Let's provide more details to my previous post.
The investigated object is [0,1].
Let's investigate some finite series:
0+
1 is some cardinality |2| series that provides
1 by one-step.
0.910+
0.0910+
0.00910+
0.00110 is some cardinality |4| series that provides
1 by one-step.
etc. ...
-----------------------------
Now let's investigate [0,1] by using countably infinite cardinality |
N|.
0.910+
0.0910+
0.00910+... (which is a |
N| cardinality series) provides
1 by one-step.
It is obvious that no series of countably finite cardinality |
n| (where
n is some natural number > 1) among
0.910+
0.0910+
0.00910+... (which is a |
N| cardinality series) provides
1.
Actually, we are using here (at the level of countably infinite cardinality |
N|) the fact that |
N|+1=|
N| in order to conclude that
0.910+
0.0910+
0.00910+...=
1 (this "+1" provides the number of the interval [1,1] (which is number
1) that enables to claim that
0.910+
0.0910+
0.00910+...=
1, exactly as done by jsfisher here:
jsfisher said:
What about Y = X union {[1,1]}. Isn't Y an infinite set? Do the members of Y not fully cover the interval, [0,1]?
(
http://www.internationalskeptics.com/forums/showpost.php?p=10296055&postcount=4183)
Actually this jsfisher's remark helped me to understand the |
N|+1=|
N| elegant transfinite trick that enables to claim that, for example,
0.910+
0.0910+
0.00910+...=
1).
-----------------------------
Now let's investigate [0,1] by using uncountably infinite cardinality |P(
N)| series.
Again, let's use, for example,
0.910+
0.0910+
0.00910... (which is a |
N| cardinality series).
It is obvious that no |
N| cardinality series among uncountably infinite cardinality |P(
N)| series provides
1, simply because of the fact that |
N|+1=|
N|<|P(
N)| (and in this framework cardinality is a factor of a given solution).
-----------------------------
So, the general conclusion in this framework is as follows:
A given series provides a given solution (by one-step), where the used cardinal numbers are factors of a given solution.
In this particular example, if a greater cardinality is also used,
1 is inaccessible to the series that is based on smaller cardinality.
-----------------------------
Let's transcend the real-line as follows:
BY using the tower of powers line |P(P(
N))| < |P(P(P(
N)))| < |P(P(P(P(
N))))| < ... there are obviously the following numbers:
Between any two irrational numbers along the tower of powers line, there are |P(P(
N))| numbers.
Between any two |P(P(
N))| numbers along the tower of powers line, there are P(P(P(
N)))| numbers.
Between any two P(P(P(
N)))| numbers along the tower of powers line, there are |P(P(P(P(
N))))| numbers.
...
etc. ... ad infinitum, where the inaccessible limit of the tower of powers line is simply the non-composed 1-dimesional space (which is not necessarily a metric space).
-----------------------------
My theorem is false only if |
n>1| < |
N| < |P(
N)| < |P(P(
N))| < |P(P(P(
N)))| < |P(P(P(P(
N))))| < ... is false.