Vector Machines
COMP 551 Applied Machine Learning Lecture 11: Supprt Vectr Machines Instructr: (jpineau@cs.mcgill.ca) Class web page: www.cs.mcgill.ca/~jpineau/cmp551 Unless therwise nted, all material psted fr this curse
More informationCHAPTER 3 INEQUALITIES. Copyright -The Institute of Chartered Accountants of India
CHAPTER 3 INEQUALITIES Cpyright -The Institute f Chartered Accuntants f India INEQUALITIES LEARNING OBJECTIVES One f the widely used decisin making prblems, nwadays, is t decide n the ptimal mix f scarce
More informationPreparation work for A2 Mathematics [2017]
Preparatin wrk fr A2 Mathematics [2017] The wrk studied in Y12 after the return frm study leave is frm the Cre 3 mdule f the A2 Mathematics curse. This wrk will nly be reviewed during Year 13, it will
More informationGAUSS' LAW E. A. surface
Prf. Dr. I. M. A. Nasser GAUSS' LAW 08.11.017 GAUSS' LAW Intrductin: The electric field f a given charge distributin can in principle be calculated using Culmb's law. The examples discussed in electric
More informationHigher Mathematics Booklet CONTENTS
Higher Mathematics Bklet CONTENTS Frmula List Item Pages The Straight Line Hmewrk The Straight Line Hmewrk Functins Hmewrk 3 Functins Hmewrk 4 Recurrence Relatins Hmewrk 5 Differentiatin Hmewrk 6 Differentiatin
More informationAIP Logic Chapter 4 Notes
AIP Lgic Chapter 4 Ntes Sectin 4.1 Sectin 4.2 Sectin 4.3 Sectin 4.4 Sectin 4.5 Sectin 4.6 Sectin 4.7 4.1 The Cmpnents f Categrical Prpsitins There are fur types f categrical prpsitins. Prpsitin Letter
More informationMAKING DOUGHNUTS OF COHEN REALS
MAKING DUGHNUTS F CHEN REALS Lrenz Halbeisen Department f Pure Mathematics Queen s University Belfast Belfast BT7 1NN, Nrthern Ireland Email: halbeis@qub.ac.uk Abstract Fr a b ω with b \ a infinite, the
More informationPhys102 Second Major-102 Zero Version Coordinator: Al-Shukri Thursday, May 05, 2011 Page: 1
Crdinatr: Al-Shukri Thursday, May 05, 2011 Page: 1 1. Particles A and B are electrically neutral and are separated by 5.0 μm. If 5.0 x 10 6 electrns are transferred frm particle A t particle B, the magnitude
More informationPlan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations
STAPLE Physics 201 Name Final Exam May 14, 2013 This is a clsed bk examinatin but during the exam yu may refer t a 5 x7 nte card with wrds f wisdm yu have written n it. There is extra scratch paper available.
More informationThermodynamics and Equilibrium
Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,
More informationThe Equation αsin x+ βcos family of Heron Cyclic Quadrilaterals
The Equatin sin x+ βcs x= γ and a family f Hern Cyclic Quadrilaterals Knstantine Zelatr Department Of Mathematics Cllege Of Arts And Sciences Mail Stp 94 University Of Tled Tled,OH 43606-3390 U.S.A. Intrductin
More informationNUMBERS, MATHEMATICS AND EQUATIONS
AUSTRALIAN CURRICULUM PHYSICS GETTING STARTED WITH PHYSICS NUMBERS, MATHEMATICS AND EQUATIONS An integral part t the understanding f ur physical wrld is the use f mathematical mdels which can be used t
More informationComputational modeling techniques
Cmputatinal mdeling techniques Lecture 2: Mdeling change. In Petre Department f IT, Åb Akademi http://users.ab.fi/ipetre/cmpmd/ Cntent f the lecture Basic paradigm f mdeling change Examples Linear dynamical
More informationMODULE FOUR. This module addresses functions. SC Academic Elementary Algebra Standards:
MODULE FOUR This mdule addresses functins SC Academic Standards: EA-3.1 Classify a relatinship as being either a functin r nt a functin when given data as a table, set f rdered pairs, r graph. EA-3.2 Use
More informationUNIT 1 COPLANAR AND NON-COPLANAR FORCES
UNIT 1 COPLANA AND NON-COPLANA FOCES Cplanar and Nn-Cplanar Frces Structure 1.1 Intrductin Objectives 1. System f Frces 1.3 Cplanar Frce 1.3.1 Law f Parallelgram f Frces 1.3. Law f Plygn f Frces 1.3.3
More informationI. Analytical Potential and Field of a Uniform Rod. V E d. The definition of electric potential difference is
Length L>>a,b,c Phys 232 Lab 4 Ch 17 Electric Ptential Difference Materials: whitebards & pens, cmputers with VPythn, pwer supply & cables, multimeter, crkbard, thumbtacks, individual prbes and jined prbes,
More informationEXERCISES FOR MATHEMATICS 205A
EXERCISES FOR MATHEMATICS 205A FALL 2008 The references denote sections of the text for the course: J. R. Munkres, Topology (Second Edition), Prentice-Hall, Saddle River NJ, 2000. ISBN: 0 13 181629 2.
More informationCHAPTER 8b Static Equilibrium Units
CHAPTER 8b Static Equilibrium Units The Cnditins fr Equilibrium Slving Statics Prblems Stability and Balance Elasticity; Stress and Strain The Cnditins fr Equilibrium An bject with frces acting n it, but
More informationPreparation work for A2 Mathematics [2018]
Preparatin wrk fr A Mathematics [018] The wrk studied in Y1 will frm the fundatins n which will build upn in Year 13. It will nly be reviewed during Year 13, it will nt be retaught. This is t allw time
More informationKinematic transformation of mechanical behavior Neville Hogan
inematic transfrmatin f mechanical behavir Neville Hgan Generalized crdinates are fundamental If we assume that a linkage may accurately be described as a cllectin f linked rigid bdies, their generalized
More information4th Indian Institute of Astrophysics - PennState Astrostatistics School July, 2013 Vainu Bappu Observatory, Kavalur. Correlation and Regression
4th Indian Institute f Astrphysics - PennState Astrstatistics Schl July, 2013 Vainu Bappu Observatry, Kavalur Crrelatin and Regressin Rahul Ry Indian Statistical Institute, Delhi. Crrelatin Cnsider a tw
More informationMedium Scale Integrated (MSI) devices [Sections 2.9 and 2.10]
EECS 270, Winter 2017, Lecture 3 Page 1 f 6 Medium Scale Integrated (MSI) devices [Sectins 2.9 and 2.10] As we ve seen, it s smetimes nt reasnable t d all the design wrk at the gate-level smetimes we just
More informationCHAPTER 24: INFERENCE IN REGRESSION. Chapter 24: Make inferences about the population from which the sample data came.
MATH 1342 Ch. 24 April 25 and 27, 2013 Page 1 f 5 CHAPTER 24: INFERENCE IN REGRESSION Chapters 4 and 5: Relatinships between tw quantitative variables. Be able t Make a graph (scatterplt) Summarize the
More informationThis section is primarily focused on tools to aid us in finding roots/zeros/ -intercepts of polynomials. Essentially, our focus turns to solving.
Sectin 3.2: Many f yu WILL need t watch the crrespnding vides fr this sectin n MyOpenMath! This sectin is primarily fcused n tls t aid us in finding rts/zers/ -intercepts f plynmials. Essentially, ur fcus
More informationLHS Mathematics Department Honors Pre-Calculus Final Exam 2002 Answers
LHS Mathematics Department Hnrs Pre-alculus Final Eam nswers Part Shrt Prblems The table at the right gives the ppulatin f Massachusetts ver the past several decades Using an epnential mdel, predict the
More informationMATHEMATICS Higher Grade - Paper I
Higher Mathematics - Practice Eaminatin B Please nte the frmat f this practice eaminatin is different frm the current frmat. The paper timings are different and calculatrs can be used thrughut. MATHEMATICS
More information1. What is the difference between complementary and supplementary angles?
Name 1 Date Angles Intrductin t Angles Part 1 Independent Practice 1. What is the difference between cmplementary and supplementary angles? 2. Suppse m TOK = 49. Part A: What is the measure f the angle
More informationComputational modeling techniques
Cmputatinal mdeling techniques Lecture 4: Mdel checing fr ODE mdels In Petre Department f IT, Åb Aademi http://www.users.ab.fi/ipetre/cmpmd/ Cntent Stichimetric matrix Calculating the mass cnservatin relatins
More informationDead-beat controller design
J. Hetthéssy, A. Barta, R. Bars: Dead beat cntrller design Nvember, 4 Dead-beat cntrller design In sampled data cntrl systems the cntrller is realised by an intelligent device, typically by a PLC (Prgrammable
More informationCHAPTER 2 Algebraic Expressions and Fundamental Operations
CHAPTER Algebraic Expressins and Fundamental Operatins OBJECTIVES: 1. Algebraic Expressins. Terms. Degree. Gruping 5. Additin 6. Subtractin 7. Multiplicatin 8. Divisin Algebraic Expressin An algebraic
More informationf t(y)dy f h(x)g(xy) dx fk 4 a. «..
CERTAIN OPERATORS AND FOURIER TRANSFORMS ON L2 RICHARD R. GOLDBERG 1. Intrductin. A well knwn therem f Titchmarsh [2] states that if fel2(0, ) and if g is the Furier csine transfrm f/, then G(x)=x~1Jx0g(y)dy
More informationCONSTRUCTING BUILDINGS AND HARMONIC MAPS
CONSTRUCTING BUILDINGS AND HARMONIC MAPS LUDMIL KATZARKOV, ALEXANDER NOLL, PRANAV PANDIT, AND CARLOS SIMPSON Happy Birthday Maxim! Abstract. In a cntinuatin f ur previus wrk [17], we utline a thery which
More informationMathematics Methods Units 1 and 2
Mathematics Methds Units 1 and 2 Mathematics Methds is an ATAR curse which fcuses n the use f calculus and statistical analysis. The study f calculus prvides a basis fr understanding rates f change in
More informationCS 477/677 Analysis of Algorithms Fall 2007 Dr. George Bebis Course Project Due Date: 11/29/2007
CS 477/677 Analysis f Algrithms Fall 2007 Dr. Gerge Bebis Curse Prject Due Date: 11/29/2007 Part1: Cmparisn f Srting Algrithms (70% f the prject grade) The bjective f the first part f the assignment is
More informationSOLUTIONS SET 1 MATHEMATICS CLASS X
Tp Careers & Yu SOLUTIONS SET MTHEMTICS CLSS X. 84 7 Prime factrs f 84 are, and 7.. Sum f zeres 5 + 4 Prduct f zeres 5 4 0 Required plynmial x ( )x + ( 0) x + x 0. Given equatin is x + y 0 Fr x, y L.H.S
More informationMath Foundations 10 Work Plan
Math Fundatins 10 Wrk Plan Units / Tpics 10.1 Demnstrate understanding f factrs f whle numbers by: Prime factrs Greatest Cmmn Factrs (GCF) Least Cmmn Multiple (LCM) Principal square rt Cube rt Time Frame
More informationCONSTRUCTING BUILDINGS AND HARMONIC MAPS
CONSTRUCTING BUILDINGS AND HARMONIC MAPS LUDMIL KATZARKOV, ALEXANDER NOLL, PRANAV PANDIT, AND CARLOS SIMPSON Happy Birthday Maxim! Abstract. In a cntinuatin f ur previus wrk [21], we utline a thery which
More informationFinite Automata. Human-aware Robo.cs. 2017/08/22 Chapter 1.1 in Sipser
Finite Autmata 2017/08/22 Chapter 1.1 in Sipser 1 Last time Thery f cmputatin Autmata Thery Cmputability Thery Cmplexity Thery Finite autmata Pushdwn autmata Turing machines 2 Outline fr tday Finite autmata
More informationA little noticed right triangle
A little nticed right triangle Knstantine Hermes Zelatr Department f athematics Cllege f Arts and Sciences ail Stp 94 University f Tled Tled, OH 43606-3390 U.S.A. A little nticed right triangle. Intrductin
More informationThe standards are taught in the following sequence.
B L U E V A L L E Y D I S T R I C T C U R R I C U L U M MATHEMATICS Third Grade In grade 3, instructinal time shuld fcus n fur critical areas: (1) develping understanding f multiplicatin and divisin and
More information1 The limitations of Hartree Fock approximation
Chapter: Pst-Hartree Fck Methds - I The limitatins f Hartree Fck apprximatin The n electrn single determinant Hartree Fck wave functin is the variatinal best amng all pssible n electrn single determinants
More informationSmoothing, penalized least squares and splines
Smthing, penalized least squares and splines Duglas Nychka, www.image.ucar.edu/~nychka Lcally weighted averages Penalized least squares smthers Prperties f smthers Splines and Reprducing Kernels The interplatin
More informationNOTE ON THE ANALYSIS OF A RANDOMIZED BLOCK DESIGN. Junjiro Ogawa University of North Carolina
NOTE ON THE ANALYSIS OF A RANDOMIZED BLOCK DESIGN by Junjir Ogawa University f Nrth Carlina This research was supprted by the Office f Naval Research under Cntract N. Nnr-855(06) fr research in prbability
More informationBootstrap Method > # Purpose: understand how bootstrap method works > obs=c(11.96, 5.03, 67.40, 16.07, 31.50, 7.73, 11.10, 22.38) > n=length(obs) >
Btstrap Methd > # Purpse: understand hw btstrap methd wrks > bs=c(11.96, 5.03, 67.40, 16.07, 31.50, 7.73, 11.10, 22.38) > n=length(bs) > mean(bs) [1] 21.64625 > # estimate f lambda > lambda = 1/mean(bs);
More informationCurriculum Development Overview Unit Planning for 8 th Grade Mathematics MA10-GR.8-S.1-GLE.1 MA10-GR.8-S.4-GLE.2
Unit Title It s All Greek t Me Length f Unit 5 weeks Fcusing Lens(es) Cnnectins Standards and Grade Level Expectatins Addressed in this Unit MA10-GR.8-S.1-GLE.1 MA10-GR.8-S.4-GLE.2 Inquiry Questins (Engaging-
More informationPart a: Writing the nodal equations and solving for v o gives the magnitude and phase response: tan ( 0.25 )
+ - Hmewrk 0 Slutin ) In the circuit belw: a. Find the magnitude and phase respnse. b. What kind f filter is it? c. At what frequency is the respnse 0.707 if the generatr has a ltage f? d. What is the
More information1 Course Notes in Introductory Physics Jeffrey Seguritan
Intrductin & Kinematics I Intrductin Quickie Cncepts Units SI is standard system f units used t measure physical quantities. Base units that we use: meter (m) is standard unit f length kilgram (kg) is
More information