renard lx of
TRANSCRIPT
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 .-