Uniform spaces -_riyas

UNIFORM SPACES RIYAS.A.R . M.Sc

Uniform Spaces

Riyas.A.R M.Sc.

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

1

CHAPTER 1

UNIFORMITIES AND BASIC DEFINITIONS

2

DEFINITION 1.1

Let (X;d) be a metric space. Then a subset E of 𝑋 × 𝑋 is said to be

an entourage if there exists 휀>0 such that for all 𝑥, 𝑦 ∈ 𝑋, 𝑑 𝑥, 𝑦 <

휀 implies 𝑥, 𝑦 ∈ E.

REMARK 1.2

Given a matric d on X and 휀> 0 , let Uε denote the set 𝑥, 𝑦 ∈

𝑋 × 𝑋/𝑑(𝑥, 𝑦) < 휀 . If X is the topology introduced by d and 𝑋 × 𝑋 the

product topology , then uε is an open neighborhood of the diagonal ∆𝑥 in

𝑋 × 𝑋. Thus every entourage is a neighbourhood of ∆𝑥 in 𝑋 × 𝑋.

RESULT 1.3

Let 𝑋; 𝑑 , 𝑌; 𝑒 be matric spaces. Let 𝑓: 𝑋 → 𝑌 be a function. Then

𝑓 is uniformly continuous (with respect to 𝑑 and 𝑒) if and only if for every

entourage 𝐹 of 𝑌; 𝑒 , there exist an entourage 𝐸 of 𝑋; 𝑑 such that for all

𝑥, 𝑦 ∈ 𝐸 , 𝑓 𝑥 , 𝑓(y ) ∈ 𝐹.

RESULT 1.4

Let {𝑥n} be a sequence in a matric space 𝑋; 𝑑 . Then {𝑥n} is a

Cauchy sequence if and only if for every entourage E there exist p∈lN such

that for all m,n≥ p, (𝑥m,𝑥n) ∈ E

RESULT 1.5

Let 𝑆 be any set and {𝑓n : 𝑆 → 𝑋} n€N be a sequence of functions into

a matric space 𝑋; 𝑑 . Then {xn} converges uniformly to a function 𝑓: 𝑆 → 𝑋

if and only if for every entourage , there exist m∈N such that for all n≥m

and 𝑥 ∈ 𝑆 𝑓n 𝑥 , 𝑓 𝑥 ∈ E

3

PROPOSITION 1.6

Let 𝑋; 𝑑 be a pseudo-metric space and U the family of all its

entourages. Then

(i) if ∆𝑥 ⊂ U for each U ∈U

(ii) if U ∈U then U-1∈U

(iii) if U∈U then there exist a V∈ U such that V◦V⊂U

(iv) if U,V∈ U then U∩V∈ U

(v) if U ∈ U and U⊂ V ⊂ 𝑋 × 𝑋 then V∈ U

If, moreover d is a matric then in addition to the above, we also have

(vi) ∩{ U /U ∈ U } =∆𝑥={( 𝑥, 𝑥)/𝑥 ∈ 𝑋}

Proof

(i)

given 𝑋; 𝑑 is a pseudometric

then 𝑑(𝑥,𝑥)=0 for all 𝑥 ∈ 𝑋

⇒ U is reflexive

We know that U is reflexive if and only if ∆𝑥 ⊂U

⇒ ∆𝑥 ⊂ U for each U ∈ U

(ii)

since d is a pseudometric d(𝑥,𝑦)= d(y, 𝑥)

⇒ U is symmetric

⇒ U = U -1

⇒ U -1∈ U

4

(iii)

suppose U ∈ U then there exist 휀 >0 such that U휀 ⊂ U

Where U휀={(𝑥,y)∈ 𝑋 × 𝑋/ d(x,y)< 휀}, let V=U휀/2

Now U◦V ={(𝑥,y)∈ 𝑋 × 𝑋/ ∃ z∈ 𝑋 such that (𝑥,z)∈ V, (z,y)∈U}

To prove V◦V⊂U , let (𝑥,y)∈ V◦V.

⇒there exist z ∈ 𝑋 such that (𝑥,z)∈V, (y,z)∈V

⇒d(𝑥,z)< 휀/2 and d(y,z)< 휀/2

⇒d(𝑥, y)< 휀

⇒ (x,y)∈U

∴ V◦V⊂ U

(iv)

let 𝛿,휀 be such that Uδ ⊂ V and U휀 ⊂ U

let 𝛼=min(𝛿,휀) then Uα⊂ U∩V

there exist (𝑥,y)∈ 𝑋 × 𝑋 such that d(x,y)< 𝛼

⇒ (𝑥,y)∈U and (x,y)∈V

⇒( 𝑥,y)∈ U∩V

⇒ U∩V is an entourage

⇒ U∩V ∈ U

(v)

given U∈ U and U⊂ V ⊂ 𝑋 × 𝑋

5

Since U∈ U , there exist 휀 > 0 such that d(x,y)< 휀 ⇒ (x,y)∈U

⇒ (x,y)∈V since U⊂V

ie, there exist 휀 > 0 such that d(x,y)< 휀 ⇒ (x,y)∈ V

⇒ V is an entourage

⇒ V∈ U

(vi)

given d is a metric,

ie, d(x,y)> 0 for all 𝑥 ≠ 𝑦

⇒∩{U:U∈ U } =∆𝑥

DEFINITION 1.7

A uniformity on a set X is a nonempty collection U of 𝑋 × 𝑋

Satisfying the following properties.

(i) ∆𝑥 ⊂U for each U∈ U

(ii) if U ∈ U then U-1∈ U

(iii) if U ∈ U then there exist V∈ U such that V◦V⊂ U

(iv) if U,V∈ U , then U∩ V ∈ U

(v) if U ∈ U and U ⊂ V ⊂ 𝑋 × 𝑋 then V∈ 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, ∩ {𝑈: U∈U}=∆𝑥. 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) 𝑑 on X

such that U is precisely the collection of all entourages of (X;𝑑) in such a

case we also say that U is the uniformity induced or determined by 𝒅.

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 S⊂P 𝑋 × 𝑋 ) be a family such that for every

U∈S the fol lowing conditions hold:

a) ∆𝑥 ⊂ U

b) U-1

contains a member of S , and

c) there exists V∈S such that V◦V⊂U, 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) ∆x ⊂ U

Let U⊂U and U⊂ 𝐕 ⊂ 𝑋 × 𝑋

Here U∈S. from the construction of U, it is clear that V∈U Now we prove that U

-1∈U, for this suppose U∈U, then there exist

U1,U2,…Un∈S such that 𝑈𝑛𝑖=1 i⊂ U

By b) each U

-1contains some 𝐕i∈S

Then V𝑛

𝑖=1 i ⊂ Uni=1 i

-1 ⊂ U-1

So 𝑉𝑛

𝑖=1 i⊂B and U-1∈U

Now we prove that if U∈U then there exist a V∈U such that V◦V⊂U

Suppose U∈U and U 1, U2,…. U n∈S such that 𝑈𝑛𝑖=1 i⊂U

Now by c) each i=1,2,….n we can find Vi ∈S such that

Vi◦V i⊂ Ui

Let V = 𝑉𝑛

𝑖=1 i

Then V◦V ⊂ (𝑉𝑛

𝑖=1 i◦ Vi)

ie, V◦V ⊂ (𝑉𝑛

𝑖=1 i◦ Vi) ⊂ U𝑛𝑖=1 i ⊂U

ie, V◦V ⊂ U

further V∈B and hence V∈U Now we prove that if U,V∈U then U∩V ∈U

let U,V∈ U , we can find U1, U2,…. Un and V1, V 2,… V n ∈ S such that

V𝑚𝑗=1 j ⊂V

8

then ( U𝑛𝑖=1 i)∩( V𝑚

𝑗=1 j) ∈B .

Since U∩V is a superset of this intersection , it follows that U∩V∈ 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 {G⊂ X: for each x∈

G, ∃ U ∈U, such that U[x]⊂G}. Then Iu is a topology on X.

Proof

Clearly X,∅ ∈ 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 ∈ Iu and suppose x∈ 𝐆 ∩H.

Then there exist U,V∈U such that U[x] ⊂ G and V[x] ⊂ H

Let W=U∩V, then W∈U Also W[x] ⊂ U[x]∪V[x]

So W[x] ⊂ G∩H

so G∩H∈ 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 {U∩(Y×Y):U∈U} 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:X→Y 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 G∈ Iv. we have to show that f -1(G) is open in X.

ie, f -1(G)∈Iu. for this let x∈ f -1(G), then f(x)∈G.

by the definition of Iv there exist V∈V such that V[f(x)]⊂G.

Let U={(z,y)∈ X ×X : (f(z), f(y))∈V },

then U∈U , since f is uniformly continuous.

Moreover, U[x] ⊂ f -1(V[f(x)]) ⊂ 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

10

CHAPTER 2

METRISATION

11

PROPOSITION 2.1

Suppose a uniformity U on a set 𝑋 has a countable base. Then

there exists a countable base {Un}∞

n = 1 for U such that each Un is

symmetric and Un+1◦Un+1◦ Un+1⊂Un for all n ∈N

Proof:

Let the given countable base be V = {V1, V2, V3…. Vn…}. Set U1=V1∩

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 V◦V◦V⊂U.

Consider the set U1∩V2 .

Here U1∩V2∈U .Then there exists a symmetric member say U2 of U

such that

U2◦U2◦U2 ⊂ U1∩V2 .Next consider U2∩ V3 .

Then we get a symmetric member U3 of U such that

U3◦ U3◦ U3 ⊂ U2∩V3.

In this manner, we proceed by induction and get a sequence

{Un:n∈N} of symmetric members of U such that for each n∈N

,

Un+1◦Un+1◦Un+1 ⊂ Un∩Vn+1 .

Then Un ⊂Vn for all n and so { Un:n∈N} is a base for U since

{Vn:n∈N} is given to be a base.

NOTE 2.2

Having obtained a normalised countable base { Un:n∈N}for U, we

can construct a pseudo-metric d on X as follows.

Set U0 = X×X. Note that Un ⊂ Un-1 for all n∈N.

12

Define d:X×X → R in such a way that for each n∈N, the set

𝑥, 𝑦 ∈ 𝑋 × 𝑋/𝑑(𝑥, 𝑦) < 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

𝑥, 𝑦 ∈ 𝑋 × 𝑋/𝑑(𝑥, 𝑦) < 2-n} will be between Un and Un-1.

We begin with a function 𝒇: 𝑿 × 𝑿 → 𝐑 defined by f(x,y) = 2-n

in case

there exists n∈N. such that (x,y)∈Un-1-Un.

If there exists no such n, it means (x,y)∈ 𝑈∞𝑖=1 n and in that case we

set f(x,y)=0.

Here f (x,y)= f(y, x) ∀ x, y∈ 𝑋 since all Un’s are symmetric. Also for

each n∈N, 𝒙, 𝒚 ∈ 𝑿 × 𝑿/𝒇(𝒙, 𝒚) < 𝟐-n}=Un-1. Now for x,y∈ 𝑋 define

d(x,y)=inf 𝑓𝑛𝑖=1 (xi,xi+1) where the infimum is taken over all possible

finite sequences {x0,x1,x2,…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,…xn.

The number 𝑓𝑛𝑖=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: 𝑋 × 𝑋 → 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

⇒ d(x,y)≤ f(x,y) for all x,y∈ 𝑋

⇒d(x,y)≥0

If x=y, then f(x,y)=0

⇒d(x,y)=0

Now f(x,y)= f(y,x)

⇒d(x,y)= d(y,x)

Now only the triangle inequality remains to be established. Let x, y,

z ∈ 𝑋. 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 s1∈S1 and s2∈S2

together determine an element of S3 by juxtaposition, which we

denote by s1+s2.

Let 𝜇 denote the length of the chain.

Now let J(S3) be the image of S1×S2 in S3 under the juxtaposition

function

+: S1 × S2 →S3. Then we have,

d(x,y)+ d(y,z) = inf {𝜇(s1) : s1 ∈ S1} + inf {𝜇(s2) : s2∈S2}

= inf {𝜇(s1) + 𝜇(s2) : (s1,s2)∈S1×S2}

= inf {𝜇(s1+s2) : (s1,s2)∈S1×S2}

= inf {𝜇(s3) : s3 ∈ J(S3) ⊂S3}

≥ inf {𝜇(s3): s3∈S3}

= d(x,z)

Hence d(x,y)≥ 0, d(x,y)=0 ⇒ x = y

d(x,y)+ d(y,z) ≥ d(x, z)

14

hence d is a pseudo-metric on X

LEMMA 2.4

For any in teger k≥0 and x 0,x 1 ,x 2 ,…x k∈X,

f(x 0,x k+1)≤2 𝑓(𝑘𝑖=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 ,…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= 𝑓(𝑘𝑖=0 xi,xi+1)

Here f(x i,x i+1) ≥0 for all i

⇒ a ≥ 0 and a=0 only if f(x i,x i+1)=0 for every i=1,2,… k

also f(x i,x i+1)≤ a

now we show that f(x 0,x k+1) ≤2a

now we make three cases

case1: a ≥1/4

then 2a ≥1/2, by the definition of f the largest value it can take

is ½, since f(x,y)=2-n

, n∈N

∴ f(x 0,x k+1)≤1/2 ≤2a

Case 2: Let a=0.

Then f(x i,x i+1)=0 ∀ 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)∈ 𝑈∞𝑛=0 n

We decompose the chain x 0,x 1 ,x 2 ,…x k ,x k + 1 into three chains, say

x 0,x 1 ,x 2 ,…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 1≤ r ≤ s ≤ 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 n∈N , (x0,xr) ∈Un , (xr,xs) ∈Un, and , (xs,xk+1) ∈Un

⇒ (x0,xk+1) ∈ Un◦Un◦Un

But Un◦Un◦Un < Un-1

There for (x0,xk+1) ∈ Un-1

So f(x0,xk+1)=0

Case 3: Let 0 < a < 1/4.

Let r be the largest integer such that f𝑟−1𝑖=0 (xi,xi=1) ≤ a/2,if

f(x0,xi)> 𝑎/2, this definition fails and we set r =0 in this case. Then

0 ≤ r ≤ k

so each of the chains xo,…,xr, and xr+1,…, xk+1 has less than k nodes.

The first chain has length ≤ 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) ≤a, f(xr+1,xk+1) ≤a

While f(xr, xr+1) ≤ a

Let m be the smallest integer such that 2-m ≤ a.

16

In particular, f(x0,xr+1) ≤2-m

⇒(x0, xr+1) ∈ Um-1

similarly (xr, xr+1) ∈ Um-1 and(xr+1, xk+1) ∈ Um-1

⇒(x0, xk+1) ∈ Um-1◦ Um-1◦Um-1 ⊂ Um-2

Hence f(x0, xk+1) ≤ 2-(m-1)

≤ 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 X×X to R

where R has the usual uniformity and X×X 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 d∈D , 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= Xd∈𝐷 d. then the evaluation function f:X→Y defined by f(x)(d)=x

for all d∈D, x∈X is a uniform embedding of (X,U) into (Y,V)

17

Proof

Proof of the first statement is clear

For ii) we let πd:Y→X be the projection for d∈D . Here for each

d∈D

πd◦f : X →Xd 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) →(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 f×f 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 d∈D. Then clearly (f×f) (Vd,r)is precisely Z∩(πd×πd)-

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 X×X 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 d∈D

uniformly continuous. So V⊂U…….(1)

Now suppose u∈U. Let Ui be the symmetric member of U contained in U.

So U0=X×X . Define U2,U3......Un......... by induction so then

18

each Un is a symmetric member or U and Un◦Un◦Un⊂Un-1 for each n∈N.

Then we get a pseudometric d: X×X→R such that for each n∈N

Un+1 ⊂{(x,y)∈X×X : d(x,y)<2-n

} ⊂Un-1. Then d is uniformly continuous with

respect to the product uniformity on X×X. So d∈D , then d generates the

uniformity U.

Put n=2, we get {(x,y) ∈ X×X: d(x,y)<1/4} ⊂U1⊂U. Let S be the defining

sub-base for V. Then {(x,y) ∈ X×X: d(x,y)<1/4}∈S⊂ V ,and so u∈V , so

U⊂V ………(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,τu) is completely regular.

proof

from proposition 3.7 it follows that (X,τu) is embeddable into a product

of pseudometric spaces. Then (X,τu) 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 X×X 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

20

CHAPTER 3

COMPLETENESS AND COMPACTNESS

21

DEFINITION 3.1

A net {Sn:n∈D} in a set X is said to be a Cauchy net with respect to a

uniformity U on X if for every U∈U there exist p∈D such that for all m ≥ p

, n ≥ pin D (Sm,Sn)∈D

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:n∈D} be a netin a uniform space (X;U) . suppose {Sn:n∈D} is

converges to x in X. let U∈U , then there exist a symmetric V∈U such that

V◦V ⊂ U. Now V[x] is a neighbourhood of x in the uniform topology on X.

so threre exist a p∈D such that for all n≥p in D, Sn∈ V[x], ie, (x,Sn)∈ V.

now for any m,n ≥p, (Sm,x) ∈ V and (x,Sn)∈ V by symmetry of V

So (Sm,Sn)∈ V◦V⊂ U, since U∈U was arbitrary , it follows that {Sn:n∈D} is

a Cauchy net.

Now assume {Sn:n∈D} is a Cauchy net. Let {Sn:n∈D} converges

to x. Then x is a cluster point .

Conversely assume that x is a cluster point of a Cauchy net {Sn:n∈D} 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

U∈ U such that U[x] ⊂G. now we find

22

a symmetric V∈U such that V◦V ⊂ U . then threre exist a p∈ D such that

for all m,n≥p in D,(Sm,Sn)∈ V.Since x is a cluster point of {Sn:n∈D} there

exist a q≥p in D such that Sq∈ V[x] . ie, (x,Sq) ∈ V. Then for all n≥q we

have (Sq,Sn)∈ V and (x,Sq)∈ V .

So (x,Sn)∈ V◦V ⊂ U.

Hence Sn∈ U[x] ⊂G.

⇒ {Sn:n∈D} 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:n∈D} be

a cauchy net in it. Then (X;U) has a cluster point. So by the above

proposition, {Sn:n∈D} is convergent. ie, every Cauchy net is convergent .

⇒( X;U) is complete

⇒ every compact uniform space is complete.

DEFINITION 3.5

Let (X; U) , (Y,V) be uniform spaces and f : X→ Y be uniform

continuous.Then for any Cauchy net S :D→ X, the composite net f◦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:n∈D} 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:n∈D} is a Cauchy net with respect to a metric d. Then

for every 휀>0, there exist n0∈D such that for every m,n∈ D, m≥n0 and n≥n0

d(Sn,Sm)<휀.

Let U∈ U , then d(Sn,Sm)<휀 ⇒ (Sn,Sm)∈ U

ie for every 휀>0, there exist n0∈ D such that for every m,n∈ D, m≥n0 and

n≥n0 , (Sn,Sm)∈ U

⇒ {Sn:n∈D} is a Cauchy net with respect to the uniformity U

Conversely suppose that a net {Sn:n∈D} is a Cauchy net with

respect to the uniformity U. let U∈ U then by the definition , for every U∈U

there exist a p∈ D such that for every m≥p, n≥ 𝑝 in D, (Sn,Sm)∈ U

⇒ d(Sn,Sm)<휀 for all 휀>0

⇒ {Sn:n∈D} 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

24

net , so every net in (X;d) is convergent ⇒ every Cauchy sequence in (X; d)

is convergent.

Conversely assume every Cauchy sequence in (X;d) is

convergent. Let {Sn:n∈D} be a Cauchy net in (X;d). Now it is enough to

prove that {Sn:n∈D} has a cluster point in X . Set 휀=2-k

, where k=1,2,…. .

Then we obtain elements p1,p2,...pk…. in D such that for each k∈N , we

have,

(i) pk+1≥pk in D and

(ii) for all m,n≥pk in D , d(Sm,Sn)<2-k

Now consider the sequence {Spk} ;k=1,2…. Since d(Spk,Spk+1)<2-k ∀

k∈N, and the series 2∞𝑘=1

-k is convergent., {Spk} ;k=1,2…. 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:n∈D}. Let 휀>0 and m∈D be given,

We have to find n∈ D such that n≥m and d(Sn,x)<휀 .first choose k∈N so

that 2-k

< 휀/2 and d(Spk,x)< 휀/2

Now d(Sn,x) ≤ d(Sn,Spk)+ d(Spk,x)

< 2-k

+ 휀/2

< 휀/2 + 휀/2

< 휀

So x is a cluster point of the net {Sn:n∈D}

⇒ every Cauchy net in (X;d) is convergent

⇒ every Cauchy net in ( X;U) is convergent

⇒ ( X;U) is complete uniform space.

25

DEFINITION 3.7

Given an induxed collection {( Xi,Ui):i∈I} of uniform spaces . We

define a product uniformality on the cartesian product X= 𝑋𝑖∈𝐼 i

RESULT 3.8

Each projection 𝜋i : X→ X i is uniformly continuous. Let S be the

family of all susets of X×X of the form 𝜃i-1

(Ui) for Ui∈ Ui ,i∈I , where 𝜃i:

X× X → X i× X i is the function 𝜋I × 𝜋I defined by

( 𝜋I × 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):i∈I}. Then a net S:D→ X is a Cauchy net in (X;U) if and

only if for each i∈I , the net 𝜋i◦S is a Cauchy net in (Xi,Ui).

Proof

We know that the projection function 𝜋i is uniformly continuous where 𝜋i :

X→ X i .

Now for any Cauchy net S:D→ X , 𝜋i◦S is a Cauchy net in (Xi,Ui) , since

𝜋i◦S is the composite net.

Conversely assume that 𝜋i◦S is a Cauchy net for each i∈I.

Let S:D→ X be a net, let S be the standard sub-base for U , consisting of all

subsets of the form 𝜃i-1

(Ui) for Ui∈Ui and i∈I. where 𝜃i: X×X → Xi× Xi is

the function 𝜋i × 𝜋I . Suppose U = 𝜃i-1

(Ui) for some Ui∈Ui, i∈I.

find p∈ D so that for all m,n ≥p in D , (𝜋i(Sm), 𝜋i(Sn))∈Ui.

26

Such a p exist since 𝜋i◦S is a Cauchy net

⇒ (𝜋I × 𝜋I)(Sm,Sn)∈ Ui.

⇒ 𝜃i(Sm,Sn) ∈ U Ui

⇒ (Sm,Sn) ∈ 𝜃i-1

(Ui)

⇒ (Sm,Sn) ∈ U ∀ m,n≥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:X→Y 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): i∈I} of pseudometric spaces and a

uniform embedding,

27

f :X→ 𝑋𝑖∈𝐼 i where the product 𝑋i is assigned the product uniformity,

each Xi being given the uniformity induced by di. for each i∈I ,

let (X i*,di

*) be a complete pseudometric space containing (X i,di) up to an

isometry.

Then 𝑋i is a uniform subspace of 𝑋i* with the product uniformities, we

regard f as an embedding of (X;U) into 𝑋i* , which is complete.

Let Z be the closure of f(x) in 𝑋i*. 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 U∈U , there exist x1,x2,…xn∈ X such that X = 𝑈𝑛𝑖=1 [xi].

Equivalently (X;U) is totally bounded if and only if for each U∈U , 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 ⇒ totally boundedness

⇒ (X;U) is totally bounded.

28

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:n∈D} is a Cauchy net in (X;U).

Let U∈ U, find the symmetric V∈U such that V◦V⊂ U.

By total boundedness of (X; U) , there exist x1,x2,…xk∈ X such that X

= 𝑉𝑘𝑖=1 [xi].

Now for atleast one i=1,2,…k, {Sn:n∈D} is eventually in V[xi] . for

otherwise the net will be eventually X - V k[xi] ∀ i=1,2,…k and hence

eventually (𝑋 − 𝑉𝑘𝑖=1 [xi])=∅.

Thus for some i , {Sn:n∈D} is eventually in V[xi]. ie, there exist p∈D such

that ∀ n≥p , Sn∈ V[xi] , ie, (xi,Sn)∈ V .

So for all m,n ≥p in D, (Sm,Sn)∈ V◦V⊂ U. Thus {Sn:n∈D} is a Cauchy net

in (X;U). Since (X;U) is complete {Sn} converges

⇒ (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 U∈U such that for each x∈ X there exist a V∈ V such that

U[x] ⊂ V

29

Proof

Given (X;U) is a compact uniform space and V an open cover of X.

Then for each x∈ X , there exist Ux∈U such that Ux[x] is contained in some

members of V. Hence there exist a symmetric Vx∈U such that (Vx◦ Vx)[x] is

contained in some members of V . The interiors of the sets Vx[x] for x∈ X ,

cover X. So by compactness of X , there exist x1,x2,…xn∈ X such that

X= 𝑉𝑛𝑖=1 i[xi] , where Vi denotes Vx i for i=1,2,…n.

Now let U = 𝑉𝑛𝑖=1 i , then U∈U . Also let x∈X , then x∈Vi[xi] for some i.

So U[x] ⊂ Vi[xi]

⊂ Vi[Vi[xi]] , since x∈Vi[xi]

= (Vi◦Vi)[xi]

⊂ some members of V

Since this holds for all x∈ 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 𝜏u on X.

Proof

If possible let V be any other uniformity on X such that 𝜏u=𝜏v

Let f : (X;U)→ (X;V) and g : (X;V)→ (X;U) be identity function. Then f and

g are continuous (in fact a homeomorphism) because 𝜏u=𝜏v .

30

Hence by result 4.15 , every continuous function from a compact uniform

space to a uniform space is uniformly continuous.

⇒ f and g are uniformly continuous

⇒ U⊂V and V⊂U

⇒ U=V

Hence the proof.

THEOREM 3.18

Let (X;U) be a compact uniform space. Let 𝜏u be the uniform topology

on X and given X×X the product topology. Then U consist precisely of all

the neighbourhoods of the diagonal ∆𝑥 in X×X

Proof

We know that every member of U is a neighbourhood of the diagonal ∆𝑥 in

X×X . Now it is enough to prove that every neighbourhood of ∆𝑥 is a

member of U. Let V be such a neighbourhood.

Without loss of generality, we may assume that V is open in X×X . Let B

be the family of those members of U which are closed in X×X. Then B is a

base for U, since each member of U is a closed symmetric member of U.

Let W=∩{ U:U∈ B}. We claim that W ⊂ V .

For this suppose (x,y)∈W, then y∈ U[x] for all U∈ B .

Since B is a base , the family { U[x]:U∈ B} is a local base at x with respect

to 𝜏u.

In particular, since V[x] is an open neighbourhood of x , there exist a U∈ B

such that U[x] ⊂ V[x]. Hence y∈ U[x] ⊂ V[x] ⇒ y∈V[x].

ie, (x,y)∈ V. Thus W ⊂ V.

31

since X is compact, so is X×X , sice V is open so X×X - V is closed.

B ∪{ X×X - V } is a family of closed subsets of X×X and its intersection is

empty.

ie, there exist finitely many members U1,U2,…Un of B such that

U1∩U2∩…∩Un∩(X×X-V)=∅

⇒ U1∩U2∩…∩Un ⊂ V

⇒ V∈U.

Hence the proof.

32

BIBLIOGRAPHY

1. JOSHY.K.D “ Introduction to General Topology ” New age

International (p) Ltd.(1983)

2. AMSTRONG “Basic Topology SPR01 edition” Springer (India)(p)

Ltd.(2005)

3. JAMES.I.M “Introduction to Uniform Spaces” Cambridge University

Press (1990)

4. JAN PACHL “Uniform Spaces and Measures” Springer (2012)

5. MURDESHWAR.M.G “General Topology” New age International (p)

Ltd. (2007)