Solutions of exercise sheet 3

Transcription

1 Topology D-MATH, FS 2013 Damen Calaque Solutons o exercse sheet 3 1. (a) Let U Ă Y be open. Snce s contnuous, 1 puq s open n X. Then p A q 1 puq 1 puq X A s open n the subspace topology on A. (b) I s contnuous, A, B are contnuous by (a). Conversely, let A, B be contnuous. Assume that A, B both are closed (replace closed by open n the ollowng or the other case). Let C Ă Y be closed. Then, snce A, B are contnuous, p q 1 A pcq, p q 1 B pcq are closed n A, B, respectvely, and snce A, B are closed n X, p q 1 A pcq C X A, p q 1 B pcq C X B are also closed n X. Then, 1 pcq pc X Aq Y pc X Bq s closed n X as the unon o two closed sets. 2. Frst, note that ϕ : C Ñ t0, 1u N, 8ÿ a n 3 n ÞÑ p1 2 a nq s a bjecton between C and the set o t0, 1u-sequences. Now we have to show that ths s both open and contnuous. Observe that the sets B n ppa k qq tpb k q : a k b k, 1 ď k ď nu orm a bass o the product topology on t0, 1u N. To show that ϕ 1 s contnuous, let ε ą 0 and choose an m P N such that 3 m ă ε. Let a pa k q kpn P t0, 1u N. Then or every b pb k q kpn P B m ppa k qq, ϕ 1 pbq ϕ 1 paq ď ÿ 2 3 k 3 m ă ε, so ϕ 1 pb m paqq Ď B ε pϕ 1 paqq. kąm To see that ϕ s contnuous, consder x, y P C such that x y ă 3 r. Then, ϕpxq a pa k q kpn, ϕpyq b pb k q kpn, let N be the smallest ndex such that a N b N. Then, 3 r ą x y ϕ 1 paq ϕ 1 pbq ě 2 3 N ÿ b k a 3 k k so r ă N and thereore b P B r paq. ě kąn 2 3 N 2 ÿ kąn 3 k 3 N, 3. (a) I T 1 s a topology on Y, then all : py, T 1 q Ñ px, T q are contnuous or every, 1 pt q Ď T 1 Ť PI 1 pt q Ď T 1. Snce Ť PI 1 pt q tsel s a subbass, the topology T t generates s the coarsest such that all are contnuous.

2 Let : pz, Vq Ñ py, T q be contnuous. Then all are contnuous as compostons o contnuous maps (exercse 4 (b) on sheet 2). Conversely, let all be contnuous,.e. or every, V Ě p q 1 pt q 1 p 1 pt qq, so 1 p Ť PI 1 pt qq Ď V. Snce Ť PI 1 pt q s a subbass or the topology T, s contnuous. Now assume that T s a topology on Y satsyng ths property. Then consder the map d : py, T q Ñ py, T q. All d : py, T q Ñ px, T q are contnuous, so, snce T satses the property, d s contnuous. By exercse 6 (a) on sheet 2, T Ď T. Conversely, consder the map d : py, T q Ñ py, T q. It s contnuous, so by the property, all d : py, T q Ñ px, T q are contnuous, so agan by exercse 6 (a) on sheet 2, 1 pt q Ď T so thereore, by the constructon o T, T Ď T. So T T. (b) The ntal topology on A wth respect to the map 0 : A ãñ X s generated by the subbass 0 1 pt q, whch s precsely the denton o the subspace topology (c) The ntal topology on X 1 ˆ X 2 wth respect to the projectons 1 : X 1 ˆ X 2 ÝÑ X 1, 2 : X 1 ˆ X 2 ÝÑ X 2 s the topology generated by 1 1 pt 1q Y 1 2 pt 2q and 1 1 pt 1q tu ˆ X 2 : U P T 1 u, 2 1 pt 2q tx 1 ˆ V : V P T 2 u. Thus, the ntal topology s contaned n the product topology. Moreover, U ˆ V pu ˆ X 2 q X px 1 ˆ V q, so actually, the topologes are equal. For an arbtrary amly on topologcal spaces, the ntal topology on ś PI X s generated by ď 1 pt q ď tu ˆ ź X j : U P T u, PI PI jpi,j whch gves the product topology on ś PI X. (d) I the topology on ś 8 X were dscrete, then all sngletons tpx nq 8 u would be open. However, all basc open sets are o the orm ś 8 X n, wth X n Ă X and X n X or all but ntely many n; n partcular, they are nnte (even uncountable) sets, so any unon o basc open sets wll be nnte, and thereore no open set consts o only ntely many ponts (let alone just one). However, we take the box topology, tpx n q 8 u ś 8 tx nu s open because or every n, tx n u s open n X. 4. (a) Snce 1 phq H, 1 py q X, and 1 po 1 X... X O n q 1 po 1 q X... X 1 po n q, 1 ` ď ď O j 1 po j q, jpj jpj T to Ď P I : 1 poq P T u s a topology on Y.

3 By denton, 1 pt q Ď T, so all : px, T q Ñ py, T q are contnuous. Moreover, let T 1 be another topology on Y such that all : px, T q Ñ py, T 1 q are contnuous. Then or every O P T 1, 1 poq P T, so O P T and T 1 Ď T. Thereore T s the nest such topology. Let : py, T q Ñ pz, Vq be contnuous. Then s contnuous as a composton o contnuous maps (exercse 4 (b) on sheet 2). Conversely, let all be contnuous,.e. or every, T Ě p q 1 pvq 1 p 1 pvqq and thus, by denton o T, 1 pvq Ď T, so s contnuous. Now assume that T s a topology on Y satsyng ths property. Then the map d : py, T q Ñ py, T q s contnuous, so all d : px T q Ñ py, T q are contnuous. We saw above that ths mples that T Ď T. Conversely, the maps d : px, T q ÝÑ py, T q ÝÑ d py, T q are contnuous, so by the property or T, d : py, T q Ñ py, T q s contnuous, so by exercse 6 (a) on sheet 2, T Ď T. Thereore T T. (b) The nal topology on X wth respect to the maps d : px, T q Ñ X s T to Ď P I : O P T u č PI T. (c) The nal topology on Y X 1 \ X 2 wth respect to the nclusons 1, 2 s T to Ď X and 2 1 2u to Ď X O X X 1 P T 1 and O X X 2 P T 2 u to 1 \ O 2 Ă X O 1 P T 1, O 2 P T 2 u. Smlary, or arbtrary amles o topologcal spaces px q PI, open sets o š PI X are o the orm š PI O, where O P T. (d) To remember the ndces, we wrte elements as px, jq P š X, where x P X. Then # 1 txu, j ptpx, jquq H j. Snce X s dscrete, all 1 ptpx, jquq are open, so by denton o the nal topology, all sngletons tpx, jqu are open. They orm a bass or the dscrete topology, so X s dscrete. š 5. (a) Recall that the gven topologes have the ollowng respectve bases:! ź 8 ) B prod : U nˇˇˇun Ă R open, U n R or almost all n, and! ź 8 B box : ) U nˇˇˇun Ă R open B u : tpy n q P R N sup y n x n ă ru r ą 0 and px n q P R N(.

4 To see whether a map s contnuous, we have to check the premages o all basc open sets are open n R. The premage o a set ś 8 U n s Ş 8 U n. In B prod, all but ntely many U n are equal to R, so Ş 8 U n Ş m k 1 U n k, whch s open, so the map s contnuous wth respect to O prod. The set ś 8 p 1 n, 1 n q s open n O box, but ts premage s Ş 8 p 1 n, 1 n q t0u, whch s not open, so the map s not contnuous wth respect to O box. The premage o a basc open set U tpy n q P R N sup y n x n ă ru s 1 puq ty P R Dɛ ą 0 : sup y x n r ɛu. Let y P 1 puq and let ɛ r sup y x n. Then or any t P py ɛ, y ` ɛq, we have sup t x n sup t y ` y x n ď t y ` sup y x n ă ɛ ` r ɛ r, hence t P 1 puq. Ths shows that py ɛ, y ` ɛq Ă 1 puq,.e. every pont y P 1 puq has an open neghborhood contaned n 1 puq; so 1 puq s open, and thereore s contnuous wth respect to O u. (b) I pa n q converges to a, then every open neghborhood o a contans all but ntely many elements o pa n q. In partcular, U Ă X s an open neghborhood o a, then the basc open set U ˆ śjpiztu X j contans all but ntely many elements o pa n q, and thereore, U contans all but ntely many elements o `pan q. Hence `pan q converges to a n X. On the other P I : `pa n q converges to a, then or any, any open neghborhood U Ă X o a wll contan all but ntely many elements o `pa n q. Let N U : tn P N pa n q R U u Ă N. Now let U be a basc open neghborhood o a,.e. U ś PI U, wth U Ă X open, and U X or all but ntely many. Note that N U : tn P N a n R Uu ď PI N U. But snce all N U are nte, and N U or all but ntely many, ths unon s actually a nte unon o nte sets, so N U s nte; so U contans all but ntely many elements o pa n q,.e. pa n q converges to a. Makng the topology ner makes t harder or sequences to converge, so a sequence n R N converges wth respect to the unorm or box topology, then t also converges n the product topology and hence t converges n all coordnates (.e. the and only mplcaton also holds or the unorm and box topology). We gve a counterexample or the other mplcaton: For P N, let pa n q : δ n. Clearly, lm nñ8 pa n q 0. However, 0 ă r ă 1, because or all n P N we have sup PN pa n q 1, the open set tpy q P R N sup PN y n ă ru contans no pont o the sequence pa n q. Thereore, pa n q does not converge wth respect to the unorm topology and, snce the box topology s ner than the unorm topology, t does not converge wth respect to the box topology ether. (c) Let B : XzA tpx n q P R N x n 0 or nntely many nu. Let U Ă X be a basc open subset,.e. U ś m k 1 U n k ˆ śn n 1,...,n m R, wth U n Ă R non-empty open. For 1 ď k ď m choose x nk P U nk, and put x n : 0 or n n 1,..., n m. Then px n q 8 P U, and x n 0 or all but ntely many n,.e. px n q 8 R B. Ths shows that there s no basc open set U wth U Ă B, and thereore B. Consequently, A XzB X.

5 (d) Let ś kpn O k be open n the box topology, so all O k are open n R. Then ` ź nź O k ι 1 n kpn k 1 s open n R n by the denton o the standard topology T n on R n (whch s the product topology). So T box Ď T nal. To see that the nal topology s strctly ner than the box topology, we nd two examples o sets whch are open n the nal topology, but not n the box topology: () (Example gven by Danel Robert-Ncoud) Consder the set O k O tpt, t, t,...q : t P Rzt0uu. Then or every n, ι 1 n poq H, so O s open n the nal topology. However, O cannot be wrtten as ś PI O, so t s not open n the box topology. () (Example gven by Ljun Dng) Consder the set nÿ O tpx k q kpn P N, x 2 k ă π2 6 u. k 1 Then or every n, ι 1 n poq tpx k q n k 1 : ř n k 1 x2 k ă π2 6 u, so O s open n the nal topology. Now observe that, snce ř kpn 1 π2 k 2 6 and the summands are postve, p 1 k q kpn P O. However, O were open or the box topology, there would be a basc open set p 1 k q kpn P ź kpnpa k, b k q Ă O. Ths s not possble snce or any p 1 k q kpn px k q kpn P ś kpn pa k, b k q, x k ě 1 n, there s an ndex k 0 such that x k0 ą 1 k 0, and then ε x k0 1 k 0 ą 0. By the convergence o ř kpn 1, there s an ndex K such that π2 k 2 6 řk k 1 1 ă ε 2, k 2 so ÿ k q kpnpx 2 ą ÿ 1 k 2 ` ε2 ą π2 6. kpn Remark. The mstake n the prevous (wrong) soluton was to assume that an arbtrary subset o R N s o the orm O ś kpn O k. However, as you can see n the examples above, ths s not always the case!