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1. The International Journal Of Engineering And Science (IJES)
|| Volume || 4 || Issue || 4 || Pages || PP.49-54|| 2015 ||
ISSN (e): 2319 – 1813 ISSN (p): 2319 – 1805
www.theijes.com The IJES Page 49
On the Construction of Cantor like Sets
Dr. S. M. Padhye1
, Satish B. Khobragade2
Associate Professor, HOD, Department Of Mathematics, Shri R.L.T. College Of Science, Akola, Maharashtra,
India1
Research Student, Department Of Mathematics, Shri R.L.T. College Of Science, Akola, Maharashtra, India2
-----------------------------------------------------------ABSTRACT----------------------------------------------------------
In this paper we construct a Cantor like set S from any sequence { } with 0< <1 with the help of
sequence { } of subsets of [0,1] such that ,m( ) = and S = ∩ with m(S) =
.Further ∑ = ∞ if and only if m(S) = 0. Cantor ternary set comes out to be a particular case of
construction of Cantor like sets by choosing = for all n. Similarly we can construct Cantor - set by
choosing = for all n. In the construction of Cantor - set the length of remaining closed intervals at each
stage are equal to , k = 1,2,3,…………… Also we can construct Cantor - set by choosing = for all n.
Here in the construction of Cantor - set the length of remaining closed intervals at each stage are equal to
, k = 1,2,3,……………
KEY WORDS: - Cantor set, Cantor like sets.
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Date of Submission: 04-April-2015 Date of Accepted: 25-April-2015
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Lemma 1:- Given any sequence { } with 0< <1,∑ = ∞ if and only if )= 0
Proof :- First step: Let = ∞
Then we have to show that )= 0 i.e. )= 0
Here we use 1- ≤ , 0 ≤ < 1
1- ≤ i =1,2,3,…………..
1- ≤
1- ≤
1- ≤
Multiplying all these inequalities we get,
(1- )(1- )…………….(1- ) ≤ . ۰۰۰۰۰۰
) ≤ ………………………(1)

2. On the Construction of Cantor like Sets
www.theijes.com The IJES Page 50
To show that )= 0, let > 0 be given. Put M = .
Since ∑ =∞ then there is N such that for n ≥ N ═> > M
>
>
>
< ………………………(2)
From equation (1) and (2) we get ,
) < for all n ≥ N.
Thus )= 0
)= 0
Conversely:-Let < ∞ i.e. ∑ < ∞ is convergent.
We show that ) ≠ 0 .
Let Pn = )
Since ≥ 0, ≠ 1 j and ∑ < ∞ .
we choose N so large that + + ۰۰۰۰۰۰< --------------------- (3)
Then using induction we prove that for all n ≥ N, (1 - )(1- )۰۰۰۰۰(1- ) ≥ [1- ( + +
۰۰۰۰۰ + )]
For n = N, the inequality is obvious. For n > N
If (1 - )(1- )۰۰۰۰۰(1- ) ≥ [1- ( + +۰۰۰۰ + )] then
(1 - )(1- )۰۰۰۰۰(1- ) (1 - ) ≥ [1- ( + +۰۰۰۰ + )] (1 - )
= [1- ( + + ۰۰۰۰ + )] + ( + + ۰۰۰۰ )
≥ [1- ( + +۰۰۰۰ + )]
Thus (1 - )(1- )۰۰۰۰۰(1- ) (1 - ) ≥ [1- ( + +۰۰۰۰ + )]
By induction the inequality holds for n ≥ N
i.e. (1 - )(1- )۰۰۰۰۰(1- ) ≥ [1- ( + + ۰۰۰۰۰ + )] for all n ≥ N ………….(4)
Now by using equation (3) we get,
(1 - )(1- )۰۰۰۰۰(1- ) ≥ [1- ] =
(1 - )(1- )۰۰۰۰۰(1- ) > ,n ≥ N ………………………. (5)
Now for n ≥ N, =
=
= > ,n ≥ N ( From equation (5))

3. On the Construction of Cantor like Sets
www.theijes.com The IJES Page 51
> ,n ≥ N ………………………. (6)
{ } ≥ ≠ 0 ………………………. (7)
Consider
- = -
- = [ 1- (1- ) ]
- = ≥ 0
- ≥ 0
{ } is monotonic decreasing and bounded below by .
{ } = ≥
Pn ≥ PN-1
Pn =α PN-1 , where α≥ 1/2
) =α {PN-1} , where α≥ 1/2
═> ) is a positive number
) ≠ 0
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Lemma 2 :-
If 0 < < 1, n ≥ 1, = + +۰۰۰۰۰۰+ and = (1- )(1- )…………….(1- ) then (1- ) ≤
≤
Proof :- Given
0 < < 1, n ≥ 1, = + +۰۰۰۰۰۰+ and
= (1- )(1- )…………….(1- )
≥ 1 – ( + +۰۰۰۰۰۰+ ) ( From equation (4) of Lemma 1)
≥ 1 -
1 – ≤ …………………………………..(1)
Now,
(1- )(1 + ) = 1 – < 1
(1- )(1 + ) < 1
(1- )<

4. On the Construction of Cantor like Sets
www.theijes.com The IJES Page 52
Similarly, (1- ) <
(1- ) <
Multiplying all these equations we get,
(1- )(1- )…………….(1- ) ≤ …………………………(2)
Now
) = 1+( + +۰۰۰+ )+( + +۰۰۰)+( +۰۰۰۰ ( +۰۰۰۰۰
) ≥ 1+ + +۰۰۰+
═> ≤
Putting in equation (2) we get,
(1- )(1- )…………….(1- ) ≤
≤
≤ ………………………………(3)
From equation (1) and (3) we get,
(1- ) ≤ ≤
* * * * * * * * * * * * * * * * * * * * * * * * * * *
Corollary 3 :-
If in addition lim = and lim = then (1 - ) ≤ ≤ .
Proof :-By lemma 2 we get,
(1- ) ≤ ≤
Given lim = and lim =
(1 - ) ≤ ≤
* * * * * * * * * * * * * * * * * * * * * * * * * * *
Preposition 4 :-
Given any sequence { } with 0< <1,there is a sequence { } of subsets of [0,1] such that
,m( ) = and S = ∩ is Cantor like set with
m(S) = .

5. On the Construction of Cantor like Sets
www.theijes.com The IJES Page 53
Proof :-
Let I = [0,1]
First stage :
We remove middle open intervals of length from [0,1]
i.e. intervals = ( , ).
The remaining two closed intervals are denoted by = [0, ] and = [ ,1 ].We get the set =
with measure ) i.e.m( ) = )
Second stage :
Now we remove two middle open intervals of length ) from the remaining two
closed intervals and i.e. we remove = ( , ) and
= ( , ).The remaining four closed intervals are denoted by
= [0, ] , = [ , ] ,
= [ , ] , = [ ,1 ].
Length of removed intervals = m( ) + m( ) = ) < )
We get the set as union of remaining four closed intervals i.e. = with measure
i.e.m( ) =
Third stage :
Now we remove four middle open intervals , , , of length from the
remaining four closed intervals , , and i.e. we remove
, = ( , ),
= ( , ),
= ( , ),
= ( , )
The remaining four closed intervals are denoted by = [ 0, ],
=[ , ], = [ , ],
= [ , ], = [ , ],
= [ , ],
= [ , ], = [ , 1 ]

6. On the Construction of Cantor like Sets
www.theijes.com The IJES Page 54
Now,
Length of removed interval = m( )+m( )+m( )+m( )
= < .
We get the set = with measure
i.e. m( ) = .
Continuing in this way we can construct , n ≥ 4 of measure
۰۰۰۰۰۰ .
m( ) = ۰۰۰۰۰۰ = )
Then S4 .
If S = then m(S) =
= )
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Conclusion :- Using this Cantor like set we can construct different Cantor like sets by varying or fixing .
REFERENCES:-
[1] G.de. Barra, “Measure Theory And Integration”, New Age International (p) Limited, 2003.
[2] Kenneth Falconer, “Fractal Geometry: Mathematical Foundations and Applications”, John Willey and Sons.1990.
[3] Kannan V. “Cantor set: From Classical To Modern”. The Mathematical Student, Vol 63,No.1-4(1994)P.243-257.
[4] E. C. Titchmarsh ,“The Theory Of Functions”, Oxford University Press, Oxford Second Edition -1987.