BOLZANO-WEIERSTRASS THEOREM FOR SETS | REAL ANALYSIS

INTRODUCTION:

Have you remembered  Bounded set ,Order completeness of Real numbers.
If not just Click to Read Real Analysis part-1 .

Bounded Set:
A set is said to be bounded if it is bounded above and below.
" I will  not explain this just read about it in the above link "

Bolzano-Weierstrass Theorem:

Statement: Every infinite bounded set of real numbers has a limit point.

Proof: Let ' S ' be any infinite bounded set of real numbers.
                 
                  m ≤ x ≤ M ,      ∀  x ∈S  ---(1)
                     
    m = infimum of S       M=supremum of S

Let us assume T is a set of Real numbers.
 
          T = { x : x is greater than only for finite number of elements } ------(2)

 x ∈ T { x ≥ only a finite number of elements of S } --(3)

{ x ≤ infinite number of elements of S         }---(4)

  [ Don't confuse with 'S' and 'T' .   S is a infinitely bounded set while T is a finitely bounded set and T is the subset of S also element 'x' is common in both sets]

From equation (1) and (4)
               
                      m ∈ T  ⇔  T ≠ Φ

From equation (1)
               
                       M ≥ x  ,     ∀  x ∈S
 
Let us assume another element 'y' from set T.
                  y ∈ T  ⇔  y ≥ only a finite number of elements of S  [From equation (3)]

    M ≥ y ,    ∀  y ∈ T  ⇔ T is bounded above.

 Now we have prove that T is bounded above

So it will follow Order Completeness Property and we know that " There exist a supremum of  a set of which set follows the completeness property ".

       Let  P be the Supremum of T .
 
Claim: ( P  is a limit point  of set S )

          Let є > 0 be arbitrary number.

 Since   P + є > P
     
          ⇒ P + є ≥ finite number of elements of S   -----(5)
   [ Because P is the Supremum of T and T is the subset of S]
 
Since     P - є < P
      
         ⇒ P - є ≤  infinite number of elements of S ----(6)

From equation (5) and (6) that  nbh
      
                ］  P - є , P + є [   of  P contains infinite number of elements of S.

Hence , P is a limit point.


Example:

1.  Z  has no limit point .{it has not bounded}

    2.   S = { -1 ⁿ(n/(n+1)  : n ∈ N}

Ans. 
           S = { -1/2 ,2/3 , -3/4 .....}
          infinite | Bounded   -1 < x < 1

hence lmit point will  be 0.

If any doubt , Comment below