CONVERGENCE OF SEQUENCES IN METRIC SPACE

                                                      CONVERGENCE 

  DEFINITION 

 WE SAY THAT SEQUENCE (Xn) CONVERGES TO x𝟄 X IF  GIVEN  𝜀 > 0,

THERE EXISTS N  𝟄   NATURAL NUMBER SUCH THAT FOR ALL n >= N , WE HAVE

Xn 𝝐   B(x,𝛆).OR d(Xn ,x)<𝛆.

QUESTIONS

1. IF Xn converges to x

prove that subsequence of Xn also converges to x

proof:   IF Xn converges to x , then from above definition;

there exists 𝛆  >0 such that d(Xn,x)< 𝛆 

let subsequence of Xn be Xnk . (here k is ≥ 1)

so we have to prove that Xnk converges to x

we have  d(Xn,x)< 𝛆  

now, from triangle inequality ,  d(Xn,x)<d(Xn,Xnk)+d(Xnk,x)   and  d(Xn,x)< 𝛆 

as 𝛆  >0 ,  we have d(Xn,Xnk)+d(Xnk,x) < 𝛆 

                  hence,  d(Xnk,x) < 𝛆 

from definition of convergence, Xnk converges to x

As Xnk is subsequence of X , 

ANY SUBSEQUENCE OF CONVERGENT SEQUENCE IS CONVERGENT TO LIMIT 

OF GIVEN SEQUENCE.

2. The limit of sequence in metric space is unique.

proof:   

let Xn converges to x

and Xn converges to y

then if x=y , above theorem is proved.

let us assume that x≠ y........................1.

so                          d(x, y ) >0

As Xn converges to x.............   d(Xn , x)< 𝛆1

also   Xn converges to y ..........d(Xn , y)< 𝛆 2

IF     𝛆  = max{  𝛆1   ,    𝛆 2    }

So        d(Xn , x) < 𝛆

            d(Xn , y)  < 𝛆

By triangle inequality

          d(x,y)< d(x,Xn) + d(Xn,y) <   𝛆  +   𝛆 

d(x,y)<d(x,Xn) + d(Xn,y) < 2𝛆

d(x,y)<d(x,Xn) + d(Xn,y) < 𝛆1...........other constant

so,      d(x,y)< 𝛆1

as    𝛆1 is any positive number,

we can take  d(x,y)=0

so x=y   ..................................CONTRADICTION TO 1.

SO, OUR ASSUMSION , X≠ Y IS WRONG 

SO, X=Y

MEANS LIMIT OF SEQUENCE IN METRIC SPACE IS UNIQUE.

 

 3.  Let    Xk 𝟄  Rn  converges to x 𝟄  Rn.

Show that ‖  Xk ‖ converges to  ‖  X ‖.

proof:

  

As  Xk converges to x

      d(Xk , x)<ε

BUT  AS Xk and x are vectors, we write;

        ‖  Xk - x ‖ <  ε

FROM FORMULA,  |   ‖  X ‖  -     ‖  Y ‖    |   ≤    ‖  X -Y   ‖

  WE GET                ,   |   ‖  Xk ‖  -     ‖  X ‖    |   ≤    ‖  Xk -X   ‖  <   ε

                                       |   ‖  Xk ‖  -     ‖  X ‖    |   <      ε

                                        HENCE

                                             ‖  Xk ‖  converges to    ‖  X ‖ 

                                         

.