Metric space
Let (X,d) be a non empty set . A mapping f:XxX-->R is called metric SP if it is satisfy following conditions..
M1:-> d(x,y)=0 <=> x=y for all x,y ∈ X
M2:-> d(x,y)= d(y,x) x,y ∈ X
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
Statement M1 shows the distance between two points is zero and conversaly if the distance between two points is zero then points will be equal.
Statement M2 shows that the distance between x to y and y to x is equal.
Statement M3 shows that lenth of one side is less or equal two others sides..
A rather trivial example of a metric on any set X is the discretemetric
d(x, y) = {0 if x = y,
1 if x ̸= y.
Define d : R × R → R by
d(x, y) = |x − y|.
Then d is a metric on R. Nearly all the concepts we discuss for metric spaces are natural generalizations of the corresponding concepts for R with this absolute-value
metric...
Definition:-
Limit point:-Let S be a subset of a X. A point x in X is a limit point of S if every neighborhood of x contains at least one point of S different from x itself... Where it is not necessary that pt x is in S but necessary is that pt x is in X.
Examples
Quasi metric space :- A metric space (X,d) is called quasi metric space if there are exist a function d:XxX-> [0,∞] such that for all x,y,z is in X then it satisfies following axioms.
A rather trivial example of a metric on any set X is the discretemetric
d(x, y) = {0 if x = y,
1 if x ̸= y.
Define d : R × R → R by
d(x, y) = |x − y|.
Then d is a metric on R. Nearly all the concepts we discuss for metric spaces are natural generalizations of the corresponding concepts for R with this absolute-value
metric...
Proof:-
[M1] for
x,y in X
from mapping if x=y ó d(x,y) = 0 and for
conversaly both.
Now if x≠y
then d(x,y) = 1
≠
0 .
[M2] for
x,y in X
d(x,y) = 0=d(y,x) and also d(x,y)
=1=d(y,x).
[M3] for
x,y,z in X.
Let
d(x,y) = 1 , d(x,z) =1 , d(z,y) =1
d(x,y) =1 < 1+1= d(x,z)+d(z,y)
ie
d(x,y)≤d(x,z)+d(z,y) for all x,y,z in X
therefore
it is matric . it is also called trival metric or discrete metric. And the
space (X,d) is called trival or discrete metric space.
Limit point:-Let S be a subset of a X. A point x in X is a limit point of S if every neighborhood of x contains at least one point of S different from x itself... Where it is not necessary that pt x is in S but necessary is that pt x is in X.
Examples
- R with the usual metric
Sets sometimes contain their limit points and sometimes do not.- The points 0 and 1 are both limit points of the interval (0, 1).
- The set Z R has no limit points.
For example, any sequence in Z converging to 0 is eventually constant.
- R2 with the usual metric
The limit points of the open disc {(x, y) R2 | x2+y2 < 1} form the closed disc {(x, y) R2 | x2+ y21}.
Any point on the boundary of the circle is a limit of a sequence of points inside the circle. - In R every real number is a limit point of the subset Q of rationals.
Proof
Every real number can be approximated arbitrarily closely by a sequence of rationals .(st-and.ac.uk)
Quasi metric space :- A metric space (X,d) is called quasi metric space if there are exist a function d:XxX-> [0,∞] such that for all x,y,z is in X then it satisfies following axioms.
M1:-> d(x,y)= 0 if x=y for all x,y ∈ X
M2:-> d(x,y)= d(y,x) x,y ∈ X
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
pseudo metric space:- (X,d) be a non empty set . A mapping f:XxX-->R is called pseudo metric Sp if it is satisfy following conditions..
Open Sphere:- let (X,d) be a metric space ,if y be any point of X,and r be a positive real number then a subset of X is called open sphere if it is define as
Sr(y)= {x∈X: d(x,y)< r) where y is a center and r is a radius of it.
Example: in a Real line [0,1) is not an open set because 0∈[0,1). If we remove 0 from this then we have a set (0,1) . this is an open set.
Definition of interior point:-let G be a subset of X.then any point of x of G is called interior point if there are exist an open sphere Sr(x) where x is the center point of Sr(x) such that Sr(x) )⊆G.
Φ is an open set :- since there is not point in set Φ . then nothing to prove. Therefore Φ is an open set.
Full space is an open set:- since X is a metrix space . so it have many points. We can take a point and setup a open sphere.
Let x∈X . then there are exist a positive real number r>0 such that Sr(x).
this open sphere will be in X. i.e. Sr(x)⊆G. or can say x∈Sr(x)⊆G.
and hence full space X and null set Φ are open sets.
Theorem:- In a metric space ,every open sphere is an open set.
Result:- In Rk , an open ball is a convex set.
Similarly we can prove finite intersection of convex sets is convex.
cheractiristic of open sets-
1. A subset of metric space is open iff it is the union of a family of open sphare.
2. in a discreate metric space ,every set is open.
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
M1:-> d(x,y)=0 => x=y for all x,y ∈ X
M2:-> d(x,y)= d(y,x) x,y ∈ X
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
M3:-> d(x,y)≤d(x,z)+d(z,y), for all x,y,z∈X
Open Sphere:- let (X,d) be a metric space ,if y be any point of X,and r be a positive real number then a subset of X is called open sphere if it is define as
Sr(y)= {x∈X: d(x,y)< r) where y is a center and r is a radius of it.
Definations:-
Where y is center and r
is radius of X. we can also show it S(y,r).
Open set:- let (X,d) be a
metric space. A subset G of X is called an open set if for every point x of G
there are exist a positive real number r such that x∈Sr(x)⊆G. that is every point of G is a center point of G which
is also in G.
In other words, every
point of G will be an interior point of G.
Example: in a Real line [0,1) is not an open set because 0∈[0,1). If we remove 0 from this then we have a set (0,1) . this is an open set.
Definition of interior point:-let G be a subset of X.then any point of x of G is called interior point if there are exist an open sphere Sr(x) where x is the center point of Sr(x) such that Sr(x) )⊆G.
Now u can understand with examples
Theorem:- In a metric
space X, the empty set and full space X are open sets.
Proof:- for this, we
have to show that,a point of that set is a center point which is also in that
set.
Φ is an open set :- since there is not point in set Φ . then nothing to prove. Therefore Φ is an open set.
Full space is an open set:- since X is a metrix space . so it have many points. We can take a point and setup a open sphere.
Let x∈X . then there are exist a positive real number r>0 such that Sr(x).
this open sphere will be in X. i.e. Sr(x)⊆G. or can say x∈Sr(x)⊆G.
and hence full space X and null set Φ are open sets.
Theorem:- In a metric space ,every open sphere is an open set.
Proof:-
let (X,d) be a metric space and let Sr(x0) is an open sphere in X.where r is a radius and x0 is a center.
To
prove:- an open sphere is an open set.
For this we will show , there are exist an open sphere Sρ(x) such that Sρ(x) ⊆ Sr(x0).
Now, let x∈ Sr(x0) be an orbitary element.
x∈ Sr(x0) => d(x,x0 )< r
=> r - d(x,x0 )>0.
Let
r - d(x,x0 )=ρ. ………………(1)
Therefore take ρ>0 as radius.and let x be a
center then Sρ(x) is an open sphere.
Let
y∈ Sρ(x) => d(y,x)<ρ
=> d(y,x) < r - d(x,x0 ) BY (1)
Now, d(y,x0 )≤
d(y,x)+d(x,x0) {by m3}
< r - d(x,x0
)+ d(x,x0 )= r
Then d(y,x0 )<r
So,
d(y,x0 )<r => y∈ Sr(x0)
i.e
y∈ Sρ(x)
=> y∈ Sr(x0)
therefore
Sρ(x) ⊆ Sr(x0) ,
for all x∈ Sr(x0).
So we can say ,there are exist
open sphere such that Sρ(x) ⊆ Sr(x0). So by the definition Sr(x0) is an open set.
Hence
proved.
Theoram:- In a metric space
, the union of an orbitary collection of open sets is open.
Proof:- let (X,d) be a metric space and let {Gλ :λ∈V} is a collection of family of all open sets.
We have to prove:- G=Uλ∈V Gλ is an open set
Condition 1:- if {Gλ} = Φ .since a null set is an open set.then nothing to prove.
Condition 2:- If {Gλ} ≠ Φ. So let x∈G.
x∈G => x∈ Uλ∈V Gλ
=> x ∈
Gλ ,for all λ∈V.
Since, Gλ is an open set
then there are exist Sr(x) ⊆
Gλ but
Gλ ⊆G => Sr(x) ⊆
G.
i.e
we
can say there are exist a positive real number r such that Sr(x) ⊆ G. therefore G is
open.
Theorem :- In a metric space,
the intersection of a finite number of open sets is open.
Proof:- let (X,d) be a metric space and let {Gi :λ∈V} is a finite collection of family of all open sets.
Then we have to prove:- G = ∩ni=1 Gi is also an open set.
Condition
1:- {Gi} = Φ and we know that each null set is an
open set then nothing to prove.
Condition2:-
{Gi} ≠ Φ.
Let
{Gi} = {G1,G2…….Gn} where n is any natural number.
Let
x∈G.
x∈G => x∈Gi , for all i=1,2,….n.
then there are exist a positive real number ri such that Sri(x)
⊆Gi ,for all i=1,2,3…n ……(a)
then
there are exist a positive real number r=min {r1,r2,…rn}
such that Sr(x) ⊆ Sri(x) for all i=1,2,…n. ……….(b)
by
(a) and (b) Sr(x) ⊆Gi for all I = 1,2…n.
=>
Sr(x) ⊆∩ni=1 Gi =G
Note:- * In a metric space , complement of each singleton set is open .
·
In a discrete mathematics each set is an open set.
Open ball:- let a∈ Rk and r>0 where a is a center and r is a radius is define
as set of all points like x∈ Rk such that B(a,r) = { x∈ Rk : │x-a│< r }
Closed ball :- let a∈ Rk and r>0 where a is a center and r is a radius is define
as set of all points like x∈ Rk such that B[a,r] = { x∈ Rk : │x-a│≤ r }
Convexity in Rk :- A subset A¥ Rk is called convex if it contains line segment between
two points in the set .let x, y∈ A then λx+(1-λ)y∈A , for all x∈A and y∈A
and 0<λ<1.
 |
Proof:- let x,y ∈ B(a,r) then │x-a│< r and │y-a│< r
We will prove :- │ {λx+(1-λ)y} - a│∈ B(a,r)
Now suppose 0<λ<1.
Now , │ {λx+(1-λ)y} - a│ = │ {λx+y-λy- a│
= │ {λx-λa+λa+y-λy- a │
= │λ(x-a)+λa-a-λy +y│
= │λ(x-a)+(1-λ) (y-a)│
≤ λ │x-a│ + (1-λ) │y-a│
< λr + (1-λ) r = λr +
r – λr = r
i.e
│ {λx+(1-λ)y} - a│<r
Therefore we can say
Therefore we can say
x,y ∈ B(a,r) => │ {λx+(1-λ)y} - a│∈ B(a,r) where 0<λ<1.
And hence B(a,r) is a convex set.
closed sphere:- let (X,d) be a metric space . x is an element of X and
r is a positive real number then a subset
of X is called open sphere if it is define as
Sr[x] = { x∈X : d(y,x)≤ r }
Where x is center and r is radius of X. we can
also show it S(x,r).cheractiristic of open sets-
1. A subset of metric space is open iff it is the union of a family of open sphare.
2. in a discreate metric space ,every set is open.
Theorem :- In a metric space, the intersection of an orbitary collection of closed sets is closed.
Proof:- let (X,d) be a metric space and let {Di :λ∈V} is a orbitary collection of closed sets.
Then we have to prove:- D = ∩ni=1 Di is also an open set.
Condition 1:- {Di} = Φ and we know that each null set is a closed set then nothing to prove.
Condition2:- {Di} ≠ Φ.
Dλ is closed for all λ∈V => compiments of Dλ is an open λ∈V.
=>so union of compliment of Dλ is open [orbitary union of open sets is open]
=>compliment of intersection of Dcompliment Dλ is open [by de morgan law]
=>so intersection ∩λ∈V Di =D is closed.
hence proved.
theoram :-In a metric space the union of a finite number of cloesed sets is closed.
for students home work.
defination of closed set is :- A set is closed if it compliment is open.
another defination of closed set :- A set D is called closed if it contains each of its limit points.
=>so union of compliment of Dλ is open [orbitary union of open sets is open]
=>compliment of intersection of Dcompliment Dλ is open [by de morgan law]
=>so intersection ∩λ∈V Di =D is closed.
hence proved.
theoram :-In a metric space the union of a finite number of cloesed sets is closed.
for students home work.
defination of closed set is :- A set is closed if it compliment is open.
another defination of closed set :- A set D is called closed if it contains each of its limit points.
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