A hidden assumption

[doronshadmi]

A contradictory result is a valid and useful one, because it tells you that you have something wrong.

You did not get it.

X and X's negation cannot "wait" for you in order to conclude that there is a contradiction because they are immediately cannot be found as terms (objects, or what's ever) of the same framework.

In other words, no terms (objects, or what's ever) no conclusion.

Classical Logic does not provide the answer to this question:

How X and X's negation can be found as terms (objects, or what's ever) of the same framework?

If you still do not get it, then please ask yourself a simple question:

How you are enable to know X and X's negation?

 

[PixyMisa]

You did not get it.

X and X's negation cannot "wait" for you in order to conclude that there is a contradiction because they are immediately cannot be found as terms (objects, or what's ever) of the same framework.

Wrong.

In other words, no terms (objects, or what's ever) no conclusion.

Wrong.

Classical Logic does not provide the answer to this question:

How X and X's negation can be found as terms (objects, or what's ever) of the same framework?

Wrong.

If you still do not get it, then please ask yourself a simple question:

How you are enable to know X and X's negation?

I take X and negate it.

 

[Complexity]

doronshadmi - Just out of curiosity, can you describe for us your relevant mathematical education and reading?

I'm amazed at how spectacularly wrong you're getting things.

 

[doronshadmi]

Wrong.
I take X and negate it.

But both of them are simultaneously known to you.

What enables this?

 

[ArmillarySphere]

Well, does anyone that uses the standard definition of things like set, complement, member and so forth find a problem with my proof? I admit, I haven't done formal proofs since my college years, and I'm no matematician. doron's objection holds no water, as he's not assigning the same meaning to the words as I do.


A note: If I understood doronshadmi, the "set of all sets" is meaningless.

And yet, it's so very easy to define: S = { x: x is a set }
How is this meaningless?

 

[doronshadmi]

doronshadmi - Just out of curiosity, can you describe for us your relevant mathematical education and reading?

I'm amazed at how spectacularly wrong you're getting things.

If you are a traditional mathematician you are unable to examine my work, because
my mathematical framework can by understood only if the mathematician uses his consciousness as an inspirable pert of his mathematical work.

Since this is not the case in traditional mathematics, it is based on hidden assumptions, because avoiding consciousness during a mathematical work is impossible, whether you like it or not.

 

[jsfisher]

If you are a traditional mathematician you are unable to examine my work, because
my mathematical framework can by understood only if the mathematician uses his consciousness as an inspirable pert of his mathematical work.

Since is not the case in traditional mathematics, it is based on hidden assumptions, because avoiding consciousness during a mathematical work is impossible, whether you like it or not.

Translation: My mathematical framework is so baseless that only the feeble-minded mathematically inept could possibly be sucked in to believe even one undefined word of it.

 

[doronshadmi]

Well, does anyone that uses the standard definition of things like set, complement, member and so forth find a problem with my proof? I admit, I haven't done formal proofs since my college years, and I'm no matematician. doron's objection holds no water, as he's not assigning the same meaning to the words as I do.


A note: If I understood doronshadmi, the "set of all sets" is meaningless.

And yet, it's so very easy to define: S = { x: x is a set }
How is this meaningless?

Because S cannot be {S} so the term "all" is meaningless, if related to set.

 

[PixyMisa]

But both of them are simultaneously known to you.

Sure.

What enables this?

Propositional logic.

 

[doronshadmi]

Translation: My mathematical framework is so baseless that only the feeble-minded mathematically inept could possibly be sucked in to believe even one undefined word of it.

A good description of mathematics that ignores the mathematician's cognition.

 

[PixyMisa]

Because S cannot be {S} so the term "all" is meaningless, if related to set.

Wrong.

 

[doronshadmi]

Propositional logic.

Popositional logic is based of your basic ability to simultaneously get X and X's negation.

Without this basic ability Popositional logic does not hold.

So what enables you to simultaneously get X and X's negation?

Please do not give us an example of some mechanical device (abstract or not) because it is no more than some agent of your cognition.

 

[doronshadmi]

Wrong.

Wrong wrong.

And both of us are using our cognitions all along this thread (something that is beyond your awarness).


Please show how S is also {S}.

 

[ArmillarySphere]

It's very simple:
1) x ∈ S if x is a set
2) S is a set
Thus, S ∈ S

Note that you're talking about a different set, i.e. S = { S } , the set which contains only itself. It seems to be undefinable though - I have difficulty seeing how it would satisfy the uniqueness axiom.

 

[doronshadmi]

It's very simple:
1) x ∈ S if x is a set
2) S is a set
Thus, S ∈ S

Note that you're talking about a different set, i.e. S = { S } , the set which contains only itself. It seems to be undefinable though - I have difficulty seeing how it would satisfy the uniqueness axiom.

S ∈ S is eqiuvalent to {...{S}...} where incompleteness is its built-in property (no one of its members is its final member).

In other words, a collection of distinct objects, is incomplete by definition.

 

[ArmillarySphere]

Not really.
{2} is also a set, so {2} , { x : x is a natural number} and S itself are all members of S.

You really need to work on your notation. The curly brackets do not denote membership.

 

[doronshadmi]

Not really.
{2} is also a set, so {2} , { x : x is a natural number} and S itself are all members of S.

You really need to work on your notation. The curly brackets do not denote membership.

The term all is meaningless because no member is the final member of S.

In other words, S is incomplete and so is N and so is R and so is power(R) etc. etc. ... ad infinitum.

The reason is very simple, no collection has the completeness of the atomic state itself, because it is the result of opposite aspects of the atomic state (total unity and total isolation).

This is clearly and simply understood only if you are aware of both invariant and variant "sides" of your cognition.

 

[ArmillarySphere]

Please define:

* completeness
* final member
* atomic state

collection is undefined too, but we can use it for now as it seems to be a class that includes sets.

 

[doronshadmi]

Please define:

* completeness
* final member
* atomic state

collection is undefined too, but we can use it for now as it seems to be a class that includes sets.

Collection is a general concept for both a set or a multiset (or any combination of them).

Completeness = atomic state = that has no parts.

The atomic state is the common basis of any given cardinality.

As a result no collection has a final member if compared to the atomic state.

The 1↔1 mapping between collections cannot provide this notion, because we are using collections in order to research collections, which is circular.

Things are changed by a paradigm-shift if ↔ is understood as the completeness that can be found between and beyond any given part.

By doing that we get the whole\parts Hilbert's organic paradigm of the mathematical science as written here http://www.internationalskeptics.com/forums/showpost.php?p=3589768&postcount=341 .

 

[jsfisher]

Completeness = atomic state = that has no parts.

The atomic state is the common basis of any given cardinality.

As a result no collection has a final member if compared to the atomic state.

Ignoring for a moment how completely wrong that is, there is no logical connection among the three statements.