Theses
1. th=
e
notion of uncountability is self contradictory.=
2. al=
eph_0
is not "a whole number larger than any natural number".
3. ac=
tual
infinity does not exist.
1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The binary tree
Consider a binary tree which has (no finite paths but only) infinite
paths representing the real numbers between 0 and 1 as binary strings. The
edges (like a, b, and c below) con=
nect
the nodes, i.e., the binary digits 0 or 1.
=
0.
=
/a
\
0 1
&nb=
sp; /b \c
/ \
&nb=
sp; 0 1 0 1
..........................
The set of edges is countable, because we can enumerate them. Now we=
set
up a relation between paths and edges. Relate edge a to all paths which begin =
with
0.0. Relate edge b to all paths=
which
begin with 0.00 and relate edge c to
all paths which begin with 0.01. Half of edge a is inherited by all paths=
which
begin with 0.00, the other half of edge a
is inherited by all paths which begin with 0.01. Continuing in this manner =
in
infinity, we see by the infinite recursion
=
f(n+1) =3D=
1 + f(n)/2
with f(1) =3D 1 that for n --> oo
=
1
+ 1/2 + 1/ 4 + ... =3D 2
edges are related to every single infinite path w=
hich
are not related to any other path. (By the way, the recursion would yield t=
he limit
value 2 for any starting value f(1).) The load o=
f 2
edges is only related to infinite paths because any finite segment of a path
with n edges will carry a load =
of
=
(1
- 1/2^n)/(<=
/span>1
- 1/2) < 2
edges. The set of paths is uncountable, but as we
have seen, it contains less elements than the se=
t of
edges. Cantor's diagonal argument does not apply in this case, because the =
tree
contains all binary representations of real numbers within [0, 1], some of =
them
even twice, like 1.000... and 0.111... . Therefo=
re we
have a contradiction:
|IR|
> &=
nbsp;
|IN|
|| =
||
|{paths}| &p=
ound;
|{edges}|
The Intercession
It is argued that the bijection is the
"natural" method to compare the number of elements of sets. Howev=
er,
it is not the only natural method. As we obtain from fences and alleys, the
interstices are equinumerous to the fencing pos=
ts or
trees, respectively (at least up to plus or minus 1 - but finite sets are n=
ot
of interest here).
Definition: Two infinite sets A
and B intercede (each other) if=
they
can be ordered such that there is an element a of A between any two elements b
and b' of B and there is an element b
of B between any two elements a and a' of A. If we furt=
her
use the "natural" definition that two infinite sets which intercede each other are equinume=
rous,
then the set IQ of all rational numbers and the set IR of all real numbers =
are equinumerous.
It can be shown that equinumerousity is =
an
equivalence relation on infinite sets of finite numbers. Cantor's definitio=
n of
a bijection is included. Every two infinite sets
which can be put in bijection can be put in
intercession. Alas, the intercession is less interesting, because there is =
only
one equivalence class for infinite sets of finite numbers like naturals,
Dedeki=
nd cuts
The set of all rational numbers q is countable. The set of all linear subsets [-oo,
q) of rational numbers is then
countable too by the bijection q <--> [-oo, q) . There are not uncountably
many linear subsets which would be required to define =
uncountably
many real numbers. It is impossible to have uncountabl=
y
many linear sets to distinguish uncountably man=
y real
numbers.
Why Cantor's famous diago=
nal
argument fails
The real numbers r are defined as (equivalence classes) of sequences=
or
series like those used in decimal representation SUM {k =3D 1 to oo) a_k * 10^(-=
k) with a a digit and k a natural number. Convergence of the infinite series is
established by the fact that 10^(-k) --> 0 for k -=
-> oo. So we have for all real numbers the condition
=
LIM
{k --> =
oo)
a_k=
* 10^(-k) =3D =
LIM {k --> oo) b_k * 10^=
(-k)
for any two digits a_k and b_k of {0, 1, 2, 3, ,..., 9}. Otherwis=
e the
infinite sequences were not defined. As an example consider the sequence of
digits of pi, but without decimal point, i.e., in the form SUM {k =3D 1 to oo=
) a_k * 10^(+=
k). Such a series is undefined.
Cantor's diagonal argument does not take into account that, but uses=
the
distinction of a_k
and b_k for
any digit of the real number. Considering 1.000... =3D 0.999... this failure becomes obvious (and corresponding replac=
ement
of digits is "forbidden") but it is clear that all other cases
=
LIM
{k --> =
oo)
a_k=
* 10^(-k) =3D =
LIM {k --> oo)
(1 + a_k) *
10^(-k)
ha=
ve
not been and cannot be excluded.
Why Cantor's famous diago=
nal
argument fails again
Cantor's argument is assumed to show by contradiction that no list of
real numbers can be complete, i.e., that there is no surjection IN --> I=
R.
If valid, the proof by contradiction must hold for every list, i.e., in eve=
ry
list the diagonal number must differ from any list entry. Now consider the =
following
list:
=
0 &=
nbsp; 0.0
=
1 &=
nbsp; 0.1
=
2 &=
nbsp; 0.11
=
3 &=
nbsp; 0.111
=
... =
.........
If we replace the diagonal digit a_kk =3D 0 by 1, then the diagonal number constructed is
0.111...
Either, this number is in the list. Then Cantor's general proof by
contradiction fails.
Or, the number is not in the list, then not all of its digits can be
indexed by natural numbers. Why?
All those series of ones the digits of which can be indexed by natur=
al numbers, are in the list - by definition.
The list has omega lines. A diagonal with omega digits, however, can=
not
exist without the existence of at least one line with at least omega digits=
. In
the list above every line has a finite number of=
digits.
The non-existence of a line with omega digits implies the non-existence of a
diagonal with omega digits. So we have the result: There is no actual infin=
ity
omega > n.
An easy proof of aleph_0 =
=3D
2^aleph_0
Oresme's proof of the divergenc=
e of
the harmonic series uses aleph_0 terms with 2^aleph_0 fractions
=
1
+ (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...
If there were not 2^aleph natural numbers to form as many fractions =
this
proof would fail.
2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Die Anzahl einer unendlichen Menge [ist] eine durch das
Gesetz der Zählung mitbestimmte unendliche ganze Zahl. (G. Cantor,
Collected Works p. 174)
... kann also omega sowohl als eine gerade, wie als eine
ungerade Zahl aufgefaßt werden. (G. Cantor, Collected Works p. 178)
Es ist sogar erlaubt, sich die
neugeschaffene Zahl omega als Grenze zu denken, welcher die Zahlen nu
zustreben, wenn darunter nichts anderes verstanden wird, als daß omega die e=
rste ganze
Zahl sein soll, welche auf alle Zahlen&nbs=
p;
folgt, d. h. größer zu nennen ist als jede der Zahlen n.<=
/span> (G. Cantor, Collected Works p. 195)
Constructible numbers
In this paragraph we understand=
by a
constructible number a number which can be constructed by a formula or a
catalogue (list), such that every digit a_k is
determined. The set of constructible numbers is countable, because all
definitions we can express as finite words of a finite alphabet are countab=
le.
And we have only a finite alphabet and can only understand (write, construc=
t,
read, store) finite expressions.
Every diagonal number is constructed. Therefore Cantor's proof shows
only that not all numbers of a countable set can be written in a list. It d=
oes
not show that there are uncountable sets in the sense of having more elemen=
ts
than any countable set.
Finite sequences
All mathematical entities are denoted by finite sequences of letters
over a finite alphabet. So the set of all entities is countable. All elemen=
ts can
be enumerated (=3D written in a list). Constructing the diagonal sequence y=
ields
a finite sequence because the sequence cannot have more positions than the
entries of the list. This, however, would be necessary, if aleph_0 was a nu=
mber
larger than every finite number.
aleph_0 > n leads to 2n £ n
Let A0 :=3D aleph_0 and=
LIM :=3D
limit n --> oo
=
&nb=
sp; =
2n
&p=
ound; A0
for every natu=
ral
number. Therefore
=
&nb=
sp; =
LIM
2n £<=
/span> A0
We have
=
&nb=
sp; =
|{2,4,6,...,2n=
}| =3D
n
such that
=
&nb=
sp; =
2n/|{2,4,6,...,2n}| =3D 2
The notion of the set of even natural numbers woul=
d be
meaningless at all, if there were numbers which could not be reached induct=
ively.
So, if the set of all even natural numbers does exist, then it is
=
&nb=
sp; =
LIM
{2,4,6,...,2n}
the set of all=
even
numbers and thus
=
&nb=
sp; =
LIM
{2,4,6,...,2n}
=3D A0
This leads to (here no division of transfinite
cardinal numbers is introduced but only the inequality a £<=
/span> b is
expressed in an equivalent manner by a/b £<=
/span> 1)
=
&nb=
sp; =
LIM
2n/|{2,4,6,...,2n}| £ 1
or
=
&nb=
sp; =
LIM
2n/n £ 1
omega<=
/span> does not count the natural numbers
There are n numbers in the set {1,2,3,...,n}
=
&nb=
sp; =
&nb=
sp; =
&nb=
sp; 2 &=
nbsp; 1,2
&nb=
sp; =
&nb=
sp; 3 &=
nbsp; 1,2,3
&nb=
sp; =
&nb=
sp; ... &=
nbsp; ...
&nb=
sp; =
&nb=
sp; n &nb=
sp; =
1,2,3,...,n
&nb=
sp; =
&nb=
sp; ... &=
nbsp; ...
&nb=
sp; =
&nb=
sp; omega =
; 1,2,3,...
&nb=
sp; =
&nb=
sp; omega + 1 &=
nbsp; 1,2,3,...,omega
&nb=
sp; =
&nb=
sp; ... &=
nbsp; ...
&nb=
sp; =
&nb=
sp;
=
&nb=
sp; =
... &=
nbsp; ...
The ordinal omega (which is not a natural number) cannot count the
natural numbers, because they count themselves: There are n numbers up to
number n. All natural numbers (without omega) are less than omega numbers. =
omega is not their maximum (which does not exist) but =
their supremum which is not reached.
Small wonder that set theorists insist on 0 being =
the
first natural number.
3 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Proven by the finiteness of the universe.
______
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