PREFACE
This book is intended as a text for the uniform space , a branch of topology,
there are no formal subject matter pre requisites for studying most of this book. I
do not even assume the reader known much set theory. Having said that , I must
hasten to add that unless the reader has studied a bit of analysis or rigorous
calculus, he will be missing much of the motivation for the concepts introduced
in this book. Things will go more smoothly if he Already has some experience
with continuous functions, open and closed sets , metric spaces and the like
although none of these is actually assumed. Most students in topology course
have , in my experience, some knowledge of the foundations of mathematics.
But the amount varies a great deal from one student to another. Therefore I begin
with a fairly through chapter on introduction of some important definitions that
is useful to study uniform space . It treats those topics which will be needed
later in the book.
CONTENT
TITLE PAGE NO.
UNIFORMITIES AND BASIC DEFINITIONS 1
METRISATION 10
COMPLETENESS AND COMPACTNESS 20
BIBLIOGRAPHY 32
2
DEFINITION 1.1
Let (X;d) be a metric space. Then a subset E of π Γ π is said to be
an entourage if there exists ν>0 such that for all π₯, π¦ β π, π π₯, π¦ <
ν implies π₯, π¦ β E.
REMARK 1.2
Given a matric d on X and ν> 0 , let UΞ΅ denote the set π₯, π¦ β
π Γ π/π(π₯, π¦) < ν . If X is the topology introduced by d and π Γ π the
product topology , then uΞ΅ is an open neighborhood of the diagonal βπ₯ in
π Γ π. Thus every entourage is a neighbourhood of βπ₯ in π Γ π.
RESULT 1.3
Let π; π , π; π be matric spaces. Let π: π β π be a function. Then
π is uniformly continuous (with respect to π and π) if and only if for every
entourage πΉ of π; π , there exist an entourage πΈ of π; π such that for all
π₯, π¦ β πΈ , π π₯ , π(y ) β πΉ.
RESULT 1.4
Let {π₯n} be a sequence in a matric space π; π . Then {π₯n} is a
Cauchy sequence if and only if for every entourage E there exist pβlN such
that for all m,nβ₯ p, (π₯m,π₯n) β E
RESULT 1.5
Let π be any set and {πn : π β π} nβ¬N be a sequence of functions into
a matric space π; π . Then {xn} converges uniformly to a function π: π β π
if and only if for every entourage , there exist mβN such that for all nβ₯m
and π₯ β π πn π₯ , π π₯ β E
3
PROPOSITION 1.6
Let π; π be a pseudo-metric space and U the family of all its
entourages. Then
(i) if βπ₯ β U for each U βU
(ii) if U βU then U-1βU
(iii) if UβU then there exist a Vβ U such that Vβ¦VβU
(iv) if U,Vβ U then Uβ©Vβ U
(v) if U β U and Uβ V β π Γ π then Vβ U
If, moreover d is a matric then in addition to the above, we also have
(vi) β©{ U /U β U } =βπ₯={( π₯, π₯)/π₯ β π}
Proof
(i)
given π; π is a pseudometric
then π(π₯,π₯)=0 for all π₯ β π
β U is reflexive
We know that U is reflexive if and only if βπ₯ βU
β βπ₯ β U for each U β U
(ii)
since d is a pseudometric d(π₯,π¦)= d(y, π₯)
β U is symmetric
β U = U -1
β U -1β U
4
(iii)
suppose U β U then there exist ν >0 such that Uν β U
Where Uν={(π₯,y)β π Γ π/ d(x,y)< ν}, let V=Uν/2
Now Uβ¦V ={(π₯,y)β π Γ π/ β zβ π such that (π₯,z)β V, (z,y)βU}
To prove Vβ¦VβU , let (π₯,y)β Vβ¦V.
βthere exist z β π such that (π₯,z)βV, (y,z)βV
βd(π₯,z)< ν/2 and d(y,z)< ν/2
βd(π₯, y)< ν
β (x,y)βU
β΄ Vβ¦Vβ U
(iv)
let πΏ,ν be such that UΞ΄ β V and Uν β U
let πΌ=min(πΏ,ν) then UΞ±β Uβ©V
there exist (π₯,y)β π Γ π such that d(x,y)< πΌ
β (π₯,y)βU and (x,y)βV
β( π₯,y)β Uβ©V
β Uβ©V is an entourage
β Uβ©V β U
(v)
given Uβ U and Uβ V β π Γ π
5
Since Uβ U , there exist ν > 0 such that d(x,y)< ν β (x,y)βU
β (x,y)βV since UβV
ie, there exist ν > 0 such that d(x,y)< ν β (x,y)β V
β V is an entourage
β Vβ U
(vi)
given d is a metric,
ie, d(x,y)> 0 for all π₯ β π¦
ββ©{U:Uβ U } =βπ₯
DEFINITION 1.7
A uniformity on a set X is a nonempty collection U of π Γ π
Satisfying the following properties.
(i) βπ₯ βU for each Uβ U
(ii) if U β U then U-1β U
(iii) if U β U then there exist Vβ U such that Vβ¦Vβ U
(iv) if U,Vβ U , then Uβ© V β U
(v) if U β U and U β V β π Γ π then Vβ U
Members of U are called entourages. The pair (X, U ) is called uniform
space.
6
DEFINITION 1.8
A uniform space (X; U ) is said to be Hausdorff or separated if for a
metric d, β© {π: UβU}=βπ₯. Then U is called Hausdorff uniformity.
DEFINITION 1.9
A uniform space (X; U ) is said to be pseudo-metrisable ( or
metrisable) if there exist a pseudo-metric (respectively a metric) π on X
such that U is precisely the collection of all entourages of (X;π) in such a
case we also say that U is the uniformity induced or determined by π .
DEFINITION 1.10
Let (X; U ) be a uniform space. Then the subfamily B of U is said to
be a base for U if every member of U contains some members of B ; while
a subfamily S of U is said to be sub-base for U if the family of all finite
intersections of members of S is a base for U.
PROPOSITION 1.11
Let X be a set and SβP π Γ π ) be a family such that for every
UβS the fol lowing conditions hold:
a) βπ₯ β U
b) U-1
contains a member of S , and
c) there exists VβS such that Vβ¦VβU, then there exists a unique
uniformity U for which S is a sub-base.
Proof
Let B be the family of all finite intersections of members of S and
7
let U be the family of all supersets of members of B. We have to prove
that U is a uniformity on X.
From condition a) βx β U
Let UβU and Uβ π β π Γ π
Here UβS. from the construction of U, it is clear that VβU Now we prove that U
-1βU, for this suppose UβU, then there exist
U1,U2,β¦UnβS such that πππ=1 iβ U
By b) each U
-1contains some πiβS
Then Vπ
π=1 i β Uni=1 i
-1 β U-1
So ππ
π=1 iβB and U-1βU
Now we prove that if UβU then there exist a VβU such that Vβ¦VβU
Suppose UβU and U 1, U2,β¦. U nβS such that πππ=1 iβU
Now by c) each i=1,2,β¦.n we can find Vi βS such that
Viβ¦V iβ Ui
Let V = ππ
π=1 i
Then Vβ¦V β (ππ
π=1 iβ¦ Vi)
ie, Vβ¦V β (ππ
π=1 iβ¦ Vi) β Uππ=1 i βU
ie, Vβ¦V β U
further VβB and hence VβU Now we prove that if U,VβU then Uβ©V βU
let U,Vβ U , we can find U1, U2,β¦. Un and V1, V 2,β¦ V n β S such that
Vππ=1 j βV
8
then ( Uππ=1 i)β©( Vπ
π=1 j) βB .
Since Uβ©V is a superset of this intersection , it follows that Uβ©Vβ U.
Thus U is a uniformity for X. But its very construction B is a base and
S is a subbase for U.
Uniqueness of U is trivial since a sub-base determines uniformity.
THEOREM 1.12
For a uniform space (X; U) , let Iu be the family {Gβ X: for each xβ
G, β U βU, such that U[x]βG}. Then Iu is a topology on X.
Proof
Clearly X,β β Iu .
Also from the very nature of the definition, it is clear that Iu is
closed under arbitrary unions.
Now we show that Iu is closed under finite intersections.
Let G , H β Iu and suppose xβ π β©H.
Then there exist U,VβU such that U[x] β G and V[x] β H
Let W=Uβ©V, then WβU Also W[x] β U[x]βͺV[x]
So W[x] β Gβ©H
so Gβ©Hβ Iu , thus Iu is a topology on X.
9
DEFINITION 1.13
Given a uniform space (X;U ) and a subset Y of X, we defined the
relativised uniformity U/Y as the family {Uβ©(YΓY):UβU} here (Y; U/Y)
is said to be a uniform subspace of (X;U ).
PROPOSITION 1.14
Let (X;U ), (Y;V ) be uniform spaces and f:XβY a function which is
uniformly continuous with respect to U and V .let Iu and Iv be the
topology on X and Y respectively as given by above theorem. Then f is
continuous with respect to these topologies.
Proof
Let Gβ Iv. we have to show that f -1(G) is open in X.
ie, f -1(G)βIu. for this let xβ f -1(G), then f(x)βG.
by the definition of Iv there exist VβV such that V[f(x)]βG.
Let U={(z,y)β X ΓX : (f(z), f(y))βV },
then UβU , since f is uniformly continuous.
Moreover, U[x] β f -1(V[f(x)]) β f -1(G).
So f -1(G) is open in X. Hence f is continuous.
DEFINITION 1.15
Let (X;U ) be a uniform space. Then the topology Iv is called a
uniform topology on X induced by U. A topological space (X,I) is said to
be uniformisable if there exist a uniformity U on X such that I=Iu
11
PROPOSITION 2.1
Suppose a uniformity U on a set π has a countable base. Then
there exists a countable base {Un}β
n = 1 for U such that each Un is
symmetric and Un+1β¦Un+1β¦ Un+1βUn for all n βN
Proof:
Let the given countable base be V = {V1, V2, V3β¦. Vnβ¦}. Set U1=V1β©
V1-1
Then U1 is a symmetric member of U .We know that for every member
U of U, there exists a symmetric member V of U such that Vβ¦Vβ¦VβU.
Consider the set U1β©V2 .
Here U1β©V2βU .Then there exists a symmetric member say U2 of U
such that
U2β¦U2β¦U2 β U1β©V2 .Next consider U2β© V3 .
Then we get a symmetric member U3 of U such that
U3β¦ U3β¦ U3 β U2β©V3.
In this manner, we proceed by induction and get a sequence
{Un:nβN} of symmetric members of U such that for each nβN
,
Un+1β¦Un+1β¦Un+1 β Unβ©Vn+1 .
Then Un βVn for all n and so { Un:nβN} is a base for U since
{Vn:nβN} is given to be a base.
NOTE 2.2
Having obtained a normalised countable base { Un:nβN}for U, we
can construct a pseudo-metric d on X as follows.
Set U0 = XΓX. Note that Un β Un-1 for all nβN.
12
Define d:XΓX β R in such a way that for each nβN, the set
π₯, π¦ β π Γ π/π(π₯, π¦) < 2-n} will be very close to the set Un.Now we
construct a pseudo-metric d on X such that for each n, the set
π₯, π¦ β π Γ π/π(π₯, π¦) < 2-n} will be between Un and Un-1.
We begin with a function π: πΏ Γ πΏ β π defined by f(x,y) = 2-n
in case
there exists nβN. such that (x,y)βUn-1-Un.
If there exists no such n, it means (x,y)β πβπ=1 n and in that case we
set f(x,y)=0.
Here f (x,y)= f(y, x) β x, yβ π since all Unβs are symmetric. Also for
each nβN, π, π β πΏ Γ πΏ/π(π, π) < π-n}=Un-1. Now for x,yβ π define
d(x,y)=inf πππ=1 (xi,xi+1) where the infimum is taken over all possible
finite sequences {x0,x1,x2,β¦xn,xn+1} in X for which x0=x, and, xn+1=y.
Such a sequence will be called a chain from x to y with n nodes at
x1,x2,β¦xn.
The number πππ=1 (xi,xi+1) will be called the lengths of the chain .
Thus d(x,y) is the infimum of the lengths of all possible chains from x
to y.
LEMMA 2.3
The function d: π Γ π β R just defined is a pseudometric on the
set X.
Proof
Here f(x,y)=0 or 2-n
d(x,y) <2-n
13
β d(x,y)β€ f(x,y) for all x,yβ π
βd(x,y)β₯0
If x=y, then f(x,y)=0
βd(x,y)=0
Now f(x,y)= f(y,x)
βd(x,y)= d(y,x)
Now only the triangle inequality remains to be established. Let x, y,
z β π. Let S1, S2 and S3 be respectively the sets of all possible chains
from x to y, from y to z and from x to z. A chain s1βS1 and s2βS2
together determine an element of S3 by juxtaposition, which we
denote by s1+s2.
Let π denote the length of the chain.
Now let J(S3) be the image of S1ΓS2 in S3 under the juxtaposition
function
+: S1 Γ S2 βS3. Then we have,
d(x,y)+ d(y,z) = inf {π(s1) : s1 β S1} + inf {π(s2) : s2βS2}
= inf {π(s1) + π(s2) : (s1,s2)βS1ΓS2}
= inf {π(s1+s2) : (s1,s2)βS1ΓS2}
= inf {π(s3) : s3 β J(S3) βS3}
β₯ inf {π(s3): s3βS3}
= d(x,z)
Hence d(x,y)β₯ 0, d(x,y)=0 β x = y
d(x,y)+ d(y,z) β₯ d(x, z)
14
hence d is a pseudo-metric on X
LEMMA 2.4
For any in teger kβ₯0 and x 0,x 1 ,x 2 ,β¦x kβX,
f(x 0,x k+1)β€2 π(ππ=0 x i ,x i+1).
Proof:
We apply induction on k.
when k=0, the result is trivial.
Assume k>0, and that the result holds for all possible chains with less
than k nodes. Let x 0,x 1 ,x 2 ,β¦x k ,x k + 1 be a chain with k nodes. The
idea is to break this chain into smaller chains and then to apply the
induction hypothesis to each of them.
Let a be the length of this chain, that is, a= π(ππ=0 xi,xi+1)
Here f(x i,x i+1) β₯0 for all i
β a β₯ 0 and a=0 only if f(x i,x i+1)=0 for every i=1,2,β¦ k
also f(x i,x i+1)β€ a
now we show that f(x 0,x k+1) β€2a
now we make three cases
case1: a β₯1/4
then 2a β₯1/2, by the definition of f the largest value it can take
is Β½, since f(x,y)=2-n
, nβN
β΄ f(x 0,x k+1)β€1/2 β€2a
Case 2: Let a=0.
Then f(x i,x i+1)=0 β i = 0,1,. . . , k.
We have to show that f(x0,x k+1)= 0
15
ie, to show that f(x0,x k+1)β πβπ=0 n
We decompose the chain x 0,x 1 ,x 2 ,β¦x k ,x k + 1 into three chains, say
x 0,x 1 ,x 2 ,β¦x r ; x r+1,...x s and x s+1,...x k,x k+1 where such that r and s
are any intiger such that 1β€ r β€ s β€ k
note that each of these three chains has a lenth zero and less than k nodes.
so by induction, f(x0, xr) , f(xr,xs)and f(xs,xk+1) are all zero.
Hence for every nβN , (x0,xr) βUn , (xr,xs) βUn, and , (xs,xk+1) βUn
β (x0,xk+1) β Unβ¦Unβ¦Un
But Unβ¦Unβ¦Un < Un-1
There for (x0,xk+1) β Un-1
So f(x0,xk+1)=0
Case 3: Let 0 < a < 1/4.
Let r be the largest integer such that fπβ1π=0 (xi,xi=1) β€ a/2,if
f(x0,xi)> π/2, this definition fails and we set r =0 in this case. Then
0 β€ r β€ k
so each of the chains xo,β¦,xr, and xr+1,β¦, xk+1 has less than k nodes.
The first chain has length β€ a/2.
Then the length of the second chain is also at most a/2 for otherwise
the chain xo,...,xr+1 will have length less than a/2 contradicting the
definition of r.
By induction hypothesis we now get, f(xo,xr) β€a, f(xr+1,xk+1) β€a
While f(xr, xr+1) β€ a
Let m be the smallest integer such that 2-m β€ a.
16
In particular, f(x0,xr+1) β€2-m
β(x0, xr+1) β Um-1
similarly (xr, xr+1) β Um-1 and(xr+1, xk+1) β Um-1
β(x0, xk+1) β Um-1β¦ Um-1β¦Um-1 β Um-2
Hence f(x0, xk+1) β€ 2-(m-1)
β€ 2a
Hence the proof.
RESULT 2.5
A uniformity is a pseudoprime if and only if it has a countable base.
RESULT 2.6
A uniformity is metrisable if and only if it is Hausdorff and has
countable base.
PROPOSITION 2.7
Let U be uniformly generated by a family D of pseudometrices on a set
X. Then
i) each member of D is a uniformly continuous function from XΓX to R
where R has the usual uniformity and XΓX has the product uniformity
induced by U. Moreover U is the smallest uniformity on X which
makes each member of D uniformly continuous.
ii) Let Y be the powerset XD. For each dβD , let Xd be a copy of the set
X and let Vd be the uniformity on Xd induced by the pseudometric d.
let V be the product uniformity on
Y= Xdβπ· d. then the evaluation function f:XβY defined by f(x)(d)=x
for all dβD, xβX is a uniform embedding of (X,U) into (Y,V)
17
Proof
Proof of the first statement is clear
For ii) we let Οd:YβX be the projection for dβD . Here for each
dβD
Οdβ¦f : X βXd is the identity function and is uniformly continuous because
the uniformity on X is U which is stronger than Vd. So by the general
properties of products f: (X,U) β(Y,V) is uniformly continuous, clearly f is
one-one. Let Z be the range of f . We have to shoe that f uniformly
isomorphism when regarded as a function from (X,U) to (Z,V/Z). now we
prove that the image of every sub-basic entourage under fΓf is an entourages
in V/Z. Take the sub-base s for U. A member of S is of the form Vd,r for
some r>0 and dβD. Then clearly (fΓf) (Vd,r)is precisely Zβ©(ΟdΓΟd)-
1(Vd,r)which is an entourage in the relative uniformity on Z
PROPOSITION 2.8
Let (X,U) be a uniform space and D be the family of pseudometrices on
X which are uniformly continuous as function from XΓX to R, the domin
being given the product uniformity induced by U on each factor. Then D
generates the uniformity on X.
proof
Let V be the uniformity generated by D By statement (i) of the last
proposition, V is the smallest uniformity on X rendering each dβD
uniformly continuous. So VβUβ¦β¦.(1)
Now suppose uβU. Let Ui be the symmetric member of U contained in U.
So U0=XΓX . Define U2,U3......Un......... by induction so then
18
each Un is a symmetric member or U and Unβ¦Unβ¦UnβUn-1 for each nβN.
Then we get a pseudometric d: XΓXβR such that for each nβN
Un+1 β{(x,y)βXΓX : d(x,y)<2-n
} βUn-1. Then d is uniformly continuous with
respect to the product uniformity on XΓX. So dβD , then d generates the
uniformity U.
Put n=2, we get {(x,y) β XΓX: d(x,y)<1/4} βU1βU. Let S be the defining
sub-base for V. Then {(x,y) β XΓX: d(x,y)<1/4}βSβ V ,and so uβV , so
UβV β¦β¦β¦(2)
From (1) and (2) we get U=V.
Thus D generates the uniformity U.
RESULT 2.9
Every uniform space is uniformly is uniformly isomorphic to a
subspace of a product of pseudometric spaces.
CORROLLARY 2.10
If (X,U) is a uniform space , then the corresponding topological space
(X,Οu) is completely regular.
proof
from proposition 3.7 it follows that (X,Οu) is embeddable into a product
of pseudometric spaces. Then (X,Οu) is completely regular.
RESULT 2.11
A topological spaces is uniformisable if and only if it is completely
regular.
19
DEFINITION 2.12
The gage of uniform space (X,U) is the family of all pseudometrics on
the set X which are uniformly continuous as functions from XΓX to R.
REMARK 2.13
In the view of the statement (i) in proposition 3.7 the uniformity is
completely characterized by its gage. Thus we have answered the two basic
questions through the result 3.6 and 3.11
21
DEFINITION 3.1
A net {Sn:nβD} in a set X is said to be a Cauchy net with respect to a
uniformity U on X if for every UβU there exist pβD such that for all m β₯ p
, n β₯ pin D (Sm,Sn)βD
PROPOSITION 3.2
A uniform space (X;U) is said to be complete if for every cauchy net in
X (with respect to U) converges to atleast one point in X (with respect to the
topology Iu)
PROPOSITION 3.3
Every convergent net is a Cauchy net. A Cauchy net is convergent if
and only if it has a cluster point.
Proof
Let {Sn:nβD} be a netin a uniform space (X;U) . suppose {Sn:nβD} is
converges to x in X. let UβU , then there exist a symmetric VβU such that
Vβ¦V β U. Now V[x] is a neighbourhood of x in the uniform topology on X.
so threre exist a pβD such that for all nβ₯p in D, Snβ V[x], ie, (x,Sn)β V.
now for any m,n β₯p, (Sm,x) β V and (x,Sn)β V by symmetry of V
So (Sm,Sn)β Vβ¦Vβ U, since UβU was arbitrary , it follows that {Sn:nβD} is
a Cauchy net.
Now assume {Sn:nβD} is a Cauchy net. Let {Sn:nβD} converges
to x. Then x is a cluster point .
Conversely assume that x is a cluster point of a Cauchy net {Sn:nβD} in a
uniform space (X; U) . we have to show that Sn converges to x in the
uniform topology. Let G be a neighbourhood of x. Then there exist a
Uβ U such that U[x] βG. now we find
22
a symmetric VβU such that Vβ¦V β U . then threre exist a pβ D such that
for all m,nβ₯p in D,(Sm,Sn)β V.Since x is a cluster point of {Sn:nβD} there
exist a qβ₯p in D such that Sqβ V[x] . ie, (x,Sq) β V. Then for all nβ₯q we
have (Sq,Sn)β V and (x,Sq)β V .
So (x,Sn)β Vβ¦V β U.
Hence Snβ U[x] βG.
β {Sn:nβD} converges to x.
COROLLARY 3.4
Every compact uniform space is complete
Proof
A compact uniform space means a uniform space whose associated
topological space is compact. We know that every net in a compact space
has a cluster point . Let (X;U) be a compact uniform space and {Sn:nβD} be
a cauchy net in it. Then (X;U) has a cluster point. So by the above
proposition, {Sn:nβD} is convergent. ie, every Cauchy net is convergent .
β( X;U) is complete
β every compact uniform space is complete.
DEFINITION 3.5
Let (X; U) , (Y,V) be uniform spaces and f : Xβ Y be uniform
continuous.Then for any Cauchy net S :Dβ X, the composite net fβ¦S is a
Cauchy net in (Y,V).
23
PROPOSITION 3.6
Let (X,d) be a metric space and U the uniformity of X induced by d.
Then a net {Sn:nβD} in X is a Cauchy net with respect to uniformity U.
Moreover ,(X,d) is a complete metric space if and only if (X; U) is a
complete uniform space.
Proof
First suppose that {Sn:nβD} is a Cauchy net with respect to a metric d. Then
for every ν>0, there exist n0βD such that for every m,nβ D, mβ₯n0 and nβ₯n0
d(Sn,Sm)<ν.
Let Uβ U , then d(Sn,Sm)<ν β (Sn,Sm)β U
ie for every ν>0, there exist n0β D such that for every m,nβ D, mβ₯n0 and
nβ₯n0 , (Sn,Sm)β U
β {Sn:nβD} is a Cauchy net with respect to the uniformity U
Conversely suppose that a net {Sn:nβD} is a Cauchy net with
respect to the uniformity U. let Uβ U then by the definition , for every UβU
there exist a pβ D such that for every mβ₯p, nβ₯ π in D, (Sn,Sm)β U
β d(Sn,Sm)<ν for all ν>0
β {Sn:nβD} is a Cauchy net with respect to the metric d.
Now we have to prove that (X; d) is a complete metrics space if and only if
(X; U) is a complete uniform space.
Since the metric topology on X induced by d is the same as the uniform
topology induced by U , convergens of a net with respect to the metric d is
same as that with respect to the uniformity U.
So to prove the above argument , it is enough to prove that every Cauchy net
in (X;d) is covergent if and only if every Cauchy sequence in (X;d) is
convergent. here a sequence is a special type of
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net , so every net in (X;d) is convergent β every Cauchy sequence in (X; d)
is convergent.
Conversely assume every Cauchy sequence in (X;d) is
convergent. Let {Sn:nβD} be a Cauchy net in (X;d). Now it is enough to
prove that {Sn:nβD} has a cluster point in X . Set ν=2-k
, where k=1,2,β¦. .
Then we obtain elements p1,p2,...pkβ¦. in D such that for each kβN , we
have,
(i) pk+1β₯pk in D and
(ii) for all m,nβ₯pk in D , d(Sm,Sn)<2-k
Now consider the sequence {Spk} ;k=1,2β¦. Since d(Spk,Spk+1)<2-k β
kβN, and the series 2βπ=1
-k is convergent., {Spk} ;k=1,2β¦. is
Cauchy sequence in (X;d)
So {Spk} converges to a point say x of X. now we claim that x is the
cluster point of the net {Sn:nβD}. Let ν>0 and mβD be given,
We have to find nβ D such that nβ₯m and d(Sn,x)<ν .first choose kβN so
that 2-k
< ν/2 and d(Spk,x)< ν/2
Now d(Sn,x) β€ d(Sn,Spk)+ d(Spk,x)
< 2-k
+ ν/2
< ν/2 + ν/2
< ν
So x is a cluster point of the net {Sn:nβD}
β every Cauchy net in (X;d) is convergent
β every Cauchy net in ( X;U) is convergent
β ( X;U) is complete uniform space.
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DEFINITION 3.7
Given an induxed collection {( Xi,Ui):iβI} of uniform spaces . We
define a product uniformality on the cartesian product X= ππβπΌ i
RESULT 3.8
Each projection πi : Xβ X i is uniformly continuous. Let S be the
family of all susets of XΓX of the form πi-1
(Ui) for Uiβ Ui ,iβI , where πi:
XΓ X β X iΓ X i is the function πI Γ πI defined by
( πI Γ πI)(x,y) =( πi(x), πi(y))
PROPOSITION 3.9
Let ( X;U) be the uniform product of a family of nonempty uniform
spaces {(Xi,Ui):iβI}. Then a net S:Dβ X is a Cauchy net in (X;U) if and
only if for each iβI , the net πiβ¦S is a Cauchy net in (Xi,Ui).
Proof
We know that the projection function πi is uniformly continuous where πi :
Xβ X i .
Now for any Cauchy net S:Dβ X , πiβ¦S is a Cauchy net in (Xi,Ui) , since
πiβ¦S is the composite net.
Conversely assume that πiβ¦S is a Cauchy net for each iβI.
Let S:Dβ X be a net, let S be the standard sub-base for U , consisting of all
subsets of the form πi-1
(Ui) for UiβUi and iβI. where πi: XΓX β XiΓ Xi is
the function πi Γ πI . Suppose U = πi-1
(Ui) for some UiβUi, iβI.
find pβ D so that for all m,n β₯p in D , (πi(Sm), πi(Sn))βUi.
26
Such a p exist since πiβ¦S is a Cauchy net
β (πI Γ πI)(Sm,Sn)β Ui.
β πi(Sm,Sn) β U Ui
β (Sm,Sn) β πi-1
(Ui)
β (Sm,Sn) β U β m,nβ₯p
So S is a Cauchy net in ( X;U)
REMARK 3.10
(X;U) is complete if and only if each (Xi,Ui) is complete.
DEFINITION 3.11
If (X;U) , (Y,V) are two uniform spaces, then the function f:XβY is
said to be an embedding, if it is one-one, uniformly continuous and a
uniform isomorphism, when regarded as a function from (X;U) onto
(f(x),V/f(x)).
THEOREM 3.12
Every uniform space is uniformly isomorphic to a dense subspace of
a complete uniform space.
Proof
Let (X;U) be a uniform space. We know that every uniform space is
uniformly isomorphic to a subspace of a product of pseudometric spaces.
Then there exist a family {(X i,di): iβI} of pseudometric spaces and a
uniform embedding,
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f :Xβ ππβπΌ i where the product πi is assigned the product uniformity,
each Xi being given the uniformity induced by di. for each iβI ,
let (X i*,di
*) be a complete pseudometric space containing (X i,di) up to an
isometry.
Then πi is a uniform subspace of πi* with the product uniformities, we
regard f as an embedding of (X;U) into πi* , which is complete.
Let Z be the closure of f(x) in πi*. then Z is complete with the relative
uniformity. Also f(x) is dense in Z.
Hence the proof.
DEFINITION 3.13
A uniform space (X;U) is said to be totally bounded or pre-compact if
for each UβU , there exist x1,x2,β¦xnβ X such that X = πππ=1 [xi].
Equivalently (X;U) is totally bounded if and only if for each UβU , there
exist a finit subset F of X such that U[F]= X.
THEOREM 3.14
A uniform space is compact if and only if it is complete and totally
bounded.
Proof
First assume that the uniform space (X;U) is compact. We have every
compact uniform space is complete. So (X;U) is complete.
Now compactness β totally boundedness
β (X;U) is totally bounded.
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Conversely assume that (X;U) is bounded and totally bounded.we have to
prove that (X;U) is compact.
We know that a space is compact if and only if every universal net in it is
convergent. {Sn:nβD} is a Cauchy net in (X;U).
Let Uβ U, find the symmetric VβU such that Vβ¦Vβ U.
By total boundedness of (X; U) , there exist x1,x2,β¦xkβ X such that X
= πππ=1 [xi].
Now for atleast one i=1,2,β¦k, {Sn:nβD} is eventually in V[xi] . for
otherwise the net will be eventually X - V k[xi] β i=1,2,β¦k and hence
eventually (π β πππ=1 [xi])=β .
Thus for some i , {Sn:nβD} is eventually in V[xi]. ie, there exist pβD such
that β nβ₯p , Snβ V[xi] , ie, (xi,Sn)β V .
So for all m,n β₯p in D, (Sm,Sn)β Vβ¦Vβ U. Thus {Sn:nβD} is a Cauchy net
in (X;U). Since (X;U) is complete {Sn} converges
β (X;U) is compact.
RESULT 3.15
Every continuous function from a compact uniform space to a uniform
space is uniformly continuous.
PROPOSITION 3.16
Let (X;U) be a compact uniform space and V an open cover of X. then
there exist a UβU such that for each xβ X there exist a Vβ V such that
U[x] β V
29
Proof
Given (X;U) is a compact uniform space and V an open cover of X.
Then for each xβ X , there exist UxβU such that Ux[x] is contained in some
members of V. Hence there exist a symmetric VxβU such that (Vxβ¦ Vx)[x] is
contained in some members of V . The interiors of the sets Vx[x] for xβ X ,
cover X. So by compactness of X , there exist x1,x2,β¦xnβ X such that
X= πππ=1 i[xi] , where Vi denotes Vx i for i=1,2,β¦n.
Now let U = πππ=1 i , then UβU . Also let xβX , then xβVi[xi] for some i.
So U[x] β Vi[xi]
β Vi[Vi[xi]] , since xβVi[xi]
= (Viβ¦Vi)[xi]
β some members of V
Since this holds for all xβ X, the result follows.
PROPOSITION 3.17
Let (X;U) be a compact uniform space. Then U is the only uniformity
on X which induces the topology πu on X.
Proof
If possible let V be any other uniformity on X such that πu=πv
Let f : (X;U)β (X;V) and g : (X;V)β (X;U) be identity function. Then f and
g are continuous (in fact a homeomorphism) because πu=πv .
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Hence by result 4.15 , every continuous function from a compact uniform
space to a uniform space is uniformly continuous.
β f and g are uniformly continuous
β UβV and VβU
β U=V
Hence the proof.
THEOREM 3.18
Let (X;U) be a compact uniform space. Let πu be the uniform topology
on X and given XΓX the product topology. Then U consist precisely of all
the neighbourhoods of the diagonal βπ₯ in XΓX
Proof
We know that every member of U is a neighbourhood of the diagonal βπ₯ in
XΓX . Now it is enough to prove that every neighbourhood of βπ₯ is a
member of U. Let V be such a neighbourhood.
Without loss of generality, we may assume that V is open in XΓX . Let B
be the family of those members of U which are closed in XΓX. Then B is a
base for U, since each member of U is a closed symmetric member of U.
Let W=β©{ U:Uβ B}. We claim that W β V .
For this suppose (x,y)βW, then yβ U[x] for all Uβ B .
Since B is a base , the family { U[x]:Uβ B} is a local base at x with respect
to πu.
In particular, since V[x] is an open neighbourhood of x , there exist a Uβ B
such that U[x] β V[x]. Hence yβ U[x] β V[x] β yβV[x].
ie, (x,y)β V. Thus W β V.
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since X is compact, so is XΓX , sice V is open so XΓX - V is closed.
B βͺ{ XΓX - V } is a family of closed subsets of XΓX and its intersection is
empty.
ie, there exist finitely many members U1,U2,β¦Un of B such that
U1β©U2β©β¦β©Unβ©(XΓX-V)=β
β U1β©U2β©β¦β©Un β V
β VβU.
Hence the proof.
32
BIBLIOGRAPHY
1. JOSHY.K.D β Introduction to General Topology β New age
International (p) Ltd.(1983)
2. AMSTRONG βBasic Topology SPR01 editionβ Springer (India)(p)
Ltd.(2005)
3. JAMES.I.M βIntroduction to Uniform Spacesβ Cambridge University
Press (1990)
4. JAN PACHL βUniform Spaces and Measuresβ Springer (2012)
5. MURDESHWAR.M.G βGeneral Topologyβ New age International (p)
Ltd. (2007)
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