không gian metric

8/2/2019 Khng gian metric

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I HC HU

TRUNG TM O TO TXA

TS. NGUYN HONG

GIO TRNH

KHNG GIANMTRIC

(CSGII TCH)

Hu - 2007

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MC LC

LI NI U ...........................................................................................................3A. KIN THC B SUNG.......................................................................................5

1 TP HP S THC.......................................................................................52. LC LNG CA CC TP HP............................................................10

B. KHNG GIAN MTRIC....................................................................................161. KHI NIM MTRIC. .................................................................................16BI TP...............................................................................................................212.TP M V TP NG..............................................................................23BI TP...............................................................................................................303. NH X LIN TC .....................................................................................32BI TP...............................................................................................................37

$4 KHNG GIAN MTRIC Y ...............................................................38BI TP...............................................................................................................505 KHNG GIAN COMPACT...........................................................................52BI TP...............................................................................................................676. KHNG GIAN LIN THNG .....................................................................69BI TP...............................................................................................................71

C. LI GII V HNG DN.............................................................................72PHN A ...............................................................................................................72PHN B ...............................................................................................................73

TI LIU THAM KHO........................................................................................87

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LI NI U

Gio trnh ny c vit da trn bi ging cho sinh vin khoa Ton trngHSP Hu trong nhng nm va qua. Hc phn ny c mc ch trang b nhng

kin thc cn bn v gii tch hin i m bt c sinh vin Ton no cng phinm c. Khc vi gii tch cin, trong ngi ta lm vic ch yu trntp IRk cc b k s thc, y cc khi nim cbn ca gii thch nh ln cn,gii hn lin tc c xt trong khng gian tng qut hn m phn t ca nc th l cc i tng tu min sao c th xc nh c khong cch giahai phn t. Ngoi mt cch bn cht v su sc nhng kin thc v gii thchcin hc trong nhng nm trc, cng nh chun b hc tt cc hc

phn tip theo nh l thuyt o, tch phn, gii tch hmCc kh nhiu sch vit v khng gian mtric, tuy nhin ngi ta thng

ch trnh by nhng kin thc dng cho mc ch ca cun sch nn chac mt gio trnh tng i hon chnh ring cho phn l thuyt ny. y, bnc s thy nhiu bi tp c a vo vi t cch rn luyn t duy v ng thicng c th xem nh bi b sung l thuyt. Phn ln cc bi tp u c li gintm tt hoc chi tit. iu ny c l s mang li li ch thit thc rt hn ch vcng c t sch gii bi tp gip cho sinh vin trong lc hc tp.

hc tt hc phn ny, v nguyn tc sinh vin ch cn nm c nhngkin thc scp v l thuyt tp hp v nh x, php qui np v cc suy lun

logic ton hc. Cn phi bit din t mt mnh bng nhiu mnh tngng vi n cng nh hiu v vn dng cch chng minh hay xy dng cc itng bng qui np hu hn. Tuy nhin c th hiu su sc v nht l lmc cc bi tp. y, ngn ng hnh hc c dng din t cc khi nimkhng gian mtric, nhng i lc c nhng vn vt ra khi trc gic v suylun ch quan thng thng. Do vi tng khi nim, ngi hc nht thit

phi hiu thu c nh ngha, t mnh tm c nhng v d minh ha cho ccnh ngha . Nh Dieudonne ni:... trc quan hnh hc, cng vi s

phng thch ng l mt ngi hng dn rt ng tin tng trong hon cnh

tng qutCun sch c chia lm hai phn. Phn kin thc b sung nu li mt cch

c h thng cc tnh cht ca tp s thcIR. Sinh vin tng cng ch n khinim infimum v suptemum ca mt tp s thc v cn s dng mt cch thnh

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tho, bin son. V khi nim lc lng tp hp, cn nm c trong trng hp noth mt tp l m c,

Phn th hai l phn chnh ca chng trnh. C nhiu con ng trnhby cc khi nim. y chng ti chn cch tip cn vi ngn ng thngdng, mt mt ngi hc d nh, mt khc phn no gii thch l do a ra

tn gi nh vy. Tuy nhin, nht thit phi c hiu theo ng nh ngha. Cckhi nim quan trng phi k n l hi t, m, ng, lin tc, y ,compact c trng phn ny l nng v suy lun hn tnh ton, hn na nhiuthut ng chng cht ln nhau lm ngi mi hc thy lng tng. V th sinhvin nn tm thm v d v hnh nh trc quan d nh. Sau khi nm c lthuyt, cc bn t mnh gii cc bi tp cn thn trc khi xem li gii. Cc bitp kh hn c nh du * dnh cho sinh vin kh, v phi c thi gian nghinngm nhiu hn.

Tc gi xin cm n cc bn trong t Gii tch khoa Ton trng HSP Hu ng vin gp khi vit cun sch ny. Mong c nhn c nhng ph

bnh ca cc ng nghip gn xa.

Tc gi

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A. KIN THC B SUNG

1 TP HP S THC

Chng ta tip xc nhiu vi tp hp s thc t chng trnh ton bcph thng. C nhiu cch xy dng tp hp s thc, chng hn dng nht ctDedekind, cc dy cbn. ca tp hp s hu tQ. y vi mc ch l hthng li nhng kin thc cn thit cho gii tch, chng ti s chn mt s mnh cbn lm tin nh ngha tp hp s thc. Cc tnh cht cn li csuy t cc tin ny.

1.1. nh ngha:

Tp hp s thc, k hiu IR l mt tp cng vi cc php ton cng + vnhn . xc nh trn , tho mn cc tin sau:

I. (IR, +) l mt nhm cng Abel, tc l vi mix, y, z thuc IR ta c:

x + y = y + x

x + (y + z) = (x + y) + z

(0 IR) (x IR): x + 0 = 0 + x= x

(x IR)((-x)IR): x + (-x) = 0

II. (IR*,.) l mt nhm phn Abel, trong IR* = IR \{0}, ngha l vi mix,

y, z thuc IR

*

, ta c:xy = yx

x( yz) = (xy) z

(( 1 IR*) : x1= 1x = x

(x IR*)( x-1IR*): xx -1 = x-1x = 1(y cho gn, ta vit xy thay cho x.y)

III. Php nhn c tnh cht phn phi i vi php cng:

Vi mi x,y thuc IR ta c: x(y + z) = xy+ xzNh th IR cng vi cc php ton cng v nhn lp thnh mt trng

IV. IR l mt trng c sp th t, ngha l trong IR c xc nh mt quanh th t tho:

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1.x y vy z ko theo x z2. x y vy z tng ngx = y3.Vi hai phn t tu x,y IRth hoc x y hoc y x4.x y ko theox + z y + z vi mi z IR5. 0 x v 0 y ko theo 0 xy

Nu x y v x y th ta vit x < y hay y > x .V. Ta gi mt nht ct trong IR l mt cp (A,B) cc tp con ca IR sao cho A,

B khc trng, AB = , IR = A B v vi mi aA, b B th a < b .

Tin Dedekink. IR l mt trng c sp lin tc, ngha l: Vi minht ct (A,B) ca tp IRu xy ra: hoc c mt phn t ln nht trong A hocc mt phn t nh nht trong B v khng th va c phn t ln nht trong A,va c phn t nh nht trong B.

Phn t ln nht trong A (hoc phn t nh nht trong B) gi l bin canht ct (A,B). Tp hp s thc cng gi l ng thng thc.

1.2. Cc tnh cht cbn:

1.2.1 Supremum v infimum :

ChoMl mt tp con khc trng ca IR. SxIR c gi l mt cn trncaMnu vi miyM th y x, sxIRgi l cn di ca M nux y vimiy M. Tt nhin nu x l cn trn (tng ng, cn di) th vi mix1 > x (t.x1 < x) cng l cn trn (t. cn di) ca tp M.

Cn trn b nht (nu c) ca tp Mc gi l supremum ca tp M, k

hiu supM. Nh vy, = supMkhi v ch khii)x M: x ii) ( < ) (x M) : < x

(iu kin ii) ni rng v l cn trn b nht nn nu ) (x M) :x


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A ={x IR : (a M)x a};B ={y IR: (aM) a < y}.

Khi A v M A; B v vi c> c th cB. Vi miz IRthhoc z A hocz B nn IR = A B. NuzAB th c a M sao choz a< z hayz < z, v l nn AB = . Hn na, nux A ,y B ta c x a < y vi

a no thucMnnx < y. Theo nh ngha, (A,B) l mt nht ct ca IR. Gim l bin ca (A,B). Khi ta s c m = sup A. Thc vy, chng hn m A ththeo nh ngha s c a M m a v M A nn m = a. Cn nu m B tha M : a < m. Nu m< m th m B tc l m A, hn na mkhng phil phn t ln nht trong A nn c m A,a M m< ma < m. Phn cnli ca nh l chng minh tng t.

Ch : Gi s M l mt tp con khc rng ca IR nhng khng c cn trnno c. Khi ta quy c sup M = + . Tng t, nu M khng c cn di, taquy c inf M = - .

1.2.2 Ta gi cc sa IR, a > 0 l s dng, a < 0 l s m v t x nux 0; x= - x nux < 0 v gi xl gi tr tuyt i ca s thcx. Sa IRgi l gii hn ca dy s (xn)n IR v k hiu axn

n=

lim nu:

(> 0)(n0)(n n0): x a <

Dy (xn)n gi l n iu tng (t. gim) nuxn xn+1 (t.xn xn+1) vi min Nb chn trn (t. di) nu tp {xn} c cn trn (t.., di) hi t nu (xn)c gii hn.

Nguyn l Weierstrass: Mi dy n iu tng (t..,gim) v b chn trn(t.., di) u hi t.Chng minh: Gi s (xn)n l mt dy n iu tng v b chn trn. Theo

nguyn l supremum, tp {xn} c mt supremum . Vi > 0 cho trc, theo iukin ii) c s nguyn n0 sao cho < xn0. Mt khc, theo tnh n iu tngca dy (xn), ta c


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N ZQ IR

Nguyn l Archimde: Cho hai s thc a, b bt k vi a > 0. Khi tn ti nN sao cho b < na.

Thc vy, do N khng b chn trn (tc l khng c cn trn) nn vi s

thc a

b

s c nN b

a < n hay b < na

1.2.4. Cc tp

(a, b) = {x IR : a < x < b } v[a,b] = {x IR : a x b}

ln lt gi l khong (hay khong m) v on (hay khong ng). Mt dyon {[an, bn]} gi l tht li nu [an+1,bn+1] [an,bn] v 0)ab(lim nn

n=

Nguyn l Cantor: Mi dy on tht li c mt phn t duy nht chung

cho tt c cc on y.Chng minh: Gi s ([an, bn])n l dy on tht li. Ta c:

a1a2 an+1 bn+1bn b1vi mi n N. Theo nguyn l Weierstrass, dy (an)n tng, b chn trn (bi b1chng hn) nn hi t v s = sup {an}. Nh than vi mi n. Nu [ano, bno] vi mt n0 no th t hn bno < . t = - bno. Khi vi n ln th - a n< - bnotc l bno < an! v l. Vy [an,bn] vi mi n. Mt khc, nu c[an,bn] vi mi n th-

bn an. Do 0 - 0)(lim =

nnnab

hay- = 0 ngha l =

1.2.5. Dy (xn) c gi l b chn nu n va b chn trn va b chndi. iu ny tng ng vi:

(a IR)(n N):xnaNguyn l Bolzano Weierstrass: Mi dy s thc b chn (xn)nu c

mt dy con hi t.Chng minh: Theo gi thit, tn ti sa sao cho vi mi n N ta c a

xna. Trong hai giai on [-a,0] v [0,a] phi c mt on cha v s cc phn

txn (nu khng, ho ra (xn)n ch c hu hn cc s hng). Ta gi on ny l[a1,b1].Chia hai on ny bng im gia c1=

a1+ b12

. Trong hai on [a1,c1] v

[c1,b1] cng c mt on cha v s cc xn, k hiu on ny l [a2,b2] v li

chia i on ny bi im gia c2 =2

22 ba + v.v... Tip tc qu trnh ta thu

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c mt dy on tht li [ak, bk] (v hin nhin [ak+1, bk+1] [ak, bk] v bk ak

=k

a

20 khi k). Theo nguyn l Cantor, dy on ny c duy nht phn t

chung . V mi on [a[kk

k

ba ,1

=I ] k, bk] cha v s cc phn txn nn ta

hy ly phn txn1 [a1, b1] ri xn2 [a2, b2] vi n2 > n1, xn3 [a3, b3], n3 >n2 khi (xnk)k l dy con ca dy (xn)n vxnkbk- ak0 (k), nghal dy (xnk) hi t v.

1.2.6 Dy s thc (xn)nc gi l dy cbn (hay dy Cauchy) nu:(> 0)(n0)(n n0)(m n0) : xn xm<

Nguyn l Cauchy: Mi dy s thc cbn th phi hi t:

Chng minh: Trc ht ta chng minh rng nu (xn)n cbn th n phi bchn. Vi = 1, tn ti n0 vi mi n n0 ta c xn xno < 1 hayxno - 1xn

xno + 1. t a = max {x1,,xno, xno+1}, khi y vi mi n th -a xn a. Do theo nguyn l Bolzano- Weierstrass, dy (xn)n c mt dy con hi t v

. By givi > 0 cho trc s c nkn

x

0 sao cho vi m, n n0 thxn xm< /2 do(xn)n cbn. Mt khc nn cng tn ti sm

knx 0 nu n m0 th | | n0 th

xn xn +

1

b-a hay b - a > 1/n. Tng t, c s nguynpp nb. Gi q l s nguyn b

nht tho mn q n, do q-1 < nb hayq-1

n< b. Lc ny a .Ngi ta chng minh c rng, cho hai tp A, B bt k bao gicng xy

ra mt v ch mt trong ba trng hp.

1. Xy BA = (tc l A, B tng ng vi nhau)

2 Xy BA < 3. Xy BA >

2.2. Tp hp m c:

2.2.1. nh ngha: Tp hp A c gi l tp hp m c nu A tngng vi tp s t nhin N. Ni cch khc, A m c nu v ch nu tn timt song nh tN ln A. Khi ta cng ni A c lc lng m c.

Gi a: N A l song nh ni trn, ta c:N na (n) = an A

Nh vy c th ni tp hp m c l mt tp m cc phn t ca n cthnh s thnh mt dy v hn.a1, a2, a3,,an,

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2.2.2. V d:1. Tp hp cc s t nhin chn, cc s t nhin lu l cc tp m c.

Tht vy, theo mc trc, card {2,4,6} = cardN, cn E = { 1,3,5,...,2n +1,}tng ng vi N nhsong nh.

N n 2n + 1 E

2. Tp Z c s nguyn l m c. chng tiu , ta xt nh xf:NZ cho bi :

n

2 nu n chn

n f(n) = 1- n

2 nu n l

D dng kim trafl song nh ta c c kt lun

3. Tp cc s hu t Q l m c. Tht vy, mt s hu t c th vit

c duy nht thnh mt phn s ti ginq

p, q > 0. Ta hy tm gi tng |p| + q

l hng ca s hu tq

p. R rng tp hp tp hp cc phn s c hng cho

trc l hu hn, v d: phn s c hng 1 l1

0= 0, hng 2 l

1

1v

1

1, hng 3

l

1

2,

2

1,

1

2,

2

1,... Hn na mi s hu tu c hng xc nh nn ta c th

nh s hu t thnh dy theo th t tng dn ca hng, tc l bt u nh scc s hng 1 ri tip theo cc s hng 2, hng 3,Vy cc phn t ca Q c thsp xp thnh dy Qm c.

Tip theo, chng ta thit lp cc nh l cbn ca tp m c.2.2.3. nh l: Mi tp v hn lun lun c cha mt tp con m c.Chng minh: Gi s M l tp v hn. Ly ra mt phn t bt ka1 M. Khi

M \ {a1} v hn nn ly tip phn ta2 M\ {a1} ri a3 M {a1,a2} v.v Qu trnh ny c tip tc mi v ta thu c tp m c A = {a1, a2,} M

2.2.4 nh l: Mi tp con ca mt tp m c th phi l tp hu hnhoc m c.

Chng minh: Gi s A = {a1, a2,} l tp m c v B l mt tp conca A. Gi an1, an2,... L cc phn t ca A thuc tp hp B theo th t tng dntrong A. Nu trong cc sn1, n2,... c s ln nht th B l hu hn. Trng hp

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tri li, cc phn t ca B c sp thnh dy v hn an1, an2,... nn B mc.

2.2.5. nh l: Hp mt h hu hn hay m c cc tp m c l mttp m c.

Chng minh: Cho A1, A2, l dy cc tp m c. Ta c th gi thit cctp ny khng giao nhau v nu khc i, ta t B1 = A1, B2 = A2\ A1, B3 = A3\ (A1U A2),... Cc tp Bi ny hu hn hoc m c, khng giao nhau v

. By gita sp xp cc phn t ca Ai

ii

i

BA

=

==

11UU 1,A2,... thnh mt bng v hn

nh sau:

A1 : a11 a12 a13 ....

A2 : a21 a22 a23 ....

A3 : a31 a32 a33 ..... . . . ...

Ta hy nh s tt c cc phn t ny theo ng cho t tri ln phatrn. Do mi ng cho c hu hn phn t nn c thnh s th t trnng cho th nht ri ng cho th hai, th ba,... nh sau:

a11, a21, a12, a31, a22, a13,

Vy tt c cc phn t ca tp c nh s thnh mt dy nn

tp A m c.

ii

AA U

=

=1

Nhn xt: Trong cch chng minh ta thy nu mt s hu hn hay mc cc tp Ai (khng phi tt c) c thay bng cc tp hu hn th kt lunca nh l khng thay i.

2.2.6. nh l: Khi thm mt tp hp hu hn hay m c vo mt tpv hn M th lc lng ca n khng thay i.

Chng minh: Gi s A l tp hu hn hay m c. K hiu N = M A.Theo nh l 2.2.3, tn ti mt tp m c B M. t M= M\B, ta c M =M B nn N = M B A. Theo nh l 2.2.5, B A l tp m c nn

tn ti song nh f gia B v B A. Ta t:g : M = M B N = M (B A)

x nux M

g (x) =f(x) nuxB

Nh th g l song nh t M ln N nn card M = Card N.

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Theo nh l ny ta thy khong (a,b) tng ng vi on [a,b]. Hn na(a,b) tng ng vi IR nn [a,b] cng tng ng vi IR.

Nhn xt: T cc nh l 2.2.3 v 2.2.6 ta thy lc lng m c l lclng b nht trong cc lc lng ca tp v hn.

2.2.7. nh l: Tp hp tt c cc dy hu hn c th thnh lp c vi

tt c cc phn t ca mt tp hp m c l tp m c.Chng minh: Gi s A = {a1,a2,...} l mt tp m c. K hiu Sm l tp

cc dy c ng m phn t ca A dng (ai1, ai2,...aim). t . Ta chng minh

S m c. Trc ht S

mm

SS U

=

=1

1 = A m c. Bng qui np, gi s Smm c,hy ly ak A v k hiu S

km+1 l tp hp tt c cc dy c dng (ai1, ai2,,aim,

ak). Gia Skm+1 v Sm c mt song nh cho bi (ai1, ai2,,aim,ak)

(ai1,ai2,,aim). nn Skm+1 m c. Mt khc v Sm+1 = nn S

k

mk

S 11

+

=U m+1 m

c theo nh l 2.2.5. Cng tnh l ny, S l mt tp m c.

2.2.8. H qu: Tp hp tt c cc a thc P(x) = a0 +a1x +...anxn (n bt k)

ly gi tr trongIR vi cc h s hu t a0,a1,, an l m c.Chng minh: Mi a thc tng ng vi mt v ch mt dy hu hn cc h

s hu t ca n. V tp Qm c nn theo nh l 2.2.7, tp tt c cc dyhu hn cc s hu t l m c nn tp cc a thc ny m c.

2.3. Lc lng continum:

Ta xt cc v d v thit lp cc nh l v cc tp hp m c. Vy c

tp hp v hn no khng phi l tp m c hay khng? nh l sau y chota cu tr li khng nh.

2.3.1. nh l. Tp hp cc s thc IR l tp v hn khng m c.Chng minh: Trong v dnh l 2.2.6 ta thy IR tng ng vi on

[0,1]. Do ch cn chng minh [0,1] khng m c. Gi s tri li [0,1] lm c. Khi cc phn t ca n c nh s thnh dy x1,x2,..xn, Chiacho [0,1] thnh 3 on bng nhau v gi on khng cha x1 l 1. Li chia tip1 thnh 3 on bng nhau na v gi 2 l on khng cha x2, Tip tc qu

trnh ny ta thu c dy on 1 2 ...vi n c di l |n| =

1

3n sao choxn n. y l dy on tht li nn theo nguyn l Cantor, tn ti

. Do phi trng vi mtx[ 1,01

=n

n

I ] no no . V n vi mi n nn

xnono. iu ny mu thun vi cch xy dng cc on n. Vy on [0,1]khng phi l tp m c.

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Nhn xt:

1. t a = {1

n: n N). R rng A l tp m c v cha trong on [0,1].

Do lc lng on [0,1] (hay IR)ln hn lc lng m c. Ngi ta gilc lng ny l lc lng continum hay lc lng c.

2. Tp hp s thc bng hp ca s hu t v s v t. Do tp s hu tmc nn tp s v t khng m c v cng c lc lng l c.

BI TP

1.Hy thit lp mt song nh gia hai tp (0,1) v [0,1]2.Chng minh tp cc im gin on ca mt hm sn iu xc nh

trn [a,b] l hu hn hoc m c.3. Gi s E l mt tp con ca tp s thc IR c tnh cht |x-y| > 1 vi mix, y

E. Chng minh E l mt tp hu hn hoc m c.

4. Gi s E l mt tp v hn. D l mt tp con hu hn hay m c caE sao cho E\D v hn. Chng minh E\D c cng lc lng vi E.5. Cho A v B l cc tp m c. Chng minh A B l tp m c.6*. K hiu E l tp hp tt c cc dy s (xn) trong xn = 0 hoc xn = 1.

Chng minh E l tp hp khng m c. (Thc ra E c lc lng c)

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B. KHNG GIAN MTRIC

1. KHI NIM MTRIC.

Php ton c trng ca mn gii tch l php ton ly gii hn. din tkhi nim ny ta phi tm cch xc nh mc xa, gn gia cc itng. Cc mcs xa, gn c tha vo mt cch kh t nhin thngqua khis nim khong cch hay mtric c chnh xc ho bi cc nh nghasau y.

1.1. nh ngha:

Gi s X l mt tp tu khc trng cho trc, mt mtric ( hay khongcch) trn X l mt hm s d: X X IR tho mn 3 tin sau y:

1) d(x, y) 0, vi mix, y X: d (x, y) = 0 khi v ch khi x = y.2) d(x, y) = d(y, x) vi mix, y X, (tnh i xng).3) d(x, z) d(x, y) + d(y, z),vi mix, y, z X (bt ng thc tam gic).Khi tp X vi mtric d cho gi l mt khng gian mtric v k hiu l

(X,d). i khi n gin v nu mtric dc xc nh r rng, ta ch k hiuX.

Bng ngn ng hnh hc, phn tx X gi l im ca khng gian X, sthc dng (hay bng 0) d(x,y) gi l khong cch gia 2 imx vy.1.2. Cc v d:

1.2.1. Gi s M l tp hp con khc trng ca tp s thc IR. Ta hy td(x,y) = |x-y | vix,y M. Khi nhcc tnh cht quen thuc ca gi tr tuyti, ta kim tra d dng (M, d) l mt khng gian mtric.

1.2.2. K hiu IRk = {(x1,...xk) : xi IR, i = k,1 } l tp hp cc b k s thc.Vix = (x1,,xk),y = (y1,...,yk) thuc IRk, ta t:

d(x,y) = i =1

k(xi - yi)2

Khi cc tin 1)-2) r rng, ta ch cn kim tra tin 3) tc l chng

minh:2

1

)zx( ik

i

i =

2

1

)yx( ik

i

i =

+ 2

1

)zy( ik

i

i =

t ai = xi y

i, bi = yi z

i khi ai+ bi = x

i- z

i

Ta li c :

16


17/88

d2(x,z) = i =1

k(ai+bi)

2 =

i =1

kai

2=

i =1

kbi

2 + 2

i =1

kai

bi

p dng bt ng thc Cauchy Schawrz cho s hng sau cng ta c:

d2

(x,z) i =1

k

ai2

+ i =1

k

bi2

+ 2 i =1

k

a2

i

i =1

k

b2

i

( i =1

ka

2i +

i =1

kb

2i )2

T ly cn hai v v trli vi k hiu c, ta c:d(x, z) d(x, y) + d(y, z).

Vy (IRk,d) l mt khng gian mtric v gi mtric ny l mtric thngthng trn IRk.

Ch :1. Khi k= 1 ta trv v d 1.2.1 vi M = IR2. Khi xt IRk m khng ni r mtric no th ta qui c l xt IRk vi mtric

thng thng.1.2.3. Gi s X l mt tp tu khc trng. Ta t

0, nux = yd(x,y) =

1, nux y

vi mix, y X. Ta hy kim tra d l mt mtric trn X.

Tin 1) v 2) c nghim ng. Tin 3 c dng:d(x, z) d(x, y) + d(y, z)i. Nux z th d(x,z) = 1 cn v sau 1ii)x = z th d(x,z) = 0 cn v sau 0Vy tin 3) cng tho mn nn (X,d) tr thnh mt khng gian mtric.

Mtric d ny gi l mtric tm thng trn X.1.2.4. K hiu tp hp cc hm lin tc

f: [a,b] IRl vi hm f,g thuc ta hy t[ b,aC ] ][ b,aC

d(f,g) = [ ] )x(g)x(fmaxb,a Vf,gl cc hm lin tc trn [a,b] nn hmf - gcng vy. Do gi tr

ln nht ca hm f - gt c trn khong ng [a,b] nn d(f,g) xc nh.Cc tin 1)-2) hin nhin. Tin 3) suy ra t

x [a,b] : f(x)-h(x)f(x)-g(x)+g(x)+h(x)

17


18/88

[ ] [ ]

)x(h)x(gmax)x(g)x(fmaxb,ab,a

+

nn

[ ] [ ] [ ])x(h)x(gmax)x(g)x(fmax)x(h)x(fmax

b,ab,ab,a+

hay d(f,h) d(f,g) + d(g,h) vi mif,g,h . Khng gian mtric ny thng

c k hiu gn l .[ b,aC ]

]

[ ]b,aC

1.2.5 Cng trn tp hp ta t[ b,aC

d(f,g) = b

a

dx)x(g)x(f

Cc tin 2)-3) d dng kim tra. Ta c d(f,g) 0. Nu d(f,g) = 0 tc l

b

a

dx)x(g)x(f = 0. Gi s f g khi y c x0[a,b] f(x)-g(x)> 0 vi

mix[,] no cha trong [a,b]. Nh vy

.0)()()()()( >=

dxdxxgxfdxxgxfb

a

iu ny mu thun. Vyf = gKhng gian metric ny c k hiu l [ ].,

LbaC

Nhn xt: Qua cc v d trn, ta thy c th cho nhiu mtric khc nhau trncng mt tp X (tt nhin s nhn c cc khng gian mtric khc nhau). Tymc ch nghin cu, ngi ta s chn mtric no ph hp vi yu cu.

1.3. Mt s tnh cht n ginGi s (X,d) l mt khng gian metric, ta c:

1.3.1 Cho x1,...,xn l cc im ca X. Khi ta c bt ng thc tam gicmrng:

d(x1,xn) d(x1,x2) +...+d(xn-1,xn)Tnh cht ny c suy t tin 3 v lp lun qui np.

1.3.2. Vi mix,y,u,v thuc X ta c bt ng thc t gic:d(x,y) d(u,v) d(x,u) + d(y,v)

Thc vy ta p dung 1.3.1 ta cd(x,y) d(x,u) + d(u,v) + d(v,y)

hayd(x,y) - d(u,v) d(x,u) + d(y,v)

Thay i vai tr ca x,y cho u,v ta li cd(u,v) - d(x,y) d(x,u) + d(y,v)

18


19/88

Nh vy c c iu phi chng minh1.3.3. Cho A,B l hai tp con khc trng trong khng gian mtric X. t

),(inf),(,

yxdBAdByAx

=

v gi s thc d(A,B) ny l khong cch gia hai tp A v B. Nu A = {a} tavit d(A,B) = d(a,B) v gi l khong cch tim an tp B. rng nu A

B th d(A,B) = 0 nhng iu ngc li ni chung khng ng.Chox,yX, vi miz A ta c

d(x,A)-d(y,B) d(x,y)Thc vy vix,yX ta c d(x,A) d(x,z) d(x,y) + d(y,z), zA. Do d(x,A) d(x,y) + ),(inf zyd

Az

hayd(x,A) - d(y,A) d(x,y)

Tng t d(y,A) - d(x,A) d(x,y). T kt quc chng minh.

1.4. Khng gian metric con v khng gian metric tch.1.4.1. nh ngha. Gi s (X,d) l mt khng gian metric v Y l mt tp conkhc trng ca X. Nu xt hm thu hp d ca hm d ln tp Y x Y : d\Y x Y th hinnhin d l mt metric trn Y. Ta gi d l mtric cm sinh bi dln Y. Vi mtriccm sinh ny, (Y,d) c gi l khng gian mtric con ca khng mtric (X, d).

1.4.2 nh ngha: Gi s (X,dx) v (Y,dy) l hai khng gian mtric tu .Trn tch Descartes X Y = {(x,y) :x X,y Y} ta t d((x1, y1),(x2, y2)) = dX(x1,

x2) + dY(y1, y2)

D dng kim tra thy rng d l mt mtric trn tp X Y. Khi khng

gian ( X Y,d) c gi l tch ca cc khng gian mtric X v Y.1.5. Shi t trong khng gian mtric:

Cc khi nim hi v gii hn trong khng gian mtric X bt kc nhngha mt cch tng t trong tpIR vi vic thay |x-y| bng khong cch giahai phn t d(x,y). Mt dy trong khng gian mtric (X, d) l mt nh x.

Ta cng dng k hiu quen thuc l dy (xn)n N. Gi s nk l mt dy tngthc s cc s nguyn dng. Khi dy (xnk)kc gi l mt dy con cady (xn).

1.5.1. nh ngha: Gi s X l mt khng gian mtric v (xn)n l mt dytrong X. Ta ni dy (xn)n hi tnxX nu khong cch giaxn vx dn n 0khi n. Lc xc gi l gii hn ca dy xn v ta s k hiu

xxn

n=

lim

hayxn x, n . Din t li, ta c

19


20/88

( xxn

n=

lim ) )0),(lim( =

xxd

nn

(> 0 n0n n0 : d(xn, x) < )

1.5.2. Cc tnh cht.

Cho (xn)n, (yn)n l cc dy trong khng gian mtric X. Ta ca. Nu dy (xn)n hi tnx X th mi d con (xnk)k ca dy (xn)n cng

hi tnx.b. Gii hn ca mt dy hi t l duy nhtc. Nuxn x, yn y th d(xn, yn) d(x,y) khi n Chng minh:a. Gi s (nk)k l dy tng thc s cc s nguyn. Cho > 0 tn ti s

nguyn n0 sao cho d(xn, x) < khi n n0. T vi mi nknk0n0 nn d(xnk,x) < ngha l dy conxnk x, k

b. Gi s xn x vx

n x

. Khi t bt ng thc tam gic ta c:d(x, x) d(xn, x) + d(xn, x)

Cho n th0 d(x, x) 0)',(lim),(lim =+

xxdxxd n

nn

n

Vy d(x, x) = 0 hayx = x.c. Theo bt ng thc t gic (1.3.2.) ta c:|d(xn,yn) d(x, y)| d(xn,x) + d(yn,y).Qua gii khi n ta nhn c kt qu.

1.5.3. Cc v d:

a. Hi t trong IRk. Trong IRkvi mtric thng thng, ta xt dy sau:(xn)n :xn= .),...,(

1 knn xx

Theo nh ngha dy (xn)n hi t vimx0 = khi v ch khi d(x),...,(1 k

nn xx n, x0) 0 (n) hay

0)(0)( 20

2/1

11

20

=

iin

kii

n xxxx vi mi i = 1,...,k

|xni xo

i| 0, vi mi i = 1,...,k xn

i xoi vi mi i = 1,...,k

Vy s hi t ca mt dy trong IRk chnh l s hi t theo to (thnhphn) ca dy. c bit, vi k = 1 th y chnh l s hi t cu mt dy s thcthng thng.

b. Hi t trong C[a,b]. Gi s (xn)n l mt dy (dy hm) trong C[a,b] hi t vx C[a,b]. Theo nh ngha, ta c:

20


21/88

d(xn, x) =[ ]

)(0)()(max,

ntxtxnbat

Din t li, ta c :(> 0)(n0)(n n0)(t[a,b]) : |(xn(t) x(t)| < Vy s hi t trong C[a,b] chnh l s hi tu ca mt dy hm trn tp

[a,b] trong gii tch cin.c. Trong CL[a,b] s hi t ca dy (xn)nn x c ngha l:

d(xn,x) = b

a

ndttxtx )()( 0 (n )

S hi t ny gi l s hi t trung bnh ca dy hm (xn)Nhn xt: Theo nh l qua gii hn di du tch phn ca mt dy hm

lin tc, ta thy rng nu xn(t) hi tu n x(t) th xn(t) hi t trung bnh nx(t) nhng iu ngc li ni chung khng ng. C th coi s gn nhau gia

cc im trong tp C[a,b] theo mtric max cht ch hn mtric b

a

BI TP

1.1. Kim tra cc tp v cc hm sau y lp thnh khng gian mtric.

a. X = IRk, d(x,y) = iiki

yx = ,...,1max

b. X =IRk, d(x,y) = =

k

i

ii yx1

trong x = (x1,...,xk),y = (y1,...,yk) IRk

c. X = M[a,b] ={ f : [a,b] IR, f l hm b chn trong [a,b]},d(f,g) =[ ]

)()(sup,

xgxfbax

d. X = C[a,b]: tp cc hm lin tc trn [a,b] vi mi f,g X, d(f,g) =

b

a

dxxgxf 2/12

))()((

e. X= C[a,b] = tp cc hm s kh vi lin tc trn [a,b]d(f,g) =

[ ])()()(')('max

,agafxgxf

bax+

1.2. K hiu c l tp hp tt c cc dy s thc hi t.Vi x = (xn)n,y =(yn)n thuc c, ta t:

[ ]nn

b,ax

yxsup

Chng minh d l mt mtric trn c.1.3. Gi s d(x,y) l mt mtric trn tp X. Chng minh cc hm sau y

cng l nhng mtric trn X.

21


22/88

a. d1(x,y) =),(1

),(

yxd

yxd

+

b. d2(x,y) = min(1, d(x,y))1.4. Cho X l mt khng gian mtric v (xn)n l mt dy trong X. Chng

minhxn x0 khi v ch khi mi ln cnx0u cha tt c cc xn ngoi tr mt s

hu hn xn. (Khi nim ln cn xem 2.1.1)1.5. Gi s(un)nl mt dy s thc, un 0 v un 0. Chng minh rng tnti v s n sao cho vi mi m n th un um

1.6* Cho (xn) l mt dy trong khng gian mtric X. Chng minh rng nuba dy con (x2n), (x2n+1) v (x3n) u hi t dy th (xn) cng hi t.

1.7. Trong khng gian C[0,1] kho st s hi t ca cc dy sau:a. xn(t) = t

n

b.xn(t) =sin nt

n

1.8. Cho X Y l tch ca hai khng gian mtric (x, dX), (Y, dY). Chngminh dy (xn,yn)n trong X Y hi tn (x,y) X Y khi v ch khi xn xtrong X v yn y trong Y.

22


23/88

2.TP MV TPNG

2.1. Cc nh ngha. Gi s X l mt khng gian mtric

2.1.1. Ln cn. Cho a l mt im ca X.

a. Ta gi hnh cu mtm a bn knh r> 0 trong X v k hiu B(a,r) l tp{x X : d (x,a) < r} cng cn gi l r- ln cn ca im a.

b. Tp U X c gi l mt ln cn ca im a nu U c cha mt r- lncn no ca a. Tp tt c cc ln cn ca a k hiu l N(a). Ni cch khc.

(U N(a)) (r> 0 : B(a,r) U)Theo nh ngha, cc r-ln cn ca a cng l ln cn ca a.

2.2.1. V tr tng i ca mt im i vi mt tp:

Cho A l mt con ca X v x l mt im ca X. C ba v tr tng i ca

im x i vi A nh sau:a. C mt ln cn ca x cha trong A. Khi x c gi l im trong ca

A (hnh 1).b. C mt ln cn ca x nm hon ton ngoi A tc l tn ti U U (x) sao

cho U A = . Lc ny x c gi l im ngoi ca A.(R rng U Ac = X \ A nn x li trthnh im trong ca phn b Ac ca A).

(hnh 2)

Hnh v trang 24

c. Bt c ln cn no ca x cng c cha nhng im ca A v nhng imca Ac, tc l vi mi U N(x): U A = v U Ac = . Khi x gi lim bin ca A. Theo nh ngha, lc x cng l im bin ca tp Ac. (hnh3)

2.1.3. Tp mv tp ng:

a. Tp m: Tp A X c gi l tp mnu A khng cha im bin noc.Cc mnh sau y tng ng vi nh ngha:i. (A m) (x A: X l im trong ca A)ii.(A m) (x A r >0 : B (x,r) A)iii.(A m) (x A, UN(x) : U A)

23


24/88

Nhn xt:

1. Theo mnh i) ta c tp X v l cc tp m2. Ta thng dng mnh ii) kim tra mt tp l m.

b. Tp ng: Tp A X c gi l tp ng nu A cha tt c cc imbin ca n.

T cc nh ngha trn ta suy ra c:a. (A ng) (Ac = X\ A l m)Tht vy, v tp cc im ca A v Ac trng nhau nn nu A cha tt c cc

im bin c n th Ac khng cha im bin no v ngc li.b. Cc tp v X cng l cc tp ng. Tht vy, v theo a) cc tp Xc =

v c = X l cc tp m.2.1.4. V d.1. Trong khng gian mtric tu mi hnh cu mu l tp m.Chng minh. Gi s B (a,r) l hnh cu m tm a bn knh r trong X. Khi

vi mi x B(a,r) ta c d(x,a) < r. t = r - d(x,y) > 0. Xt nhnh cu mB(x,). Ta chng minh B(x,) B(a,r). Nu y B(x,) th d(x,y) < . Khi d(y,a) d(x,y) = d(x,a) < + d(x,a) = r

Nn y B (a,r). Vy B(a,r) l tp m.2. K hiu B(a,r) l tp hp { x X: d(x,a) r} vi r l s dng kv gi

n l hnhcu ng. Ta c B(a,r) l tp ng v bng l lun tng t v d 1 tathy X\ B(a,r) l tp m.

3. Tp gm mt im trong bt k khng gian mtric no cng l tp ngv lun lun cha cc im bin ca n.

4. Gi s a, b l hai s thc. Cc tp (a,b), (a,+ ) l m: cc tp [a,b], [a,+] l ng trong IR.

Lu : Trong mt khng gian mtric tu X ta c:1. (A m) (Acng)2. C th c nhng tp khng mm cng khng ng.3. C nhng tp va m, va ng (chng hn, cc tp , X)

2.2. Cc tnh cht ca tp mv tp ng.

2.2.1. nh l. Trong mt khng gian mtric bt k X ta c:

a. Hp mt h tu cc tp ml tp m.

b. Giao mt h hu hn cc tp ml tp mChng minh:

24


25/88

a. Gi s l mt h cc tp m. t A = .Iii)A( UIi

iA

Nu x A th tn ti

i0 I io Aio. V Aio mnn c s dng r sao cho B(x,r) Aio. Khi B(x,r) .Vy A l tp m.U

IiiA

b. Nu Ai,,An l cc tp mta t A = . Vi xI

n

iiA

1= A ta c x Ai vi

mi i = 1,...,n. Mi Ai l tp mnn tn ti cc s dng ri sao cho B(x,ri) Ai.t r = min {r1,,rn} > 0, khi B(x,r) B(x,ri) Ai vi mi i = 1,...,n. Do

B(x,r) hay A l tp m.In

iiA

1=

2.2.2. nh l: Trong mt khng gian mtric bt k ta c:a) Hp mt h hu hn cc tp ng l tp ng

b) Giao mt h tu cc tp ng l tp ngChng minha. Gi s F1, F2,,Fn l cc tp ng.Khi cc tp 1

cF ,, cnF l m. Theo

cng thc De Morgan, (1

n

i

i

F=

U )c =

1

n

i=I ciF . p dng nh l 2.2.1 ta suy c

(1

n

i

i

F=

U )c l tp mnn

1

n

i

i

F=

U = ((1

n

i

i

F=

U )c )c l tp ng.

b. Chng minh tng t a).Ch : Giao mt h v hn cc tp mni chung cha chc l mt tp m.

Chng hn, ta xt h Gn= (-1

n ,1

n ) cc khong mtrong tp mtrong IR. Khi y

={0} li l tp khng m. Tng t, hp mt h bt k cc tp ng cha

chc l tp ng. (Ly v d, chng hn xt h F1

ni

G

=I

n= = (-cnG , -1/n] [1/n; + ))

2.3 imt, im dnh.

2.3.1. nh ngha. Cho A l tp con ca X. Ta gi im xX l im tca tp A nu bt k ln cn no caxu c cha v sim ca tp A.

2.3.2. V d.

1. Trong IR cho tp A = { 1, 12

, 13

, 1n

,}. Khi y A c im t duy nht l

im 0. Mi im thuc A u l im dnh ca n nhng khng phi l im tca A.

2. Mi im ca tp B = (0,1] u l im t ca B.

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26/88

2.3.3.nh l. imx X l im t ca tp hp A khi v ch khi bt kln cn no caxu c cha mt im ca A khc vix.

Chng minh. iu kin cn l hin nhin. Ta chng minh u kin . Gis bt k ln cn caxu c cha mt im khc vi x. Cho U l mt ln cnca x, ta chng minh trong U c cha v s cc phn t ca A. Theo nh ngha

ca ln cn, tn ti s dng r1 sao cho B(x,r1) U. Gi x1 A B(x,r1), x1 x. Ly s dng r2 < d(x,r1). Xt hnh cu mB(x,r2). Chnx2 A B(x,r2),x2 x. Hin hinx2 x1. Bng qui np, ly s dng rn < d(x, xn1) v chn cxn A B(x, xn), xn x vi mi n N. Ta thy rng l vi n n

thxn xn.Nh th trong U c cha v s phn txn ca A. Vy theo nh ngha,x l imt ca tp A.

2.3.4. nh ngha. im x X c gi l im dnh ca tp A X nubt k ln cn no caxu c cha mt im ca A.

2.3.5. Nhn xt.

1. im t hoc im dnh ca tp hp A th khng nht thit phi thucA.

2. Nu x l im t ca tp A th x l im dnh ca A. Ngc li nichung khng ng.

3.x l im t ca A khi v ch khi tn ti mt dy (xn) ca A vixn xnkhi n n, hi t vx.

Chng minh.

iu kin : Gi sU l mt ln cn cax. Khi tn ti r > 0 : B(x,r) U. Doxnx nn vi r > 0trn tn ti n0 xnB(x,r) vi mi n n0 . V

n n

thxn xnnn trong Ucha v s cc im ca A.iu kin cn: Lp lun nh trong chng minh iu kin ca nh l

2.3.2 bng cch chn rn 0 B(x,r) A = . Do Ac mtc l A ng.

H qu sau y c dng thng xuyn kim tra mt tp hp l ng.

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2.3.7. H qu. Tp A ng khi v ch khi vi mi dy (xn) A mxn xthx phi thuc A.

Chng minh. Suy trc tip tnh l 2.3.6 v nhn xt 3,4 mc 2.3.5.

2.4. Phn trong v bao ng ca mt tp.

2.4.1. Phn trong. Cho A l mt tp con ca X. Lun lun c mt tp mcha trong A, chng hn tp .

a) nh ngha. Hp tt c cc tp mcha trong A c gi l phn trong

ca A; k hiu l0

A hay int A.

Hin nhin0

A A

Nh th0

A l tp mln nht cha trong A theo nh ngha nu G l tp

mv G A th G 0

A

Tnh ngha ta c ngay:

(A m) (A =0

A)

b. nh l: Phn trong0

ca tp A l tp hp ca tt c cc im trongca A.

Chng minh. Gi sx 0

A . V0

A mnn0

A lmt ln cn cax do xl im trong ca A. Ngc li nu x l im trong ca A th c r < 0 hnh

cu mB(x,r) A. Theo nhn xt sau nh ngha th B(x,r) 0

A . Vy x0

A .

2.4.2. Bao ng. Nu A X th c t nht mt tp ng cha A (V d X A)

a) nh ngha. Giao tt c cc tp ng cha A c gi l bao ng catp A. K hiu l A .

Hin nhin A l tp ng b nht cha A.

b) nh l. Bao ng ca tp A bng hp ca A v tp tt c cc imbin ca A.

Chng minh: K hiu A l tp tt c cc im bin ca A A . Ta chng

minh A = A A. Nhn xt rng vi mi tp ng F A th tng ng vimi tp mG = Fc Ac v ngc li. Do :

I UIAdongF AmoG

c

AmoG ccGA

=== )(GF c

Vy

27


28/88

x A x U , GG m Ac

x khng l im trong ca Ac

x A hay x A.T: A = A A

H qu.1. ( A ng) (A = A )2. A l tp hp tt c cc im dnh ca A.3. (x A)((xn) A : xnx)

2.4.3. Cc v d.1. Gi sa, b l hai s thc. t A = (a,b] khi

0A = (a,b), A = [a,b],

0

A =(a,b).2. Bao ng tp cc s hu t trong IR chnh l tp IR.

3. Trong khng gian mtric bt k ta u c r)(a,B B

(a,r).2.5. Tp hp tr mt khng gian kh ly.

2.5.1. nh ngha. Gi s A,B l hai tp con trong khng gian mtric X.Nu B A th ta ni tp A tr mt trong tp B.

2.5.2. Nht xt.1. Tnh ngha, ta thy (A tr mt trong B) (Vi mi x B, x l

im dnh ca A). iu ny tng ng vi tn ti dy (xn) A,xn x.2. Nu A Bth BA . Do , nu A tr mt trong B; B tr mt trong

C th A tr mt trong C.Tht vy, ta c C B v B A nn C B AA = 3. Nu A X v

A = X th tp A c gi l tp tr mt khp ni (trong X)V d: Trong IR, tp s hu tQ tr mt khp ni.2.5.3. nh ngha. Mt khng gian mtric X c gi l kh ly nu tn ti

mt tp hp hu hn hoc m c tr mt khp ni.C th chng tp (Q l tp s hu t) l m c v tr

mt trong IR

44 344 21

lank

k Q...QQ =

k

. Do IRk

nh l mt v d v khng gian mtric kh ly.

2.6. Tp mv ng trn ng thng thc.

2.6.1. nh l. Mi tp mtrong IR bng hp mt s hu hn hay mc cc khong mkhng giao nhau.

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Chng minh. Gi s G l mt tp m trong IR . Vi x G tn ti r > 0:B(x,r) = (x- r, x+ r) G. K hiu x l hp tt c cc khong mcha trong Gv c cha x. Ta chng minh x l mt khong m.

Tht vy, t p = infx, q = sup x (p,q c th bng (- , +). Vi mi y x th p < y < q v trc ht r rng ta c p y q.

Nu y = p th c mt khong mchax v cha c p nn mu thun vi p= infx. Tng t y khng th bng q. Vy x( p,q). Ngc li nu y ( p,q),gi s p < y < x. Theo nh ngha ca infimum .

y xt

tn ti t : p < t y < x. Do c mt khong mchax v cha lun c t. Vth y thuc khong mny tc l y

x

x. Vy x = (p,q).

qp

By gita xt tt c cc khong xng vi cc im x G. Hin nhin G= . Nhn xt rng nu zUGx

x

x th x x (V z l khong m ln nht

cha x). Cho nn vi 2 khong mx v y th hoc xy = hoc x = y(v nu c z xy th x = y = z ).

Vy G bng hp ca nhng khong m ri nhau. Trong mi khong m ta chn 1 s hu t. V tp cc s hu tm c nn s cc khong mlpthnh G l hu hn hay m c. nh l c chng minh xong.

Do mi tp ng l phn b ca tp mnn ta c:

2.6.2. H qu. Mi tp ng trn IR l phn cn li sau khi rt khi IR mts hu hn hay m c cc khong mri nhau.

Cc khong mny c gi l cc khong k ca tp ng .

2.7. Tp mv tp ng trong khng gian:

Gi s X l mt khng gian mtric, Y l khng gian con ca X v A l mttp con ca Y. rng, nu A l mt tp m (hay ng) trong Y l chachc A l m(hay ng) trong X. Tuy nhin ta c:

2.7.1. nh l.iu kin cn v l tp A m trong khng gian mtriccon Y l tn ti tp mG v X sao cho A = G Y.

Chng minh. K hiu BX(a,r), BY(a,r) ln lt l cc hnh cu mtrong Xv Y tng ng. Nu a Y th BY(a,r) = {y Y : d(a,y) < r} = Y B(a,r). Gis A l tp mtrong Y, khi vi mi x A tn ti rx > 0 sao cho BY(x,rx) Y.t G = x

i A

BU (x,rx), tc l G bng hp ca mt h cc tp m(trong X) nn n

l tp mtrong X. Hn na,

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GY)r,x(BYY)r,x(B)r,x(BA xAx

XxAx

XxAx

Y =

===

UUU Ngc li,

cho A = G Y vi G l tp mtrong X. NuxG A th do G mnn tn ti r> 0 sao cho BX(x,r) G. Thnh ra B Y(x,r) = BX(x,r) Y G Y = A hay A mtrong Y.

2.7.2. nh l. iu kin cn v tp A ng trong Y l tn ti mttp ng F trong X sao cho A = Y F.

Chng minh. Tp A ng trong Y khi v ch khi Y \ A l mtrong Y. Theonh l 2.7.1, tn ti tp mG trong X sao cho Y \ A= G A. Khi

A= Y (X\G) = Y Fvi F = X\G l tp ngT cc nh l trn ta d dng suy ra h qu sau.2.7.3. H qu. mi tp con A Y m(t. ., ng) trong Y cng l m

(t. ., ng) X, iu kin cn v l Y l tp m(t. ., ng) trong X.

BI TP

2.1. Gi s X l khng gian mtric, A X v x X.a) Chng minh rng x l m dnh ca A khi v ch khi d(x,A) = 0Suy ra:

(A ng) (d(x,A) = 0 x A)

b) Cho > 0 chng minh{x X : d (x,a) < } l tp m{x X : d (x,a) } l tp ng

2.2. Cho F1, F2 l hai tp ng trong khng gian mtric X sao cho F1 F2=

a. Chng minh tp G = {x X : d (x,F1) < d (x,F2)} l tp m, ng thiF1 G, G F2 = b. T a suy ra c cc tp mG1, G2 sao cho F1 G1, F2 G2 v G1 G2 =

2.3*

Mt tp A trong khng gian mtric X c gi l mt tp kiu G vtp mkiu F2.4. Gi s A, B l cc tp con ca khng gian mtric. Chng minh:a. int(int A) = intA, A = A .

b. Nu a B th0

0

B

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c. int (a B) =0

A . Int (AB )0

B 0

A 0

B

d. BA = A B , BA A B 2.5. Chng minh rng mi khng gian con ca khng gian mtric kh ly l

kh ly.2.6. K hiu c0 l tp hp tt c cc dy s thc hi t v 0. Ta xem c0 nh

l khng gian con ca khng gian c (bi tp 1.2). Chng minh c0 l khng giankh ly.

2.7. Gi s X l khng gian mtric v Y l khng gian con ca X sao choY = U V vi U, V l cc tp m, khc trng trong Y v U V = . Chngminh tn ti cc tp mA, B trong X, A B = v U = A Y, V = B Y.

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3. NH X LIN TC

3.1. nh ngha v cc tnh cht chung.

Cho hai khng gian mtric (X,d1) v (Y,d

2). Nu khng s nhm ln, ta

dng k hiu d ch c d1 ln d2. Gi s f l mt nh x t X vo Y v x0 l mtim ca X.

3.1.1. nh ngha.

1. nh x fc gi l lin tc ti x0 nu mi > 0 cho trc, tn ti > 0sao cho d(f(x), f(x0)) < vi mi x X m d(x, x0) < .

nh ngha ny thng gi l nh ngha v tnh lin tc bng ngn ng,.

2. nh x fc gi l lin tc A X nu f lin tc ti mi im x A.

Mt tiu chun tng ng vi nh ngha trn thng dng kho sttnh lin tc mt cch c hiu qu nh sau:

3.1.2. nh l. (Tiu chun qua dy). nh x f lin tc ti x0 X khi vch khi mi dy (xn)n X, nuxn x 0 th dy f(xn) f(x0).

Chng minh:iu kin cn: Gi s f lin tc ti x0 v (xn) l dy trong X sao choxn x 0.

Ta hy chng minh f(xn) f(x 0) trong Y. Cho > 0, v f lin tc ti x0 nn c > 0.

d(f(x),f(x0)) <

khi d(x,x0) 0 trn, c n0 d(xn, x0) < khi n n0.Nhng lc th

d(f(x),f(x0)) < Vy f(x) f(x0).

iu kin : Gi s f khng lin tc ti x0. Khi tn ti > 0 sao chovi mi > 0 tn tix X : d(x,x0) < m d((f(x),f(xo)) . Ly ln lt bng 1,1

2,, 1

n, s c x1, x2,,xnthuc X tho mn d(xn,x0) >=

000

0 < d (x,x0) < d(f(x), l) < )3.1.4. nhl. Cho X, Y l hai khng gian mtric v f : X Y l mt nh

x. Cc mnh sau y l tng ng.

a) f lin tc trn X.b) vi mi tp ng F Y th f-1(F) l tp ng trong X.c) vi mi tp mG Y th f -1(G) mtrong X.d) )A(f)A(f vi mi tp A X

Chng minh.

a) d). Gi s y f( A). Khi tn ti x A y = f(x). Theo tnh chtca bao ng, tn ti dy (xn) A:xn x v v f lin tc nn f(A) f(xn) f(x)= y. Vy y ( )f A .

d) b). Gi s F ng trong Y. t A = f-1

(F). Khi y ta c f(A) F vth ( )f A F. Mt khc, rng nu E l tp con ca X ta lun lun c E f-1(f(E)). Do ly E = A , ta c.

A f-1(f(A )) f-1 ( ( )f A ) =AVy A = A = f-1(F) l tp ng

b) c). Nu G mtrong Y th Y/G = G c. T f -1(Y/G) = X\f-1(G) ngtrong X nn f-1(G) l m(trong X).

c) a). Gi s x X v > 0 bt k. Do B(f(x 0),) m trong Y nn f-

1(B(f(x0),) l tp m trong X cha x0. V th c s > 0 B(x0,) f-

1

(B(f(x0),)). iu ny c ngha l nu x X sao cho d(x,x0) < hayx B(x0,)nn f(x) B(f(x0,) hay d(f(x), f(x0)) < tc l f lin tc ti x0 theo nh ngha.Vy nh l c chng minh y .

3.1.5. nh l. Gi s X, Y, Z l ba khng gian mtric, f : X Y lintc tix0, g : Y Z lin tc tiy 0 = f(x0). Khi nh x hp h = g o f : X Zlin tc tix0.

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Chng minh: Gi s (xn) X vxn x0. Do f lin tc ti x0 nn f(xn) f(x0)= y0 v lc y g lin tc tiy0= f(x0) suy ra g(f(xn)g(y0) = g(f(x0). Ni cch khc(g o f)(xn)(g o f)(x0).Vy h = g o f lin tc tix0.

3.2. nh xng phi.

3.2.1. Php ng phi. Cho X, Y l hai khng gian mtric. Gi s f :X Y l mt song nh sao cho f v f-1u l cc nh x lin tc th fc coil mt php ng phi t X ln Y. Hai khng gian mtric c gi l ng phivi nhau nu c php ng phi t khng gian ny ln khng gian kia.

V d.1. Ly X = (a,b), Y = (0,1) l hai tp con ca tp s thc IR, khi X, Y

ng phi vi nhau nhphp ng phi.

f(x) =ab

a

ab

x

2. Cho X = IR, Y = (0,1) cng vi mtric thng thng th chng ng phi

vi nhau nhphp ng phi.F(x) = 1

actg x 1

2.

Nhn xt.1.Theo nh l 3.1.4, mt php ng phi bin mt tp m(t.., ng) trong

khng gian ny thnh tp m(t.., ng) trong khng gian kia.2. C th chng minh d dng rng cc nh ngha v ln cn, im t, im

chnh, bao ng, phn trong, tp tr mt, bt bin qua php ng phi, nghal cc tp A X cc im x X c tnh cht k trn th qua nh xng phi f,

cc tp f(A), cc im f(x) cng c tnh cht . Cn nhng khi nim v hnhcu, khong cch, bn knh, khng phi bt bin qua php ng phi.

3.2.2. Php ng c.

Cho X, Y l hai khng gian mtric. Mt song nh f t X ln Y gi l mtphp ng c nu vi mix, x X ta c d(f(x), f(x)) = d(x,x). Hin nhin lc f -1 : YX cng l php ng c v ta gi X, Y l hai khng gian ng c vinhau.

Nhn xt.1) Nu f l php ng c t X ln Y th r rng f l php ng phi gia X

v Y.2) Cho X l mt khng gian mtric, Y l mt tp bt k. Gi s c mt song

nh f : Y X . Khi nu t d *(y, y) = d(f(x), f(y) th d* l mt mtric trn Yv hn na X, Y l hai khng gian mtric ng c.

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3) Theo quan nim ca khng gian mtric, nu X v Y ng c th chngc ng nht vi nhau.

3.2.3. Mtric tng ng.

Cho d1,d2 l hai mtric trn cng mt tp X. Khi ta c hai khng gianmtric khc nhau (X,d1) v X,d2) c chung tp nn X.

Hai mtric d1,d2c gi l tng ng tp nu nh xng nhtid: X X

x a xl mt php ng phi t khng gian (X,d1) ln (X,d2)

Nu tn ti cc s dng m, M sao chomd(x,y) d2(x,y) Md1(x,y)

vi mi x, y Y th d1,d2c gi l hai mtric tng ng u.Nhn xt.1. Tnh ngha ta suy ra nu d1, d2 tng ng th chng s tng ng

tp nhng iu c ngc li ni chung khng ng.2. Hai mtric tng ng tp th cc tp m(t..,ng) trong hai khnggian ny trng nhau. Tt nhin cc khi nim khc dn xut t tp m cngtrng nhau.

Hai mtric tng ng u th thm na l cc tnh cht nh tnh linquan n khong cch cng s bt bin.

3.3. Suy rng cc nh x lin tc.

Gi s X, Y l cc khng gian mtric v f l nh x t X vo Y. Nu f lintc vi mi A X, thu hp ca f ln A, k hiu fA : A Y f A(x) = f(x) cng l

nh x lin tc trn A. Ngc li cho h : A Y lin tc th vi iu kin no tnti nh x. F : X Y lin tc, duy nht v f A = h?

Trc ht ta thit lp cc nh l suy ra tnh duy nht ca suy rng.

nh l 3.3.1. Gi s f, g l hai nh x lin tc t X vo Y. Khi tp hpA = {x X : f(x) = g(x)}

l tp ng trong AChng minh: Gi s x0 A . Khi tn ti ti dy (xn) A sao chox n x0.

Theo tiu chun qua dy ta c f(xn) f(x 0)v g(xn) g(x0). V xnA nn f(xn)= g(xn) vi mi n N nn f(x0) = g(x0) do gii hn ca mi dy hi t l duynht. Vyxn A hay A = A , c ngha l A ng

3.3.2.H qu. Gi s f, g l hai nh x lin tc t X vo Y. Nu f(x) = g(x)vi mixX

Ta cX = A D = D A

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Vy D = X hay f(x) = g(x) vi mi xX

3.3.3. nh l. cho X, Y l hai khng gian mtric, A l tp con tr mttrong X v f l nh x lin tc t A vo Y

iu kin cn v tn ti nh x f : X Y lin tc, tho mn f A = fl tn ti vi mi x X . Khi nh x)z(flim

Az

f duy nht

Chng minh. Trc ht ta din t li khi nim gii hn nh nh ngha3.1.3, nhdy, sau y.

( = l))z(flimAz

((zn) A : (zn a) (f(z n) l).

iu kin cn. Gi s tn ti f lin tc v f A = f. Khi x X v (zn) A sao cho zn x th f (zn) f (x). Nhng v f(zn) = f (zn) nn f(zn) l =f (x) tc l gii hn tn ti vi mi xX.)z(flim

Az

iu kin . Vi mi x X t f (x) = . Nu x A hin nhin)z(flimAz

f (x) = f(x) tc l f A = f. Ta chng minh f lin tc.Gi s x X v (xn)n l mt dy trong X hi tnx. Theo cch t, ta c

f (xn) = . Do , theo iu din t li ni trn, vi mi n N tn tiz)z(flimAz

n

A sao cho d(zn, xn) 0 s tn ti

n

[ baC ,

0 sao cho vi mi m, n n0 v vi mi t [a,b] ta c

| xn(t) xm(t)| < (1)Cho m(1), ta c | xn(t) xn(t)| < khi n n0 v vi mi t [a,b].Vy xn(t) hi tu n x(t) trn [a,b], lin tc trn [a,b], tc l x(t) ,

ng thi x[ ]b,aC

n x. Do l khng gian y .[ b,aC

4. Khng gian khng y [L

baC ,

Chng minh. Ta xt trng hp [a,b] = [0,1] v xt dy xn(t) nh sau: (hnh5)

++

+

=

ntkhintn

tn

khi

tkhi

txn

2

1

2

1

2

121

12

1

2

10

2

1,01

)(

Vi m, n N, (m > n), ta c

d(xn,xm) = = dttxtx mn1

0

)()( dttxtxn

mn+

2

1

2

1

2/1

)()(

V |xn(t) xn(t)| 2 nn d(xn,xm) 01 n

khi m, n

Vyxn(t) l mt dy cbn

Tuy nhin ta chng minh rng dy xn(t) khng hi t trong ] . Tht vy

x(t) l mt hm bt k trong . Xt hm s gin on trn [0,1] nh sau[L

baC ,

[ ]LC 1.0

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=

1,2

1,0

2

1,0,1

)(

t

t

ty

Nh thx(t) y(t) nn phi c t0 21,0 chng hn y(t0) x(t0). Hn

na, trn

2

1,0 c hai hmx(t) vy(t) cng lin tc nn l lun nh v d 1.2.5,

ta c:

0 sao cho d(xn,xm) 0 B(x,r) M. Vy bt k hnh cu no cha trong B(x,r) ucha trong M nn phi c giao vi M khc trng.

4.3.3. nh ngha. Cho A l tp con ca khng gian mtric X. Tp A cgi l tp thuc phm tr I trong X nu tn ti dy cc tp tha M1,M2, sao

cho A =1

i

i

M

=

U .

Nu A khng phi l tp thuc phm tr I th A gi l tp thuc phm tr II.

4.3.4. nh l. (Baire) Gi s X l mt khng gian mtric y . Khi Xtp thuc phm tr II.

Chng minh. Dng phn chng. Gi s X thuc phm I, khi tn ti dy

tp tha An X sao cho X =1

n

n

A

=U . Do A1 tha nn c hnh cu ng B1 bn

knh r1< 1 sao cho B1 A1 = . Cng vy, v A2 tha, tn ti hnh cu ng B2

B1 bn knh r2 0 ( ni chung ph thuc vo x0 v ) sao cho d(f(x),f(xo)) < mi khid(x,x

0) < . Nu s dng khng ph thuc vo mi imx

0ta c khi nim

lin tc u.

4.4.1 nh ngha. nh x f : X Y c gi l lin tc u trn X nu vimi > 0 tn ti > 0 sao cho vi mix, x

X m d(x,x) < th (d(f(x),f(x))

0 cho trc tu , do flin tc u trn A nn tn ti s dng sao cho (d(f(z),f(z) < khi d(z,z) < vi miz, z A.

Vzn x nn dy (zn)n l dy cbn : vi > 0 trn tn ti s nguyn n0 (zn,zm) < vi mi m, n n 0. Nhng khi d(f(zn),f(zm)) < . Nh th (f(zn))nl mt dy cbn trong Y. Hn na, Y y nn tn ti (f(zlim

xn)) = l. Vy ta

c c nh xf: X Y mrng lin tc duy nht ca nh x f.

Cn li ta chng minh f lin tc u trn X. Li ng dng tnh lin tc u

ca f trn A nh trc, gi sx0 vx0

l hai im trong X vi d(x0,x0

) < . Xthai dy (zn) v (zn) trong A ln lt hi tnx0 vx0

. Chn n0 ln ta thyrng nu n n 0 th

d(zn,zn) d(zn,x0) + d(z

n,x0)

2

+

2

= .

Do vy:

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d( f (zn), f (zn)) = d(f(zn),f(zn

)) < Cho n ta c d( f (x0), f (x0

)) .Vy f lin tc u trn X.

4.5.Nguyn l nh x co.

4.5.1. nh ngha. Cho nh x f t tp X bt k vo chnh n. Phn t x

X sao cho f(x) =xc gi l im bt ng ca nh x f.Vic tm im bt ng ca mt nh x l vn c nhiu ng dng trong

gii tch, c bit trong l thuyt phng trnh (vi phn, tch phn) v mtim bt ng ca nh x f l mt nghim ca phng trnh f(x) =x.

By gicho X l mt khng gian mtric v f l mt nh x t X vo X, fc gi l nh x co nu tn ti mt s[0,1] sao cho vi mix, y X ta c:

d(f(x),f(y)) d(x,y).Tnh ngha ta thy ngay mi nh x co l lin tc u.

4.5.2. nh l (nguyn l nh x co Banach). Gi s X l mt khng gianmtric y f : X X l mt nh x co. Khi f c mt im bt ng duynht.

Chng minh. Ly mt im tu x0X. tx1= f(x0),x2= f(x1)= f(f(x0)),...,xn = f(xn-1) = f(f...f)(x0),...

N lnTa chng t (xn) l mt dy cbn trong X.V f l nh x co ln nu n 1 th

d(xn,xn+1) = d(f(xn-1),f(xn) d(xn-1,xn)= d(xn-2,f(xn-1))

2d(xn-2,xn-1)) nd(x0,x1)Vi [0,1]Lc vi mi s nguyn n vp,t (*) ta c

d(xn,xn+p) d(xn,xn+1) +... + d(xn+p,xn+p)

(2+ n+1++

n+p-1) d(x0,x1) 1

n

d(x0,x1)

Khi n ln vp ty ta c d(xn,xn+p) 0 (n ) suy ra (x n)n l dy cbn trong khng gian y X nn tn ti gii hn x*= lim

nxn .

Cng t (*), ta cd(xn,x+1) d(xn, xn+1) nd(x0,x1).

Cho n v nhrng cc hm d v f lin tc,ta c 0 d(xlim

xn,f(xn)) = d(x

*,d(x*) 0

hay d(x*,f(x*)) = 0Vy f(x*) = x* tc l x*l im bt ng ca f.

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Nu cy X m f(y) =y thd(x*,y) = d(f(x*), f(x) d(x *,y

*)hay (1-)d(x*,y*) 0

Suy ra d(x*,y) = 0 tc l x*= y. Do im bt ng x* l duy nht

4.5.3.V d.

a.Chng minh phng trnh1

2arctgx x + 3 = 0

c mt nghim thc duy nht

t f(x) = 12

arctgx + 3 l hm t IR vo IR, f l nh x co v f(x) f(y) = (x-

y)2

1

2(1 )+, vi nm gia x v y theo nh l lagrange, nn

1( ) ( )

2

f x f y x y . Do f c mt im bt ng vi duy nht x*. Ni cch

khc, phng trnh 12

arctgx + 3 =x c nghim duy nht l x*.

b) Xt phng trnh vi phn

(1) dxdy

= f(t,x) vi iu kin ban u

(2)x(t0) = x0trong f(t,x) l hm lin tc trong tp mG IR2, (t0,x0) G v f l tho mniu kin Lipschitzx, ngha l c mt s dng ksao cho

f(t,x1) f(t,x2) Kx1 - x2) (3)vi mi (t,x1),(t,x2) GTa chng minh nh l Picard: Trn mt on t t0 r no phng

trnh (1) c mt nghim duy nht tho mn iu kin ban u (2).

Phng trnh (1) vi iu kin (2) tng ng vi phng trnh tch phnsau y

x(t) =x0 + (4)t

t

d))(x,(f

0

Do G l tp m cha (t0,x0) nn c hnh trn (hnh cu) tm (t0,x0) cha

trong G. GiD = {(t,x)G :t t0 a, x x0 b}

l mt hnh ch nht ng bt k ni tip trong hnh cu . V f(t,x) lin tctrn D nn ( , )f t x L vi mi (t,x) D vi L l s dng no . Ly 0 < r 0 tn tixn

X d(xn,x

*

n) < 1

n. Khi

d(xn,xm) d(xn,xn*) + d(xn

*,xm

*) + d(xm*,xm)

< 1n

+ 1m

+ d(xn*,xm

*)

Vy d(xn,xm) 0 (m, n) nn (xn) l dy cbn trong X tc l (xn) X.tx* l lp cha dy (xn) thx

* X*, ta c:d(xn

*,xn) d(xn

*,xn) + d(xn,x

*)1

n+ d(xn,x

*)

V (xn,xn,...) x+

n v (x1, x2,...) x*

nnd(xn,x*) = d(x

mlim n,xm)

Vy

d(xn*,x

*) 1n

+ d(xm

lim n,xm)

Cho ta c d(xm n*,xn) 0 hayx

*n x

*

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Nh th X* l y 5. Tnh duy nht (sai khc mt php ng c) ca y ho X* ca X.Gi s X* l mt khng gian mtric y v c cc tnh chti) X X* (X ng c vi mt khng gian con ca X*)ii) X = X*

Ta hy chng minh X*ng c vi X*Lyx* X* khi y tn ti (xn) X sao choxn X*. t : X* X*

X* x*

y l php ng c phi tm. Tht vy l ton nh v nu y* X* thy*

= yn

lim n, yn X nny* =(y*) vi y*= y

nlim n,y

* X*

Mt khcd(x*,y*) = (x

nlim n,yn) trong X

*

v d(x*,y*) = (xn

lim n,yn) trong X*

nnd(x*,y*) =d(x*,y*) = d((x*),(y*)

Vy l n nh v t cc iu va chng minh trn ta suy ra l phpng c. nh l c chng minh y .

4.6.3.V d.1. Tp hp Q cc s hu t l mt khng gian mtric vi khong cch:

d(x,y) = x y ;x, yQPhng php xy dng tp s thc IR bng dy cbn cc s hu t chnh l

y ho khng gian mtric Q nh trnh by trn. Tuy nhin trong IR tacn phi xy dng cc php ton + v . ng thi phi kim nghim li IR l mttrng lin tc.

2. Gi s M l tp con ca khng gian mtric y X. Khng gian y ho ca khng gian con M chnh l M.

3. Ta c l khng gian mtric khng y . Khng gian y ho

ca n k hiu l L[L

b,aC ]

[a,b]l tp hp cc hm o c xc nh v kh tch theo

ngha Lebesgue trn [a,b] (l thuyt tch phn Lebesgue s hc phn L thuyto v tch phn).

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BI TP

4.1.Cho (X,dX) v (Y,dY) l hai khng gian mtric.Chng minh rng khnggian mtric tch X Y y khi v ch khi cc khng gian X v Y l nhng

khng gian mtric y .4.2.K hiu m l tp hp tt c cc dy s thc b chn. Vix = (xn)m, y

= (yn) mtd(x,y) = nn

n

yxsup

Chng minh

a) d l mt mtricb) Khng gian mtric (m,d) l y

4.3.* Trong khng gian mtric C[0,1] hy xy dng mt dy cc tp ng, b

chn Fn sao cho Fn Fn+1 nhng = I

=1nnF

4.4.* Cho X l khng gian mtric y , Yi l mt dy cc tp con mtr

mt khp ni trong X. Chng minh tp hp cng l tp tr khp ni trong

X.1

i

n

Y

=I

4.5. K hiu s0 l tp hp tt c cc dy s thc (xn) sao choxn bng 0 tt c

ngoi tr s hu hn n. Ta xem s0 l khng gian mtric con ca khng gian m.Chng minh s0 l khng gian mtric khng y .

4.6.Chng minh rng hm lin tc f : IR IR xc nh bi

f(x) =2

+ x arctgx

khng c im bt ng mt d n tho mn bt ng thc( ) ( )f x f y x y <

Vi mi x, y IR, x y

4.7.*

Cho F l h cc hm lin tc trn IR c tnh cht( )( 0)( ) : ( )x xx R M f f f x M >

Chng minh rng tn ti mt tp mkhc trng U IR v mt s dng Msao cho

( )f x M

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51/88

Vi mi f F v x U4.8. Cho f: IR IR l mt hm s lin tc u. Chng minh tn ti cc s

, 0 sao cho vi mi x IR, ta c:( )f x x +

4.9*.Trn tp s thc IR ta t

d1(x,y) = | arctgx- arctgy|v

d2(x,y) = | ex - ey|

Chng minha) d1, d2 l cc mtric trn IR

b) (IR, d1),(IR, d2) l cc khng gian mtric khng y c) y ho cc khng gian(IR, d1),(IR, d2) l g?

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5 KHNG GIAN COMPACT

Trong gii tch cin, khi lm vic vi cc tp con ca IR ta c bit ch n khong ng [a,b], a, b IR. Khong ng ny c cc tnh cht tt sau

Mi dy (xn) [a,b] bao gicng c mt dy con hi t trong [a,b]Hm s lin tc trn [a,b] th lin tc u trn on Hm s lin tc trn [a,b] th t c gi tr ln nht v gi tr b nht trn

on ny.Tng qut ho s kin ny cho khng gian mtric, ngi ta a ra khi

nim quan trng l tp compact v khng gian compact.

5.1.Tp hp b chn v hon ton b chn.

5.1.1. nh ngha. Cho X l khng gian mtric v A X.

t (A) = d(x,y) (A) = + nu tp hp {d(x,y),x, yA} khng c

cn trn) v gi (A) l ng knh ca tp AAyx ,

sup

Nu (A) < + th A gi l tp b chn, cn (A) = + th A gi l khngb chn.

Ta c mt s tnh cht n gin saua) Tp hp A l b chn nu v ch nu A c cha trong mt hnh cu no

Tht vy, gi s(A) = d(x,y) = r< + . LyxAyx ,

sup 0 ty thuc A, lc

vi mi y A, ta c d(x0,y) rhay A B(x0,r) vi r > r. Ngc li, A B(x0,r) th x,y A c

d(x,y) d(x,x0) + d(y,y0) 2rDo (A) 2r< + nn A b chn

b. Hp mt s hu hn cc tp b chn l tp b chn.

5.1.2. nh ngha. Tp A X gi l hon ton b chn (hay cn gi l tincompact) nu vi mi r> 0, A c cha trong hp mt s hu hn hnh cu c

bn knh r. Lc ta cng ni tp A c ph bi mt s hu hn hnh cu bnknh r.

Mnh 1. Mt tp hon ton b chn th b chn

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Chng minh. Thc vy, cho A hon ton b chn nn vi r= 1, ta c hu hn

hnh cu A1

( ,1)n

i

i

B x=

U . Ly x A s c i x B(ai,1)

Lc d(x,a1) d(x,ai) + d(ai,a1)

d(x,ai) + =

n

ii n)a,a(d

11

Vy A nm trong B(a1,n). Ngc li, ni chung khng ng. Tuy nhintrong IRn ta c.

Mnh 2: Gi s A l mt chn trong IRn (vi mtric thng thng), khi A hon ton b chn.

Chng minh. Gi B(x0,d) l hnh cu cha A. Ta ly mt hnh lp phngcnh bng 2dsong song vi cc trc to v ngoi tip hnh cu ny. Cho r> 0ta chia hnh lp phng ny thnh cc hnh lp phng nh bi cc mt phng

song song vi mt hnh lp phng ln v cch nhau mt on rn

. S cc

hnh lp phng nh ny l hu hn. Hin nhin A c cha trong hp hu hncc hnh cu bn knh r, ngoi tip cc hnh lp phng nh. Vy A hon ton

b chn.

Ch rng ),( raB B(a,r) v nu r > r th B(a,r) B(a,r) do trongcc nh ngha v tp b chn v hon ton b chn, ta c th s dng tu cchnh cu mhoc ng

Mnh 3. Nu A X l hon ton b chn th bao ng A ca A cng lmt tp hp hon ton b chn

Chng minh. Gi srl s dng cho trc. Nu A hon ton b chn th

c hu hn hnh cu ng B(ai,r) (i = 1, )n sao cho . Nhng do

l mt tp ng nn

Un

ii raBA

1

),(=

Un

ii raBA

1

),('=

A ngha lUn

ii raB

1

),('=

A cng hon ton

b chn.

5.2.Tp hp compact v khng gian compact:

5.2.1.nh ngha. Cho X l khng gian mtric v A X. Tp A c gi lcompact nu dy (x

n) A lun lun tn ti mt dy con (x nk) (x n) hi t mtim x A

Nu bn thn X l compact th khng gian X c gi l khng giancompact

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Gi s A X m bao ng A l compact th A c gi l tp compacttng i. iu ny tng ng vi:

Mi dy tu (xn) A s c mt dy con hi tn mt im trong X(khng i hi thuc A).

Tht vy, iu kin cn r. Ngc li nu (xn)n l dy trong A vi mi x

N ta chnyn A sao cho d(xn,yn) < 1n . Theo gi thit tn ti dy con (ynk) (yn)hi t vy0 X. V (ynk) A nny0A .

Lc

d(xnk,y0) d(xnk,ynk) + d(ynk,y0) 0sao cho khng th phc A bng mt s hu hn hnh cu bn knh . Lyx1 A. Hnh cu B(x1,) khng phc A nn c x2A sao cho d(x2,x1) .Cc hnh cu B(x1,), B(x2,) cng khng phc A nn c x3 A sao chod(x3,x1) v d(x3,x2) . Tip tc cch ny bng quy np, ta xy dng cdy (xn) A vi d(x n,xm) vi mi n m. Bt c dy con no ca dy (xn) ukhng phi l dy cbn (v khong cch gia hai phn t ca dy con ny )nn khng th hi t. iu ny mu thun vi nh ngha compact. Vy ta chng minh nu A compact th A ng v hon ton b chn.

2) By gigi s X y , A ng v hon ton b chn trong X. Ly mtdy (xn) bt k trong A. V A c ph bng mt s hu hn cc hnh cu bnknh bng 1 nn mt trong cc hnh cu , k hiu B1 cha v s cc phn t

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ca dy (xn). Gi dy con ca dy (xn) trch ra t v s phn t l ( ). Tp A

cng phc bng mt s hu hn hnh cu bn knh

1nx

1

2nn c mt cu k hiu

B2 c v s cc (x1n) v dy con ca ( ) cha trong B

1nx 2 l ( ). Tip tc qu

trnh ny ta thu c mt dy cc dy (x

2nx

n), ( ),( ),...m cc dy sau l dy

con ca dy i trc ng thi dy con ( ) cha trong hnh cu B

1nx

2nx

knx k c bn knh

1

k. Ly mt phn t xn1 trong ( ). V ( ) v hn nn chn c n

1nx

2nx n2 trong cc

( ) sao cho n2nx 2 > n1 v ri chn xn3 trong ( ) vi n3nx 3 > n1 Nh th ta c

c dy con (xnk) (xn). y l dy cbn v vi k,l N bt k, gi s k l th

( ) ( ) nn xknx1nx nk, xnl cng thuc hnh cu Bl. Do d(xnk, xnl)

2

l 0 khi k,

l . V X y , dy (x

nk) hi tn x0 X. Hn na do A ng nn x0 A.

Vy A compact.5.2.4 nh l. Nu X l mt khng gian compact th X y v kh ly.Chng minh.

a) Gi s (xn) l dy cbn trong X. V X compact nn c dy con (xnk) (xn),xnk x0 X. Nhng khi chnh dy (xn) cng hi t vx0nn X l khnggian y .

b) Vi mi n N, tp A c ph bng mt s hu hn hnh cu B(xnj,1

n),

j = 1,...,mn :

X = B(xn1

nm

j=U j, 1n ) (*)

t A = {xnj : j = 1,...,mn, n = 1, 2,...} l tp hp tt c cc tm cc hnh cu

ny. A l tp m c. Mt khc x X vi > 0 bt k ta ly n ln 1n

0 c thc tm cho A mt -li hu hn N (tc l tp N c hu hn im).

b) H qu. Trong khng gian mtric y X, tp ng A X l compactnu vi mi > 0 ta c th tm cho n mt -li compact.

Chng minh. Cho > 0. Theo gi thit, tn ti2

-li compact B v cho A

v do B compact nn tn ti2

-li hu hn N cho B. Khi N chnh l -li

cho A v vi x A tn ti bB d(x,b) 0 tn tix,x A vi d(x,x) < nhng d(f(x), f(x)) . Cho ln lt

cc gi tr

1

n, n = 1, 2, ta thu c hai dy (xn)

v (xn) trong A c tnh cht

d(xn,x

n) < 1/n v d(f(x), f(x)) . V A compact, dy (x) c dy con ( )

x

'nkx

0 A. T bt ng thc:

)k()x,x(dn

)x,x(d)x,x(d)x,x(d 'nkk

'nk

'nk

''nk

''nk +


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Qn(x) = cn(1 x2)n 0, n = 1,2,... (1)

trong cn l cc s dng sao cho1

1 Qn(x) dx = 1, n =1, 2, (2)

Ta hy nh gi cc i lng cn. Ta c:

1

1 (1 x

2)n dx = 2 (1 x1

0

2)n dx 2

1

0

x

(1 x2)n dx

2

1

0

x

(1nx2)

ndx =

n3

4>

1

n(3)

Do cn < n ( , ta dng bt ng thc Bernoulli: (1 x2)n 1- nx2.)

T (3) vi bt k > 0 ta cQn(x) = cn(1 x

2)n n (1-x2)n n (1- 2)n (4)vi mix thox 1. Do Qn(x) hi tu v 0 trn tp [-1,-] [,1] (< 1).

Tip theo, t

Pn(x) =1

1 f(x + t)Qn(t)dt,x [0,1].

i bin sx + t =y, ta c

Pn(x) = f(x + t) Q1 x

x

n(t)dt =

1

0

f(y)Qn(y - x)dy

1

0

f(y) Qn(y-x)dy =1

0

f(y) (1-x)2)ndy

Khai trin biu thc (1 - (y - x)2)n v sp xp theo lu tha tin ca x v dngtnh cht tuyn tnh ca tch phn ta c c Pn(x) l mt a thc i vix. Cnli hy chng minh Pn(x) hi tu v f(x)

Cho > 0, chn 0 < < 1

2

)x(f)y(f khi x - y 0 cho trc. Theo nh l WeierstrassI, tn ti a thc g(x) = a0 + a1x +...+ anxn sao cho

d(f,g) =[ ] 210

)x(g)x(fmax *

,

Chn cc s hu tb0, b1,..., bn gn a0, a1,...,an sao cho

0

n

i= i ib a 2

,v t

g*(x) = b0 + b1x ++ bnxn

Khi vi mix [0,1] ta c*

0 0 1 1( ) ( ) ( ) ) ( )n

n ng x g x a b a b x a b x = + +

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Ch . Cc kt qu ca nh l Weierstrass I v cc H qu 5.5.2, 5.5.3 vncn ng trong khng gian C[0,1] vi a < b, a, b IR. Tht vy, khng gian C[a,b]ng c vi C[0,1] qua php ng c.

C[a,b] f

f* C[0,1] , f*(t) = f )

ab

at(

v mi a thc bin thnh mt a thc qua php bin i nn cc tnh cht caC[0,1]c bo ton trong C[a,b].

5.5.4.nh l Weierstrass II. Vi mi hm s f(x) lin t trn IR tun hontheo chu k 2 v vi mi s dng s tn ti mt a thc lng gic.

sn(x) = 02

a +1

n

k= (akcoskx + bksinh kx)

Sao cho vi mi x IR ta c( ) ( )nf x s x <

Chng minh. Xt cc hm s

2

)x(f)x(f)x(

+=

( ) ( )( )

2

f x f xx

= sin x

y l cc hm s chn, lin tc, tun ho theo chu k 2. Chox bin thintrn on [0,] v tx = arccost, ta thy rng, cc hm s:

(t) = (arccost), (t) = (arccosst)lin tc trong on [-1,1]. Dng nh l Weierstrass I, tn ti cc a thc g(t),h(t) xc nh trn [-1,1] sao cho vi mi t [-1,1] ta c

(t) g(t) < /4(t) h(t) < /4

V v (x), (x), cosx l nhng hm s chn, tun hon theo chu k 2.Nn cc bt ng thc sau cng cng ng vix IR.

Theo cch t trn, ta cf(x)sinx = (x)sin x + (x) nn( ) sin ( osx) sinx - h(cosx)f x x g c

( ) ( osx) s inx + (x) - h (cosx) / 2x g c <

Nu gi u(x) =g(cosx)sinx + h(cosx) th u(x) l mt a thc lng gic v

ta c ( ) s inx - u(x) / 2f x <

Bng cch tng t vi hm s f(2

- x), ta cng c

22

/)x(vxsin)x(f


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Vi v(x) cng l mt a thc lng gic. Thay2

-x bix, ta c

22

/)x(vxcos)x(f


64/88

X = .Uk

1ixii )r,B(x

=

t n0 = max{n(x1),,n(xk). Khi vi n n0 vi x X ta s tm c i,(i = 1,,k sao cho x B(xi,rxi). Nh th

g(x) fn(x) g(x) fn0(x) g(x) fn(xi)(x) Vy nh l c chng minh

5.6.nh l Ascoli Azelas

Vic kho st tnh compact ca cc tp khng gian mtric tu ni chungl kh. Tuy nhin trong khng gian C[a,b] ta c tiu chun kh c th. Trc htta cn cc khi nim sau. Gi s A l mt tp con ca C[a,b].

1. Tp A gi l b chn ti x0[a,b] nu c K > 0 sao cho vi mi f A,f(x0) K. Tp A gi l b chn ti tng im trn [a,b] nu A b chn ti miim x [a,b] .

2. Tp A gi l ng lin tc ti x0 [a,b] nu vi mi > 0 tn ti > 0sao cho vi mi y [a,b], f A, y y0 0 saocho vi x, y [a,b], fA, nu x - y


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( )i if x K vi mi f A

t K = max{K1,...,Kn} khi vi mi f A, khong cch t im(f(x1),,f(xn)) n im 0 = (0,,0) trong IR

n l

Kn)x(fn

i

i =1

2

iu ny c ngha l tp (a) b chn trong IRn nn n phi hon ton b

chn. Do tn ti hu hn im j = ( , j = 1,,m sao cho (a)

c ph bng hu hn hnh cu tm

),...,( jmj 1

j v bn knh6

r. Ta gi thit (a) v cc

hnh cu B(j ,6

r) c giao khc trng v nu tri li ta b hnh cu i. Vi mi

hnh cu B(i,6

r) IRn ta hy chn mt hm fi A sao cho (fi) B(i ,

6

r). Gi

f1,...,f

ml cc hm s va chn ny. By givi f

A ta c

jiii

jiiiii )x(f)x(f)x(f)x(f +

n) nn d(xn,xm) 0 (m,n). Gi

s dy ny hi tnx0 = (x01, x0

k,0,) s0. iu ny v l v nu n > k

d(xn,x0) =

iin

Ni

xxsup 0 d(xn,x0) = 20 1

1

)k(xxsup iin

ki +=

> 0 khi n. Vy

s0khng y

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4.6 Do phng trnh arctgx =2

khng c nghim trong IR

4.7. t An,f= {x IR : |f(x)|n}. Do f lin tc nn An,f l tp ng. Khi

An = cng l tp ng. Mt khc x IR theo gi thit Mx > 0 : f F :

|f(x)| Mxn vi mt sn no nnx An. Vy IR tc l IR = .

Khng gian IRy nn thuc phm tr II, vy tn ti n0 N

IFf

f,nA

I

=1nnA I

=1nnA

0

A . Hn

na, An0 ng nn ta c00

00 nnAA =

Gi U l mt hnh cu mcha trong An0. Khi x U x An0f F :|f(x)| n0 = M

4.8. f lin tc u trn IR nn vi = 1 tn ti > 0 sao cho vi mix, x IR|x x||f(x) f(x)| < (1)By gicho x IR, gi sx > 0 s tn ti snNx [(n - 1), n] nn ta c th vit

|f(x)||f(x) f(n 1)| + | f(n 1)) f((n 2))| + +| f() f(0)| + |f(0)|

Theo (1) ta c|f(x)| n + |f(0)|

hay

|f(x)| )(fn 0+

(x + )

+ |f(0)|

=

+ |f(0)| + = x +

Khi x < 0 ta cng l lun tng t vi x [-n, -(n 1)], vi n N. Vytrong hai trng hp ta c

|f(x)| |x| + 4.9. a) c gi t kim tra d1, d2 l cc mtric trn IR

b) Xt tp X =

2,

2vi mtric thng thng trn IR : d(x,y) = |x y|.

Khi nh x f : IR X cho bi f(x) = arctgx l mt php ng c t (IR ,d1) ln(X,d1) v ta c

x,y X : d1(x,y) = |arctgx arctgy| = d(f(x),f(y))

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Mt khc

2,

2l khng gian khng y . (chng hn, dy xn =

n

1

2

cbn trong X nhng khng hi t trong X) nn (IR,d1) cng khng y

. Chng minh tng t i vi d2 ta c (IR,d2) ng c vi Y = (0,) vi

mtric thng thng, trong ta cng thy ngay Y khng y .

c) y ha ca X l

2,

2. Do y cng l y ha ca (IR,d1)

(sai khc mt php ng c). Tng ty ha ca (IR,d2) l [0, )

5.1 Gi s A1,,Ak l cc tp con compact ca X. t A = . Nu (xn)n

l mt dy trong A th s c t nht mt tp Ai0 cha v hn cc (xn) v gi dycon lp t v hn (xn) ny l . Nhng do Ai0 compact nn dy ( ) c dy

con ( )l hi t trong Ai0. Hn na ( )l cng l dy con ca (xn) hi t trongA. Vy A compact.

Uk

iiA

1=

knA

knx

nlkx nlkx

5.2 Nu (xn,yn)n l mt dy bt k trong X Y, th (xn), (yn) ln lt l ccdy trong X v Y. V X compact nn dy (xn) c dy con . Dy

( ) Y compact nn cng c dy con y0 trong Y. Khi ( , ) l

dy con ca (xn,yn) hi t v (x0,y0) X Y nn X Y l khng gian compact.

Xxxkn

0

kny

klny

nlkx

nlky

5.3 Gi s d(K,F) = d(x,y) = 0. khi c cc dy (xn) K, (yn) F

sao cho d(xn,yn) 0. K l compact nn c dy con ( ) ca xn sao cho.T

Fy,Kx

inf

knx

Kxxkn

0d( ,x0) d( , ) + d( ,x0) 0 (k)

kny

knx

kny

knx

nn x0. Mt khc do F ng nnx0 F ngha l K F x0. iu ny mu

thun vi gi thit. Vy d(K,F) > 0.kn

y

5.4 Cho > 0 bt k. Do f lin tc u nn > 0 nux,y X d(x,y) < th d(f(x),f(y)) < . t B ={x1,,xk} l mt -li hu hn cho tp A. Lc f(B) = {f(x

1),,f(x

k)} l mt - li hu hn cho f(A), v nuy f(A) s tn ti

xA y = f(x). Hn na gi sxi0B d(x,xi0) < th c ngay d(f(xi0),f(x)) 0 : B(x,rx) G.

(B(x,r/2))xX l mt ph m ca X nn tn ti ph con hu hn X =

. t r=

UI

G

U )/r,x(B xii 2 021

>=

)/r(minix

k,i. Xt B(x,r) vix bt k thuc X. Lc

i = 1,,k : x B(2

ixi

r,x ). Nu y B(x,r) th d(y,xi) d(y,x) + d(x,xi) < r+

2ix

r nn B(x,r) B(xi, ) G .

ixr

ixr

5.7 Ly dy (xn) trong m (hoc c), c dngx1 = (1,0,,0,)x2 = (0,1,0,,0,)

x1 = (0,0,, ,0,))(

1n

Dy ny thuc vo hnh cu ng B(0,1) trong m (hoc c), vi 0 =(0,,0). Tuy nhin d(xn,xm) = 1, m,n, m n nn khng th c dy con nocbn c, tc l khng rt ra c dy con hi t.

5.10 t g(x) = d(x,f(x)), x X. Khi g(x) l hm lin tc xc nh trnkhng gian compact X nn t c gi tr b nht l : d(x, f(x)) 0 (dod(x,f(x)) lun 0). Nh vy cx0X d(x0,f(x0) = . Nux0 f(x0) tc l >0 th t g(f(x0)) = d(f(x0), f(f(y0))) < d(x0,f(x0)) = mu thun vi l gi tr bnht ca g(x). Do f(x0) = x0. Vyx0 l im bt ng ca f.

Nu y0 X ,y0 x0 f(y0) =y0th td(x0,y0) = d(f(x0), f(y0)) < d(x0,y0)

dn n v l. Vy ch c im bt ng duy nht

5.9 Gi sa,b X . ta1= f(a), a2 = f(a1),, an = f(an-1) (1)b1 = f(b), b2 = f(b1),, bn = f(bn-1) (2)

Do X compact nn dy (an) c dy con ( ) hi t. Dy ( ) tngng cng c

dy con ( ) hi t. Ni cch khc, t hai dy (1), (2) ta rt ra c hai dy con tngng l ( ) v ( ) cng hi t. Nh th vi > 0 tn ti n0n > n0 th

nka

nkb

mnkb

nla

nlb

d( , ) < 0n

la nla

Lc ta c

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d(a,ap) d(a,ap+1) )a,a(d)a,a(dnnnn lllpl 000

=+ <

v d(b,bp) d(b,bp+1) )b,b(d)b,b(dnnnn lllpl 000

=+ <

vi p = ln -0n

l

iu ny c ngha l f(X) tr mt trong X ng thi

d(f(a),f(b)) d(a1,b1) d(ap,bp) d(ap,a)+ d(a,b) + d(b,bp) < 2 + d(a,b)

vi mi > 0 nn d(f(a),f(b)) = d(a,b). Ty suy ra f lin tc v n nh. Hn na, doX compact nn f(X) compact. V vy f(X) = )(Xf = X hay f ton nh. Nh th f : X

X l php ng c.5.10 Do f nn theo nh l Weierstrass I, tn ti dy a thc gm(x) = a0m +

a1mx + + anmx

[ 10,C ]n hi tu v f trn on [0,1]. Khi ta c f(x)gm(x) hi tu v

f2(x) trn on [0,1]. Mt khc

dx))x(fxa...)x(xfa)x(fa(dx)x(g)x(fn

nmmmm +++= 11

0

1

00

==

1

010dx)x(fxa i

n

iim

theo gi thitp dngnh l qua gii hn di du tch phn, ta c

0

1

0

1

0

2

== dx)x(g)x(flimdx)x(f mn

Vy nn f01

0

2 =dx)x(f 2(x) = 0 hay f(x) = 0

5.11 a) 1tsin vi mi t [0,1] v [1,2] nn {x(t)} b chn tng im (vu). Hn na, t

222

)'tt(sin

)'tt(cos'tsintsin

+=

'tt'tt'tt 222 vi mi [1,2]. Nh vy {x(t)} ng lin

tc (u). Theo nh l 5.6.2 th tp {x(t)} compact.b) Khng compact (xem bi tp 1.7)6.1 Gi s A B khng lin thng. Lc cc tp mkhc trng M, N trong A

B sao cho M N = v A B = M N. Ly x BA . Khi x M N.

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TI LIU THAM KHO

[1] Dieudonn, Csgii tch hin i, T.1 (bn dch ting Vit), NXB i hc v

Trung hc chuyn nghip, H ni 1973.

[2] Cnmgrp, Phmin, Csl thuyt hm v gii tch hm, T.1,(bn dchting Vit), NXB Gio dc, H ni 1971.

[3] Hong Ty, Gii tch hini, T.1, NXB Gio dc, H ni 1978.

[4] Kirillov, A. Gvichiani, Thormes et Problmes danalyse fonctionelle, Mir,Moscou 1982.

[5] Phan c Chnh, Gii tch hm, T.1, NXB i hc v Trung hc chuynnghip, H ni 1978.

[6] B. A. Tpe, . . capec, T. C. Coa, aa ypa-e

yaaa, 1984.

[7] Yu. S. OTran, Bi tp l thuyt hm s bin s thc, (bn dch tingVit), NXB i hc v Trung hc chuyn nghip, H ni 1977.

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Chu trch nhim ni dung:

Ts. Nguyn vn ha

Bin tp:T cng ngh thng tin

Phng kho th - m bo cht lng gio dc

không gian metric

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