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The International Journal Of Engineering And Science (IJES)
|| Volume || 3 || Issue || 7 || Pages || 61-62|| 2014 ||
ISSN (e): 2319 – 1813 ISSN (p): 2319 – 1805
www.theijes.com The IJES Page 61
On Uniformly Continuous Uniformity On A Topological Space
1
Dr. S.M.Padhye , 2
Ku. S.B.Tadam
1,2
Shri R.L.T. College of Science, Akola
------------------------------------------------------------ABSTRACT:---------------------------------------------
Given a topological space (X, ) we define uniformity on X such that (X, is
uniformly continuous uniform space and it becomes the smallest uniformly continuous uniformity
with respect to which set of continuous functions is larger than set of –continuous functions.
Keywords: Pseudo-metric space, Uniform space, Uniformly continuous function.
Subject code classification in accordance with AMS procedures: 54E15.
---------------------------------------------------------------------------------------------------------------------------------------
Date of Submission: 12 April 2014 Date of Publication: 20 July 2014
---------------------------------------------------------------------------------------------------------------------------------------
INTRODUCTION
Let X be any non empty set. Let be the family of all pseudo metrics on X. Then since discrete metric
on X is a pseudo metric. For any subfamily of define the uniformity on X whose subbase is the family of all
sets
Vp,r = {(x,y) XxX /p(x,y)<r}, p and r > 0
/ is a base for uniformity on X.
Now let (X, ) be a topological space. Let (X) be the set of all continuous complex valued functions on X. For
every f (X), define df :XxX as df (x,y) = I f (x ) – f( y) l. Then df is a pseudo -metric on X. With the
help of pseudo-metric df on X define an open sphere Sr(x), x X, r > 0 as Sr (x) = { y: df (x,y) < r }
As the intersection of any two open spheres contains an open sphere about each of its pts , the family { Sr (x) /
x X } is a base for topology for X. This topology is the pseudo-metric topology for X generated by pseudo-
metric df on X say .
Let P be the family of pseudo metrics df on X as f (X). Then P defines a unique uniformity on X such that S
= { Vf,r / f (X) and r > 0 } forms a subbase where Vf,r = {(x,y) XxX / df (x,y) < r} We denote this
uniformity on X by
Definition: Uniformly continuous uniform space:
A uniform space is said to be Uniformly continuous uniform space if every real valued
continuous function is uniformly continuous.
Theorem1: Let (X, ) be a topological space and be the uniformity defined on X as above then (X, ) is
a Uniformly continuous space.
Proof: For proving this theorem we require following two lemmas.
Lemma1. Every -continuous mapping is -continuous.
Proof: Let f: X be any -continuous mapping, where is the above defined uniformity on X. Then we
show that f is -continuous. Let G be any open set in . Then f-1
(G) is – open in X. i.e. where
= {T⊂X/ T U 𝒰 s.t. U[x]⊂T} . Now we show that ⊆ .
On Uniformly Continuous Uniformity…
www.theijes.com The IJES Page 62
Suppose T and T. Then we have U 𝒰 s.t. U[x]⊂ T. Since U 𝒰 by defn
of f1,f2,…..fn (X)
and r1,r2,…….rn all positive such that . Then .
But V[x] = { y: (x,y) }
={ y: (x,y) }
={ y: (x,y) }
={ y: } }
={ y: y ( } }
={ y: y
Now for each i, is an open subset of and each fi is -continuous
hence is – open. is – open. is – open ie.
– open set V[x] such that This proves that T . Hence ⊆ .
Lemma2: If f is –continuous then f is 𝒰-uniformly continuous function.
Proof: Since f is -continuous function, df(x,y)= l f(x)-f(y) l ,x,y X defines a pseudo metric on X and hence
df P. Then for any r >0 Uf,r ={(x,y)/df(x,y)<r} is a subbase member of the uniformity 𝒰. i.e. Uf,r 𝒰 such that
(x,y) Uf,r l f(x)-f(y) l < r.Thus f is 𝒰-uniformly continuous function.
Proof of Theorem 1: Suppose f is - continuous complex valued function on X. By lemma1 f is –
continuous. By Lemma 2 f is – uniformly continuous function. Hence (X, ) is uniformly continuous
uniform space.
Note: From lemma1 But from lemma2 every –continuous function is 𝒰-uniformly
continuous and hence it is –continuous function.
ie. ⊆ =
Theorem2: Suppose (X, ) is a uniformly continuous uniform space such that ⊆ then ⊂ where
is the uniformity on X determined by –continuous functions on X. i. is the smallest uniformly continuous
uniform space such that ⊆
Proof: To show that ⊂ , let U . Then there are –continuous functions
f1,….fn on X and such that W = U ……(1) .
Since each –continuous function fi is –continuous and is a uniformly continuous uniform space, each fi is
- uniformly continuous . Hence for each such that
. But (by definition
of uniform space) i.e. … is the smallest uniformly continuous uniform space such that ⊆
Corollary: Suppose (X, ) is a uniformly continuous uniform space such that
then ⊂ .…
References:
[1]. J.L.Kelley, General Topology, Van Nostrand, Princeton, Toronto, Melbourne, London 1955.
[2]. W.J. Pervin, Foundations of General Topology,Academic Press Inc. New York, 1964.
[3]. Russell C. Walker, The Stone-Cech Compactification, Springer-Verlag Berlin Heidelberg New York 1974

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H03702061062

  • 1. The International Journal Of Engineering And Science (IJES) || Volume || 3 || Issue || 7 || Pages || 61-62|| 2014 || ISSN (e): 2319 – 1813 ISSN (p): 2319 – 1805 www.theijes.com The IJES Page 61 On Uniformly Continuous Uniformity On A Topological Space 1 Dr. S.M.Padhye , 2 Ku. S.B.Tadam 1,2 Shri R.L.T. College of Science, Akola ------------------------------------------------------------ABSTRACT:--------------------------------------------- Given a topological space (X, ) we define uniformity on X such that (X, is uniformly continuous uniform space and it becomes the smallest uniformly continuous uniformity with respect to which set of continuous functions is larger than set of –continuous functions. Keywords: Pseudo-metric space, Uniform space, Uniformly continuous function. Subject code classification in accordance with AMS procedures: 54E15. --------------------------------------------------------------------------------------------------------------------------------------- Date of Submission: 12 April 2014 Date of Publication: 20 July 2014 --------------------------------------------------------------------------------------------------------------------------------------- INTRODUCTION Let X be any non empty set. Let be the family of all pseudo metrics on X. Then since discrete metric on X is a pseudo metric. For any subfamily of define the uniformity on X whose subbase is the family of all sets Vp,r = {(x,y) XxX /p(x,y)<r}, p and r > 0 / is a base for uniformity on X. Now let (X, ) be a topological space. Let (X) be the set of all continuous complex valued functions on X. For every f (X), define df :XxX as df (x,y) = I f (x ) – f( y) l. Then df is a pseudo -metric on X. With the help of pseudo-metric df on X define an open sphere Sr(x), x X, r > 0 as Sr (x) = { y: df (x,y) < r } As the intersection of any two open spheres contains an open sphere about each of its pts , the family { Sr (x) / x X } is a base for topology for X. This topology is the pseudo-metric topology for X generated by pseudo- metric df on X say . Let P be the family of pseudo metrics df on X as f (X). Then P defines a unique uniformity on X such that S = { Vf,r / f (X) and r > 0 } forms a subbase where Vf,r = {(x,y) XxX / df (x,y) < r} We denote this uniformity on X by Definition: Uniformly continuous uniform space: A uniform space is said to be Uniformly continuous uniform space if every real valued continuous function is uniformly continuous. Theorem1: Let (X, ) be a topological space and be the uniformity defined on X as above then (X, ) is a Uniformly continuous space. Proof: For proving this theorem we require following two lemmas. Lemma1. Every -continuous mapping is -continuous. Proof: Let f: X be any -continuous mapping, where is the above defined uniformity on X. Then we show that f is -continuous. Let G be any open set in . Then f-1 (G) is – open in X. i.e. where = {T⊂X/ T U 𝒰 s.t. U[x]⊂T} . Now we show that ⊆ .
  • 2. On Uniformly Continuous Uniformity… www.theijes.com The IJES Page 62 Suppose T and T. Then we have U 𝒰 s.t. U[x]⊂ T. Since U 𝒰 by defn of f1,f2,…..fn (X) and r1,r2,…….rn all positive such that . Then . But V[x] = { y: (x,y) } ={ y: (x,y) } ={ y: (x,y) } ={ y: } } ={ y: y ( } } ={ y: y Now for each i, is an open subset of and each fi is -continuous hence is – open. is – open. is – open ie. – open set V[x] such that This proves that T . Hence ⊆ . Lemma2: If f is –continuous then f is 𝒰-uniformly continuous function. Proof: Since f is -continuous function, df(x,y)= l f(x)-f(y) l ,x,y X defines a pseudo metric on X and hence df P. Then for any r >0 Uf,r ={(x,y)/df(x,y)<r} is a subbase member of the uniformity 𝒰. i.e. Uf,r 𝒰 such that (x,y) Uf,r l f(x)-f(y) l < r.Thus f is 𝒰-uniformly continuous function. Proof of Theorem 1: Suppose f is - continuous complex valued function on X. By lemma1 f is – continuous. By Lemma 2 f is – uniformly continuous function. Hence (X, ) is uniformly continuous uniform space. Note: From lemma1 But from lemma2 every –continuous function is 𝒰-uniformly continuous and hence it is –continuous function. ie. ⊆ = Theorem2: Suppose (X, ) is a uniformly continuous uniform space such that ⊆ then ⊂ where is the uniformity on X determined by –continuous functions on X. i. is the smallest uniformly continuous uniform space such that ⊆ Proof: To show that ⊂ , let U . Then there are –continuous functions f1,….fn on X and such that W = U ……(1) . Since each –continuous function fi is –continuous and is a uniformly continuous uniform space, each fi is - uniformly continuous . Hence for each such that . But (by definition of uniform space) i.e. … is the smallest uniformly continuous uniform space such that ⊆ Corollary: Suppose (X, ) is a uniformly continuous uniform space such that then ⊂ .… References: [1]. J.L.Kelley, General Topology, Van Nostrand, Princeton, Toronto, Melbourne, London 1955. [2]. W.J. Pervin, Foundations of General Topology,Academic Press Inc. New York, 1964. [3]. Russell C. Walker, The Stone-Cech Compactification, Springer-Verlag Berlin Heidelberg New York 1974