MthT 430 Notes Chap 7d Bolzano - Weierstraß Theorem
MthT 430 Notes Chap 7d Bolzano - Weierstraß Theorem
For this discussion, we shall assume:

(P13-BIN) Binary Expansions Converge. Every binary expansion represents a real number x: every infinite series of the form
c1 2−1 + c2 2−2 + …+ ck 2−k + …,    ck ∈ {0,1},
converges to a real number x in [0,1] and write the binary expansion of x as
x
= .binc1 c2 ….
In http://www.math.uic.edu/ lewis/mtht430/chap7b.htm#BIN, there was an indication of the proof of
The Bolzano-Weierstraß Theorem

Theorem (Bolzano-Weierstraß). Let {xn}n=1∞ be a sequence of points in. [0,1]. Then there is an x in [0,1] which is a limit point3 of the sequence {xn}n=1∞.
The proof will construct a binary expansion for x.
Now - either infinitely many terms of the sequence are 1, in which case x = 1 = 1.bin0 is the desired limit point OR
Ask the question: For infinitely many k, is it true that xk ∈ [0,[1/(21)])?
If YES, let
c1
= 0,
a1
= 0 = 0.bin0,
b1
= 1
2
= a1 + 1
21
 .
s1
= a1.
Then
b1 − a1
= 1
21
 ,
Infinitely many xk are in [a1,b1).
If NO, let
c1
= 1,
a1
= 1
2
 = 0.bin1,
b1
= 1 = 1.bin0
= a1 + 1
21
 .
s1
= a1.
Then
b1 − a1
= 1
21
 ,
Infinitely many xk are in [a1,b1).
Now continue, …
x
=
lim
n → ∞ 
 sn
=
lim
n → ∞ 
sn + 1
2n
Note that 0 ≤ x − sn = |x − sn| ≤ [1/(2n)].
Fuller Proof of Bolzano-Weierstraß)
Let's be more precise.

Theorem (Bolzano-Weierstraß). Let {xn}n=1∞ be a sequence of points in. [0,1]. Then there is an x in [0,1] which is a limit point3 of the sequence {xn}n=1∞.
The proof will construct a binary expansion for x.
Now - either infinitely many terms of the sequence are as close as desired to 1, in which case x = 1 = 1.bin0 is the desired limit point.
Otherwise -
We will try to construct a sequence ck ∈ {0,1} such that if
ak
= .binc1 …ck,
bk
= ak + 1
2k
 ,
Ik
= [ak,bk),
then Ik is a decreasing sequence of intervals such that for every k, Ik contains infinitely many terms of the sequence {xn}.
Let
a0
= 0,
b0
= 1 = a0 + 1
20
 ,
I0
= [a0,b0);
I0 contains infinitely many terms of the sequence {xn}.
Divide the interval I0 into two halves, [a0,a0 + [1/(21)]) and [a0 + [1/(21)],b0) - call them the left half and right half - at least one contains infinitely many terms of the sequence {xn}. Pick it (in case of a tie, choose either as you desire). If the left is chosen, let
c1
= 0,
a1
= a0 = .binc1,
b1
= a0 + 1
21
 = a1 + 1
21
 ,
I1
= [a1,b1).
If the right is chosen, let
c1
= 1,
a1
= a0 + 1
21
 = .binc1,
b1
= b0 = a1 + 1
21
 ,
I1
= [a1,b1).
I1 contains infinitely many terms of the sequence {xn}.
Now continue, by recursion to choose ck.
If c1, …. ck, have been chosen so that for 1 ≤ j ≤ k,
aj
= .binc1 …cj,
bj
= aj + 1
2j
 ,
Ij
= [aj,bj),
I1 ⊇ … ⊇ Ik, and for 1 ≤ j ≤ k, Ij contains infinitely many terms of the sequence {xn} -
Divide the interval Ik into two halves, [ak,ak + [1/(2k+1)]) and [ak + [1/(2k+1)],bk) - call them the left half and right half - at least one contains infinitely many terms of the sequence {xn}. Pick it (in case of a tie, choose either as you desire). If the left is chosen, let
ck+1
= 0,
ak+1
= ak = .binc1 …ck+1,
bk+1
= ak + 1
2k+1
 = ak+1 + 1
2k+1
 ,
Ik+1
= [ak+1,bk+1).
If the right is chosen, let
ck+1
= 1,
ak+1
= ak + 1
2k+1
 = .binc1 ... ck+1,
bk+1
= bk = ak+1 + 1
2k+1
 ,
Ik+1
= [ak+1,bk+1).
Ik+1 contains infinitely many terms of the sequence {xn}.
The desired limit point x is
x
=
lim
k → ∞ 
 ak
=
lim
k → ∞ 
 (bk)
Note that 0 ≤ x − ak = |x − ak| ≤ [1/(2k)].

Footnotes:

3A point x is a limit point of the sequence if for every ε > 0, infinitely many terms of the sequence are within ε of x. Alternately, there is a subsequence which converges to x. A more informal idea is to say that infinitely many terms are as close as desired to x.
3A point x is a limit point of the sequence if for every ε > 0, infinitely many terms of the sequence are within ε of x. Alternately, there is a subsequence which converges to x. A more informal idea is to say that infinitely many terms are as close as desired to x.

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On 01 Nov 2006, 14:41.