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Continuum hypothesis refutation
Dmytro Tarasiuk
2022
July
Abstract
Shown set that refutes Georg Cantor’s Continuum hypothesis.
Powerset
The set of all subsets of a set, or powerset, is the set of all possible combinations
of the elements of the set without repetitions, where the order of the elements does
not matter. For finite set Mthe powerset of the set Mis
|P(M)|= 2|M|.
Continuum hypothesis is wrong
”There is no set whose cardinality is strictly between that of the natural and the
real numbers.”
This hyposesis is wrong. Let’s find such set.
Let T(M) is subset of set Mwhich is corresponded to combination Cb|M|/2c
|M|, where
Ck
nis binomial coefficient and |M|is cardinality of M. Number of elements in combi-
nation is (n
k) = Ck
n=n!
k!·(n−k)! .
So, cardinality of T(M) is
|T (M)|=Cb|M|/2c
|M|=|M|!
b|M|/2c · (|M| − b|M|/2c)!
For example, M={a, b, c, d, e}. Then |M|= 5, Cb5/2c
5=C2
5= 10.
T(M) =
{a, b},{a, c},{a, d},{a, e},
{b, c},{b, d},{b, e},
{c, d},{c, e},
{d, e}
,|T (M)|= 10.
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Let n=|M|. For big |M|
Cbn/2c
n=n!
n
2!·n−n
2!=n!
n
2!2
By using Stirling’s approximation n!∼√2πnn
en
Cbn/2c
n≈
√2πnn
en
√πnn
2e
n
2!2=r2
πn ·2n
Hence,
|T (N)|= lim
n→∞ Cbn/2c
n= lim
n→∞ r2
πn ·2n=r2
πℵ0·2ℵ0
Let’s compare |T (N)|and |N|:
|N|
|T (N)|= lim
n→∞
n
q2
πn ·2n
=rπ
2·lim
n→∞
n3/2
2n=rπ
2·lim
n→∞
(n3/2)00
(2n)00
=3
4·ln22·rπ
2·lim
n→∞
1
2n·√n= 0
It means that ℵ0<|T (N)|.
Let’s compare |T (N)|and |P(N)|:
|T (N)|
|P(N)|= lim
n→∞
Cn/2
n
2n= lim
n→∞ q2
πn ·
2n
2n=r2
π·lim
n→∞
1
√n= 0
It means that |T (N)|<ℵc.
Consequently
ℵ0<|T (N)|<ℵc
The set T(M)refutes Continuum hyposis. O.E.D.
Notes
Powerset P(M) is a set of all subsets of M. Each such subset is corresponded to
combination of elements of set Mwhere cardinality is Ck
|M|, k = 0, ..., |M|.
Coefficients Ck
|M|specify Pascal’s triangle Fig. 1, where each row corresponds to
each possible cardinality of set M.
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|M|1P|M|
k=0 Ck
|M|j|M|
2kCb|M|/2c
|M|
1→1 1 21= 2 b1/2c= 0 C0
1= 1
2→1 2 1 22= 4 b2/2c= 1 C1
2= 2
3→1 3 3 1 23= 8 b3/2c= 1 C1
3= 3
4→1 4 6 4 1 24= 16 b4/2c= 2 C2
4= 6
5→1 5 10 10 5 1 25= 32 b5/2c= 2 C2
5= 10
6→1 6 15 20 15 6 1 26= 64 b6/2c= 3 C3
6= 20
7→1 7 21 35 35 21 7 1 27= 128 b7/2c= 3 C3
7= 35
Fig. 1
It is well-known fact that each row of Pascal’s triangle has the following property:
|M|
X
k=0
Ck
|M|= 2|M|
P(M) can be thought as one row of the Pascal’s triangle. Infinite row can represent
cardinality ℵcof set R.
The idea to find set Awhich cardinality ℵ0<|A|<ℵcand refutes Continuum
hyposesis is based on using only part of each row of Pascal’s triangle. The underlined
coefficients in Fig. 1 are corresponded to combinations for T(M). In case A=N,
T(N) is the searched set.
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