Please look at the two following 0;1 infinite trees:

* 0

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

/ \ / \

0 1 0 1

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

0 1 0 1 0 1 0 1

/ \ / \ / \ / \ / \ / \ / \ / \

/ \ / \ / \ / \ / \ / \ / \ / \

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

/ \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

. . . . . .

The left tree is the complete set of 0;1 paths, such that given any 0;1 path, its complement is included in the set.

The right tree is exactly half of the left tree, such that given any 0;1 path of the right tree, its complement is not included in that set.

The following arbitrary set that starts by

**0**1101...

1**1**100...

10**1**01...

110**0**1...

0011**1**...

...

is an arbitrary set of 0;1 paths, which quarter of it is taken from the left half of the complete left tree, and the other quarter of it is taken from the right half of the complete left tree, so the two arbitrary quarters define a mixed uncountable set of 0;1 paths that **start by** 0 or 1 bits (as seen in the example above).

This arbitrary uncountable set of 0;1 paths has complements (for example: path **10010**... ) that are not included within it, **exactly as the right uncountable tree has complements that are not included within it**.

There is no bijection from **N** to this arbitrary uncountable set of mixed 0;1 paths if the complements are not ignored.

There is a bijection from **N** to this arbitrary uncountable set of mixed 0;1 paths if the complements are ignored, as follows:

1 --> **0**1101...

2 --> 1**1**100...

3 --> 10**1**01...

4 --> 110**0**1...

5 --> 0011**1**...

...

In other words, it is shown that **N** can have uncountable number of members, which means that there is no strict distinction between countably infinite cardinality and uncountable cardinality (as defined by modern mathematics).

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More details:

I am talking here about a bijection from **N** to the arbitrary uncountable mixed set of distinct paths, **if the complements of this mixed set are ignored**.

The right tree (which is exactly half of the left tree) is an uncountable set such that given any 0;1 path of that set (which starts with bit 0), its complement (that starts with bit 1) is not included in that set.

The arbitrary set is constructed by exactly two quarters of the left tree, where one quarter is taken from the left side of the left tree (therefore every path of it starts with bit 0), and the other quarter is taken from the right side of the left tree (therefore every path of it starts with bit 1).

Also in this arbitrary mixed set (that includes paths that start with bit 0 AND paths that start with bit 1), given any 0;1 path of that set, its complement (that starts with bit 0 OR bit 1) is not included in that set, yet this set (which is constructed by exactly two quarters of the left tree) is an uncountable set, exactly as the right half tree is an uncountable set.

So, I prove that there is bijection from **N** to the arbitrary uncountable mixed set of distinct paths, if the complements of this mixed set are ignored.

Conclusion: There is no strict distinction between countably infinite cardinality and uncountable cardinality, exactly because the cardinality of **N** is not fixed (it can be countably infinite OR uncountable, which is a tautology).

Let's look at it by using further important details:

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My argument is very simple.

It uses unbounded logical trees as the logical basis of the place value number system.

I logically construct (by using unbounded logical trees) two types of numbers, which are:

**a.** An unbounded path with a radix point along it.

**b.** An unbounded path without a radix point along it.

Here is an example radix point usage along the unbounded binary tree:

*

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0 1 Integers

.---Radix point--.

|\ |\

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| \ | \ Fractions

| \ | \

0 1 0 1

|\ |\ |\ |\

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| \ | \ | \ | \

0 1 0 1 0 1 0 1

|\ |\ |\ |\ |\ |\ |\ |\

| \ | \ | \ | \ | \ | \ | \ | \

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

...

*

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0 1

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| \ | \ Integers

| \ | \

| \ | \

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0 1 0 1

. . . .--------- Radix point

|\ |\ |\ |\

| \ | \ | \ | \

| \ | \ | \ | \

| \ | \ | \ | \

| \ | \ | \ | \ Fractions

0 1 0 1 0 1 0 1

|\ |\ |\ |\ |\ |\ |\ |\

| \ | \ | \ | \ | \ | \ | \ | \

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

...

etc.

By (**a.**) one logically defines the natural numbers "above" the radix point, and the rational and irrational numbers "below" the radix point.

By (**b.**) one logically defines unbounded numbers, where each one of them is bigger than any natural number, as logically defined by (**a.**).

By using type (**b.**) numbers as the cardinality of natural numbers (which are type (**a.**) numbers), one logically realizes that there is no one and only one, so called, transfinite cardinality (notated as aleph0), but there are logically infinitely many type (**b.**) numbers (some examples of these numbers are: 1000... > 01000... > 001000... > ...) where each one of them is an optional cardinal number of an infinite set of natural numbers.

In this case also 2^aleph0 has no accurate logical basis (since the exponent is aleph0), so the whole notion of the Cantorian transfinite number system has no sufficient logical basis.

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If one rejects type (**b.**) numbers and accept only type (**a.**) numbers, then there are only finitely many natural numbers simply because no natural number has infinitely many bits (there is a logical linkage between being bounded by the amount of bits (as a common property among natural numbers) and the logical fact that there are only finitely many natural numbers.

In this case no mathematical induction (as done in case of ZF Axiom Of Infinity) logically provides a set of infinitely many natural numbers (it simply "pushes" the radix point "downward" along the unbounded logical tree, but since no natural number has infinitely many bits, there is logically only finite amount of natural numbers.

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So no matter how you look at it, Standard Set Theory has no rigorous logical basis.

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Moreover, similarly to what is shown here, the fact that |S| < |P(S)| by Cantor's theorem does not prevent the fact that S and P(S) **are already uncountable**, exactly as **N** and the arbitrary mixed set of distinct paths **are already uncountable**. So in both cases we can ignore the fact that we can define an element that is not paired with some **N** (or some S) element.

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The arrangement of an infinite set, whether it is arranged as a tree or as a list, has no influence on its cardinality, which can be (by using the standard terminology) countably infinite or uncountable.

First, I prove that the fact that a list has complements that are not included in it, does not prevent the list to be uncountable.

Second, I prove that an infinite set of natural numbers has more than one infinite cardinality, so the Cantorian notion of one and only one infinite cardinality (called aleph0) has no logical basis, exactly because I use an infinite binary tree as the logical basis of the issue at hand.

Since aleph0 has no logical basis, so is the case about 2^aleph0.

In other words, if one defines numbers by also not using radix point along the paths of the infinite binary tree, then those numbers (except number 000...) are infinite spectrum of infinite cardinalites, where some example of such infinite cardinalites is 1000... > 01000... > 001000... > ...

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The infinite binary tree is used here without loss of generality, which means that any n>2 valued infinite logical tree can be used in my argument, but instead of complements that are not included in a given infinite set, there are simply distinct paths that are not included in a given set.

In other words, my argument logically holds by using any n>1 valued infinite logical tree, and it holds exactly because I define numbers by directly use logical trees, where logical trees are, by definition, can't be illogical.