J={ Infinite. if txty EX. yet U. properties X={ Example. infinite. Def. t ±Y X. top. Example : X = Rmk : X is To but not T, yet. FZCX finite set.
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1 ( TO Tr 71 Husdorft Seprtion xiom l is To if txty E 7 set U such tht xeu y U or ye U xcf U c y is T if t ±Y 7 set U such tht xe U yet U h is Tz ( Husdorff V such tht xe U if ttxt Y E 9 V 7 sets U nd UV u v Rmk Ti Tz re properties T To but To Ti # Tz Exmple b } xnx III c Iii * Tt } so is To but not T Exmple Infinite set J F F C finite } U 4 } Finite complement ology V F f ±y c U Y } in e U yet U is T For ny U C U Vt $ U V Fz F FZC finite set UV infinite set F UFZ t is NOT Tz T finite t
2 We S ocx ocxl EIU OC lemm T xe x } is closed ( so In Ti nd Tz txe s Rpyotgg x } is closed # finite set lwys closed x } tx tye } 7 U C x } ye U C * ty 7 UC YEU El Exmple ny metric Tz ( Hndoff txty E U B ( x V dcx y > 0 0 B ( y EU YEV unv & Exmple know tht 0 x is n embedding IT ( clim Husdorff # OC C closed ( x 4 E Ux V EU ye ( x y c xx # ±y Uxvc xx # UnV$ OC C is closed # o ( is fc x g e xx oc 7 ( x g EUV C U UC re xty # eu YEV # UV U V is Huidorft ( Tz
3 CTO ( then Exmple I ] Tz pt ouv #jsg prntjftee?fjocto ] to fitot Hot g c U ge V Blfltol to h lhltoḻ fctoi ( B ( 9h01 to sf Un V p 72 Connected Rmk UB clled seprtion it NB nd B C re both UB seprtion B so B re both nd 2 if is both nd closed U seprtion is if UB seprtion 134 or B subset Y C clled if Y is in the sub ology lemm is ( C both nd closed then or
4 2 [ lr[_ F F Fz Since B 3 Ex 12 } J i } z } } I }U 2 } NOT I d i } } Ex 2 n Infinite set with the finite complement UB BC re to # B Fi Fz finite sets B UFZ t hs no nontrivil! seprtion finite is Ex 3 C o U[ 0 [ oco U [ o n n 1 C o U En o 1R[ NOT ionnecled ni Ex 4 01 ] suppose C [ 01 ] both nd closed nd OE let sup x > o [ 0 x ] C } Then oesl Clim 1 > 0 OE which is C [ e n[ 01 ] F E > 0 C > > 0 [ 0 e by Clim 2 [ 0 ] <Y< C t definition 7 > Y ye [ o ] C [ 0 C ( 7 n incresing sequence xn } C [ o limit point of [ o u nd closed e
5 Obvious let B which Then B Contrdiction f clim 3 l suppne Oscl E which 7 EZO such tht ( E te C we tht [ 0 proved ] C [ o t E ] C This contrdicts the definition of Thus we hve [ 01 ] 101 ] is In the usul ology Rmk In generl It cn be ( b ] ( b tht Intervls proved such s [ b ] etc re ( 1 IR I ( 011 so lr lso Thn ( The intermedite Vlue Theorem suppne is lr continuous f ( i c b f ( 1 for some i E Then for ce ( b 7 ny E such tht ftxec suppose fixltc t e Then f ( cc xe fn > c } co 1 ftp# B f ( s cl UB re B ZE IEB This opertion of is Lemm UB is seprtion ( re nd n 1301 Y C is sub since YC or YCB Y ( YIUIYB seprtion of Y Thm ( Properties of is in is C B C B is
6 let c so Z On i NI then since since (2 i tie I to UI IEI IEI (3 is f 7 Y continuous f( (4 Y re not empty Then Y > Y re both seprtion of B let B C UD be sub we C B ssume tht my C C BCTCE BE B the other hnd C C B is closed ( nd subset BNE the closure of C in B which C itself BNE Bc c B D This mens tht B is ( 21 T UI IEI U B be opertion i C T sub C or i CB Hi suppose I io e I io C Then tie I in > intti >?± * 4 i c ti TUi C i E I T is (3 let ZfC the restriction f Z is surjective nd continuous let Z u B be seprtion of Z f ( Uf ( B is opertion of which is f( or f (B since f in surjective It implies tht fc Z or B Z
7 x confected Intimn t C Y of i IEIRW 141 txe YEY Ty x y to By Fixxoex (2 x x xy I xouty y y( Y is xo #Ih Y o y # Rmk By Induction n i TII is i tlsisn one cn prove How bout the? product Tcl L Thm IT is # d is TLET * T ( in the product let us prone for exmple tht IRw TIIR i i in the product Consider IPT lr~ IR co x i o } I ( x K 1 Ro UlfEntin n1 ~ IR is by 121 of i o fizn } the Theorem previous clim 11 (Rw ( hence IRW proof f x C x k E IRW I c U for some set U CIRW By the product n U TIU ; ITIR ii co IRW we tke U s my wher Vi C IR is inti t Ki en
8 Given Risui co Then clerly Un Itf Vi IT 01 ± $ intl so Un lr t $ The clim proved ( The Theorem bout IT lmost the sme LET 73 Pth x Ye pth in from to Y is continuous ( [ b ] CIRC 7 mp y [ b such tht r( Kb Y given the usul ology nd we usully tke [ b ] [ o I ] for convenience in clled pth 7 pth In from to y if t ty ttg # I Thm is pth is o t Y I try [ 0 I ] pth Ho o 811 y Exmples 11 so is ryt on ] is gdgiusknses ryk D # ll re pth 1R 0 } C 1122m lso pth
9 fx Connected # t txtye Exmples 2 V C IRLu convex region ( ie V tosdsl Then V is pth the point dxt YEV yy ye V 8 [ 01 ] V is pth from to Y t H ( I x+ty Et The unit bll B ± e IRI u II i l f1 } CIR is convex pth Clim IRIN Rsul ( not homeomorphic suppose f 1122m lrusu homeomorphism Then It induces homeomorphism I 1122 cool } ~> 112 flo } D o_ th Nttconnecled contrdiction Exmple # Pth ( x sintx o< xsi } clrlsu y ( 01 ] pth Prove i ( sintx to} EhD U isti but NOT pth #
10 Then } By Sn E ( 01 IQ such tht sn 0 7 the 101 such tht rctn Sn th ti tz We fly It for tzg nother esier is pth 0 exmple ;lgl ;D U to # knxt ;D? C hence U ( o y oeyei } 1 is but NOT Rin p 8 qgg g > 0 / But 1 It cool } CBCB I pth f [ 01 ] B pth such tht f 1 o suppose } 1 1 x 11 P fci of let 9 IR IR be the projection j gof [ 01 ] IR is pth ( stisfying cL 1 the Intermedite vlue theorem sequence Thus f ( tn Sn } 0 } C th choose my subsequence of th still written TE [ 01 ] then fctn s tn Sn } o } o } o } fit o } } Et B contrdiction So B is NOT pth # Exmple C[ 01 ] is pth 1 tfge CEO ] define y [ 01 ] Cio 1 ] r is continuous? Just show if ti t re t it ctt f closed tg ilose! then Ht Htr re do ( tct Htzl Mx / ( I fly + ti 914 ( 1 tz fly ( y / oey El Mx ( t ( g ( y tzldlf 91 osys 1 y is continuous Ho f y g CEO D pth
11 t the equivlence clsses re clled the ( components of [ becuse c i ~ ciil ~Y Y~ These re obvious 74 Components nd Pth Components Given we shll brek It Into pieces tht re or pth x YE define x~y if I U C reltion on This n equivlence YEU Equivlence reltions is clled ~ n equivlence reltion on if t x~y x YE } in C xx stisfies cii xnx y~x ~Y Y~z x~z ~ E [x] YE y~x } ciilxry Then t x YE ] n[ or y ] $ [ ][ y ] so there set of RC such tht representtives U [ ] ( is disjoint unions of equi 9696 ER clsses Ex 1122 ~ I 1 # where txitxif if HHr [ x] is c IR yityi r } g Ucr Cr IR re Rzo Rmk o In the definition of components ~y n equivlence reltion ciiil xny y~z xnz? let nd B re
12 B Cd EU Cj Invrint In subs of such tht x YE Y ZEB YEB to x ze UB x~z So U Cd disjoint unions of components LET C T& ] for some E LET Lemm 1 C is BC Cd for some 2 suppne xe BG ± $ xry contrdiction YEBCB to s x~&~\ p ~Y dtf lemm 2 HLET Choose point x U C oe Cd t ECL Ux C Cd ~ so 7 C UU xtcd xoeux tx C is lemm 3 Cd closed Cd is closed lso closed Ccd C is closed Prop ny disjoint union of ( components ech of which nd closed Rmk The number of components Exmple 1 IR 0 } C 0 UCO C Rc two components
13 This ine 1/23} The Exmple } B 11 4 } 12 } 13 }} } /3} 14 } 12/4} 1 34 } } wht ne the components? Sy component nd IE 1 23 } nd 1 23 } U 1 } closed i 23 } or two components is y if It pth in from to Y equivlence clsses re clled the pth components of Rmk in n equivlence reltion c i ~ c Ii ~Y Y~ trivil ( iiil ~y Y ~Z ~Z y [ o ] rz [ 12 ] 0 i 1 1 y 1 it y 2 l z Thin r [ 02 ] ntl t 1 oeee jzlt 1st < 2 pth from to Z let UL disjoint union of equivlence clsses t xd DET Then cil B C for some E 42 ET pth Bc d for some L ciil s is pth The TLET ftpygnl?pnnniiiotxnyonnicny
14 Q bsis EV since HECCU 74* Locl ness is clled loclly t if t I nbhd U of I ( nbhd V of such tht V C U U is n nbhd of I if mbhd V of such tht VCU is loclly if is loclly t txe is clled loclly pth t x if t ( nbhd U of 7 pth ( nbhd V of such tht VCU is loclly pth is if loclly t pth fxe Ex El 0 UC 01 ] C lr is loclly nd Rmk If hs B t BE B B is loclly pth loclly ( pth is loclly ( pth Ex f ( 01 ] S C 1122 usul t 1 ( t sin tt 55 U ] } t 5 NOT pth t ( 001 Ex C IR NOT nor loclly Thm loclly For ny U C ech component of U is In U C is C C U component 7 nd subset V C U C is component xe V C C C
15 P xe U Suppose tht components of sets in re is n nbhd of let C C U be component tht contins C nd loclly t It cn Similrly be proved is loclly pth C For ny set U of ech pth component of U is In Thm is l Ech pth component of lies in component of ( 21 If is loclly pth components nd the pth components re the sme EC P (2 1 let C be Suppose P C of xe P component of is PCC pth component txe C \P is contined is some pth component R of xe R C C let Q denote the union of such pth components Then C P UQ since is loclly pth ech pth component of In Thus C PUQ is seprtion of C C is xep PC contrdiction
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