3. I

Lasttime . Closed subsets of a topological space• E - S definition of continuity for maps between metric spaces .

defo f : (X , Tx ) - (Y, Ty ) is continuous⇐ tf U E Y open

,f

- "

(u ) EX is open .

• subspace topology for Y E X induced by a topology Tx on X: smallest topologyIT on Y so that the inclusion i :(Y, IT ) - (X, Tx ) is continuous

.

Explicitly U C- IT ⇐ FUN EX open so that U = On Y.

• (X , Tx ) topological space , B E Tx is a basis # t U E Tx is a union of

elements of B.

Renard for any topological space (X,T) the identity map idx : LX , Tx ) - CX, Tx ) ,id× (x) -- a tr t X is continuous

. ( why ? I

Lemma# The composite of two continuous map is continuous.

Prof Suppose f :X , Tx ) - (Y,Ty )

, g :X , Ty ) - LZ ,Tz ) are two continuous

maps. We want to show go f : H , Tx ) - (Z, Tz ) is continuous.Now given Ut ATE , cg.gg

'

ful -- f"

I g- ' (UH -

Since g is continuous, gyu) is open .

Since f is continuous, f' '

(g- ' lull is open . D.

Lemma-3.sn Let X. TNT, ) , ( Y, T.FI) be two topological spaces, B

E Ry a basis of Ty ,

f : (X , Tx ) - (Y, try ) is continuous ⇐ A Bf B,f' '1B) is open .

Proof ⇐ I suppose f is continuous.

Since each B EB is open , f-'

(B) is open .

⇐) Conversely suppose U try .

Then 7 A e-B so that U = U B,i.e

.

BEAU is a union of a collection of elements of B .

Hence

f-'(ul -- f

' "

( Bff B) = U f-'

IB) ,which is open since each f'

'

(B) is open . D

Bet

3. 2

Notation Let X be a set, A EPIX) a subset, ie A is a collection of subsets

of the set X . ThenUt : = Alfa A -

- 3 next at A for some A TAG.

Lemma3.is Let X be a set,BE Phd a set of subsets of X so that

CD VB = X

H t Bi , Bz EB , Bin Bz is a union of elements of B .

Then ⇐ l U EX l U = US for some A EB } is a topology on Xand B is a basis of NT.

Remark Lemma 3.3 gives His an easy way to define a topology on a given set X.

proofof3.jo 0 is the union of an empty collection of elements of B - X -

- U B.

Hence 0, X t NT.

• If U -- yep Ba ,V =p!fBp for some subsets { Balata

,I Bp Spec of B then

Un V = U Ban Bp .

But each Ban Bp is a union of elements of TB bya , p

assumption on B .

⇒ Unr c- IT.

• Suppose I Usher C- IT.

Then tret 7Ar EB with Ur -- UAr . Therefore

yer Ur -- tf, AH -- U l Ar ) c- AT since yer Ar E B-

D

Definition Let H ,IT ) be a topological space . A subset T of NT in a subbases of RT

if B = l Si,n - - n Siu l k c- IN

,Si

. . . - Siu ESS is a basis of IT. That is,S is a

sub basis of IT if every open set U C- HT is a union of finite intersections of

elements of

T.com/lary3.4-Let X be a set,S e PG) with US = X

.Then b is a

sub basis for a topology IT on X .

Proof Let OB -

- { BEX l B -

- Si,n . -

n Sia for some kao,Si

. . - Siu c- It

3.3

Since US -- X,U B -- X as well

.

Moreover it B -- Si

,n

. .-n Sir

,

B' = Si,n - -

n Sjn for some Si, . . . Sir , Sj , , . - Sjn E b , then

B n B' = Si, n .

n Sin n Sj , n .- n Sjn c- B

.

Hence B is a basis for a topology AT on X by Lemma 3.3.

It's easy to see that I is a subbasis for TF is

Atopologyonasetxgenerratedbyasubsetfof.PL#

Lemma3 Let X be a set, A e- PhD. There exists a topology ATA on X so that

i) b E TRAii ) it AT ' is another topology on X with A e th

'then ITS EAT!

Definition.6 The topology PTA of Lemma 3.5 is called the topology generated by A .

To prove 3.5 we need an easy lemma .

Lumina Let Heber be a family of topologies on a set X .Then D=!Tr

is also a topology on X .

Proof . 0 C- The tr. → 0 t rn Ty --IT X Ept tr. ⇒ X C- IT.

• It U,V t IT then U

,V t at tr. Since each it is a topology Un VE Rr tr.

⇒ UnV E NT.

If IUplpep e-RT then ftp.spe.BE INT tr.→ ftp.up c- ATR tr.

⇒ p¥Up c- ANTI =D. D

proofof3.SI Let f -- l HT ⇐ PH) ) IT in a topology on X containing Al .

By 3.4, TAA : = Af in a topology on X. Since A c. IT for all Tf t b,

A Ehf = at . If IT'

w any topology on X containing A thenIT ' e 8

.

⇒ Then f E IT! D

Exercise Let X be a set,B EPH) a subset with H U B -- X and

3. 4

F Bi , Bz EB , Bin Bz is a Union of elements of B .

Then the topology at generated by B ( definition 3.61 is

IT : = l U E X l U = US for some A E B 's

( see Emma 3.3 )

Hint show that ATB E MT and T E TB .

Exercise Let X. TNT, ) , ( Y, T.ME ) be two topological spaces,

I EAT,a subbasis of Ty ,

f : (X , Tx ) - (Y, try ) is continuous ⇐ A Bf S,

f-' '1B) is open .

Hihti see 3.2 .

Droducttopology Let (X ,Tx ) , CY , Ty ) be two topological spaces . What is the"

right " topology on the product Xx Y ?

The Cartesian product Xx Y comes with two projections px : Xx Y - X, Px kiss - x ;

Py : Xx Y - Y , pycx.my and the followinguniuersalproperty-i.frany set Z and any two functions f-× : 2- →X

, fy : Z -Y Fumiof : Z - X x 4 defined by f GI -- Ix Cz), fy HI ) .

That is ,

Px ( f-CHI -- fxlzl , py ( f tell = fu, HI . Equivalently the diagrams

I ! f

z ----→ Xx Y,f- Y

Pyf,§¥Y

. .

So the"

right topology " IT on XXY is the one so that px :(XxY, KT ) -(X , Tx ) ,Py :(XxY, RT ) → IT , Ty ) are continuous and so that the following universal

property holds : for any space ( Z ,Tz ) , for any pairfx : ft , Tz ) - CX , Tx ) , f-y i IZ , Tz ) - LY, Ty ) of continuous mapsF ! continuous map f : ft , Tz ) - 4x4

,T) so that p× of = Ax , Py of = fy .

→

unique we'll construct it next time .-