- #1

- 1,208

- 0

An edited version:

General:

We say that

Theorem:

Proof:

Step 1 (injection between set

Let us define injection between each member of

We define f(x) where f(x)={x}, for example: 1 <--> {1} , 2 <--> {2} , 3 <--> {3},… and so on.

At this stage we can say that

Step 2:

Suppose there exists a bijection between

We will show that this assumption leads us to contradiction.

There exists member

1 <--> {2,3} , 2 <--> {2,3,4} , 3 <-->{6,7} , 4 <--> {4,5,6} , 5 <--> {8,9}, …

In this example

-----------------------------------------------------------------------------

A proof that a non empty member

2^0=P({})={ {} }=1

2^1=P({0})={ {},{0} }=2

2^2=P({0,1})={ {},{0},{1},{0,1} }=4

2^3=P({0,1,2})={ {},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2} }=8

...

Therefore {} not in N but {} in P(N)

N members are the natural numbers.

P(N) members are the empty set + subsets of N + N itself.

It is easy to show an injection between any n of N to any singleton member {n} of P(N).

When we define this injection than P(|N|)>=|N|.

If P(|N|)=|N| then D is the member in P(N) includes the members of

N, that have no image in the member of P(N), which they are mapped with.

n in D because {} not in N, therefore D is not empty.

Q.E.D

-----------------------------------------------------------------------------

Now, let us say that there exist some member in

In this case we can ask: is

Options:

1)

2)

and so on, and so on.

As we can see, both options lead us to logical contradiction.

Therefore, there cannot be a bijection between

Q.E.D

=================================================

=================================================

Let us look at this proof from another point of view.

As we all know, any set includes only unique members or no members at all.

Now, let us examine options 1 and 2 again.

Option 1:

All the members which included in

Any member of

Therefore

Therefore

Option 2:

If we want to keep

In this case we also find that

Therefore we cannot ask nor answer to anything that is connected to member

The general idea behind this point of view is the power of existence of member

Member

Member

Organic

General:

We say that

**B**>**A**if**A**not=**B**but there is an injection between**A**and subset of**B**.Theorem:

**>***P(X)**X*Proof:

Step 1 (injection between set

**A**and subset of set**B**):Let us define injection between each member of

**to each member of***X***, which includes a singleton as its content.***P(X)*We define f(x) where f(x)={x}, for example: 1 <--> {1} , 2 <--> {2} , 3 <--> {3},… and so on.

At this stage we can say that

**>=***P(X)***.***X*Step 2:

Suppose there exists a bijection between

**and***P(X)***.***X*We will show that this assumption leads us to contradiction.

There exists member

**in***S***which includes ALL members of***P(X)***that have no image Included in the***X***members which they are mapped with, for example:***P(X)*1 <--> {2,3} , 2 <--> {2,3,4} , 3 <-->{6,7} , 4 <--> {4,5,6} , 5 <--> {8,9}, …

In this example

**={1,3,5,…}.***S*-----------------------------------------------------------------------------

A proof that a non empty member

**exists:***S*2^0=P({})={ {} }=1

2^1=P({0})={ {},{0} }=2

2^2=P({0,1})={ {},{0},{1},{0,1} }=4

2^3=P({0,1,2})={ {},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2} }=8

...

Therefore {} not in N but {} in P(N)

N members are the natural numbers.

P(N) members are the empty set + subsets of N + N itself.

It is easy to show an injection between any n of N to any singleton member {n} of P(N).

When we define this injection than P(|N|)>=|N|.

If P(|N|)=|N| then D is the member in P(N) includes the members of

N, that have no image in the member of P(N), which they are mapped with.

n in D because {} not in N, therefore D is not empty.

Q.E.D

**=n***t***=N***X***=D***S*-----------------------------------------------------------------------------

Now, let us say that there exist some member in

**(let us call it***X***) which is mapped with***t***(***S***<-->***t***).***S*In this case we can ask: is

**in***t***or***S***not in***t***?***S*Options:

1)

**in***t***, but by***S***definition***S***cannot be in***t***.***S*2)

**not in***t***, in this case by***S***definition,***S***must be in***t***, but by (1)***S***can’t be in***t***.***S*and so on, and so on.

As we can see, both options lead us to logical contradiction.

Therefore, there cannot be a bijection between

**and***P(X)***and we can conclude that***X***>***P(X)***.***S*Q.E.D

=================================================

=================================================

Let us look at this proof from another point of view.

As we all know, any set includes only unique members or no members at all.

Now, let us examine options 1 and 2 again.

Option 1:

All the members which included in

**, are different from each other.***S*Any member of

**can be mapped with some member of***X***, once and only once.***P(X)*Therefore

**is different from each member in***t***, therefore***S**t***MUST BE INCLUDED**in**.***S*Therefore

**cannot be defined***S***BEFORE**we check our assumption in step 2 of Cantor's proof.Option 2:

If we want to keep

**as an existing member, we***S***MUST NOT INCLUDE****in***t***.***S*In this case we also find that

**cannot be defined***S***BEFORE**we check our assumption in step 2 of Cantor's proof.Therefore we cannot ask nor answer to anything that is connected to member

**.***S*The general idea behind this point of view is the power of existence of member

**and member***t***.***S*Member

**is simpler than member***t***.***S*Member

**existence depends on objects like***S***, therefore we have to check***t***by***S***as we did here, and not***t***by***t***as Cantor did.***S*Organic

Last edited: