## Cantor Diagonal Argument disproof

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### Re: Cantor Diagonal Argument disproof

Gofer wrote:Right, which obviously is the main theorem, and is NOT what you stated ~(A and B).JeffJo wrote:Cantor states he is seeking "a proof of the proposition that there is an infinite manifold, which cannot be put into a one-one correlation with the totality [Gesamtheit] of all finite whole numbers."

A="S is countable" = "S can be put into that one-one correlation."

B="S=T."

Cantor: "the proposition that there is a T where A and B can't be true at the same time."

Maybe you need to brush up on logical equivalences.

Have you noticed something? What you are arguing about is the form Cantor's used. Not what forms he could have used. The assumption of ~P, for the P you now describe (as opposed to what you described before), never occurs.But do you notice something? The main theorem has exactly the form ~P, for which proof of negation is particularly well suited, making Cantor's Lemma unnecessary.

No, I'm "hung up" of the fact that the deduction doesn't follow the form "To prove ¬ϕ, assume ϕ and derive absurdity," which you say it does. But can't show that it does, even though you keep changing how you claim it is.Jeff seems to be hung up on the fact that the deduction doesn't fit the form of ~P because of the logical connective ->; this is quite wrong however.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:The "ingredients" in this case is the set of infinite strings; and the "snake oil" is the belief that this set is countable; the "cure" is Cantor's proof.robert 46 wrote:Indeed he didn't, but this is just like the snake-oil salesman who claims his patent medicine cures what-ails-you, but glosses over what the ingredients are and how they work.

There is no explanation as to how there can be infinite sets, infinite strings, infinite processes or infinite agents. It is entirely fantasy.

Yes there is, like we have been telling and yelling at you: FUNCTIONS. The function (2*) doesn't go to any limit, yet is still valid "simultaneously" for all numbers.I agree entirely that my method for producing strings of finite significant characters cannot get to the limit of producing strings of infinite significant characters because getting to the limit is an impossibility- there is no limit. [...]

Which is impossible because there is no explanation as to how Cantor's method of negating characters can get to the infinite limit either sequentially or all-at-once.

2*n=n+n, but what is 2*infinity? There is an abrupt change of character if there is a limit: it is not infinity+infinity.

You could do that; but it still has nothing to do with what Cantor does.Well then in my fantastic world, creating strings of finite significant characters presto-changeo becomes creating strings of infinite significant characters at the limit.

Cantor claims that there in an anti-diagonal ~D which has infinite characters. If there is ~D then it is complete, yet that completion cannot be demonstrated to be accomplished.

So then you believe that the function (2*) only has a finite domain?Don't need to: the world of infinite sets is an obvious delusional fantasy which makes no more sense than Carroll's Alice in Wonderland- hedgehogs for croquet balls and flamingos for mallets. Just try to play a game of croquet under those conditions.

2*n does not extend to 2*infinity

And dare we never speak of the "numbers as a whole" for which it is defined?

For any finite n, 2* is defined. 2* does not act on the "numbers as a whole".

What do you mean with "show that the method actually works"? Cantor's shows that the anti-diagonal of any enumeration can't be part of the enumeration;All Cantor does is claim to have a method, but conveniently fails to show that the method actually works: it cannot merely be defined to work- that is a con-job for the easily duped. Perhaps Cantor was the greatest snake-oil salesman ever to come to mathematics.

It applies for finite supersets: meaning that the finite list always appears to be missing an element per the diagonal method even if the list is actually complete. This is because the diagonal method cannot examine the entire list: the depth of examination is limited by the width, which is always less than the depth for supersets- finite or infinite.

if it were, we'd have the contradiction Cantor mentions, "that a thing would be both part and not part of T.", paraphrased.

Cantor is not questioning the completeness of T, a superset, but rather the completeness of a list S. He claims that any S will not contain all of T. This is patently false for finite supersets. And Cantor cannot actually get to a limit to say anything definite about infinite supersets.

- robert 46
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### Re: Cantor Diagonal Argument disproof

Robert, say someone comes up to you and says they have a formula/algorithm that turns any natural number N into a function from N to {0,1}, and that for every such function f there is a corresponding n in N such that the formula gives f for n.

How can you prove to this person that it can't possibly be true? And notice that you can't argue infinite sets here, because all this person has claimed is a formula.

How can you prove to this person that it can't possibly be true? And notice that you can't argue infinite sets here, because all this person has claimed is a formula.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Jeff still refuses to let himself comprehend that while ~(A & B) and (B->~A) certainly imply one another, they clearly are not the same. The former says that it is absurd to have both a proof of A and B, whereas the latter says that there's a proof that B implies that implying A leads to absurdity.

And then he continues to be under the mistaken impression that proof of negation can't be used in a logical formula involving implications. However, consider the two formulas above. In a proof of (B->~A)->~(A & B), proof of negation is used. Even Wikipedia uses it in proving contraposition. But Jeff of course ignores all that.

And he also seems to be claiming that Cantor's main theorem is ~(A & B); but if that were true, there's no need to use contraposition, as Jeff claims, since ~(A & B) follows directly from the Lemma using, you guessed it, proof of negation.

And then he continues to be under the mistaken impression that proof of negation can't be used in a logical formula involving implications. However, consider the two formulas above. In a proof of (B->~A)->~(A & B), proof of negation is used. Even Wikipedia uses it in proving contraposition. But Jeff of course ignores all that.

And he also seems to be claiming that Cantor's main theorem is ~(A & B); but if that were true, there's no need to use contraposition, as Jeff claims, since ~(A & B) follows directly from the Lemma using, you guessed it, proof of negation.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:Robert, say someone comes up to you and says they have

Why should I be interested?

a formula/algorithm

Which is it?

that turns any natural number N

N is a natural number.

into a function from N to {0,1}, and that for every such function f

There are many functions, generically named "f"?

there is a corresponding n in N

Now N is the set of natural numbers, and n is a natural number.

such that the formula gives f for n.

What formula (not an algorithm), which is undefined? Which gives a function, now specific, not generic? Based on the parameter "n" to the formula, which is a natural number?

The gibberish Gofer has written is nothing more than a diversion to evade the important issues raised about the inadequacy of Cantor's diagonal method to accomplish anything.

How can you prove to this person that it can't possibly be true?

All that has been, and need be, proved is that what you have said is pure gibberish and general garbage. This should be obvious to anyone who isn't basically an intellectual-zombie.

And notice that you can't argue infinite sets here, because all this person

Clearly, "this person" is you; and you are the intellectual-zombie who can't express thoughts coherently.

has claimed is a formula.

...unidentified in any way. What you have said is absurdly vague, and I believe deliberately so as a smoke screen to cover up your inability to respond to my prior post in any relevant manner.

Please leave, and not come back until you learn how to think.

Note: Fictional robots characteristically refer to themselves in the third-person. The classic example is when the robot Johnny Five says (in the Short Circuit movie), "Johnny Five is ALIVE!" An interesting quirk here is that both Gofer and JeffJo, whom I have characterized as programmed-automatons, occasionally indirectly refer to themselves in the third-person.

- robert 46
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### Re: Cantor Diagonal Argument disproof

Robert blows a gasket and cries because he can't comprehend a simple English sentence. So here it is again, in mathematical notation: ∃g∈{ℕ→F}(∀f∈F∃n∈ℕ((g n)=f)),F≝{ℕ→{0,1}}

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:So here it is again, in mathematical notation: ∃g∈{ℕ→F}(∀f∈F∃n∈ℕ((g n)=f))),F≝{ℕ→{0,1}}

Viewers can clearly see that mathematical symbolism is an arcane code used by members of a monastic religious order to communicate with each other in secret. So be it.

If anyone has anything sensible to say, I invite you to participate.

- robert 46
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### Re: Cantor Diagonal Argument disproof

Viewers can clearly see that Robert is too lazy to bother to learn something new. So be it.

Here's a translation of the above:

There exists a g in the set of functions from the natural numbers N to F such that for all functions f in F there exists an n in N such that g applied to n yields f, and where F is defined to be the set of functions from N to the set of 1 and 0.

Clearly, it's much easier to read mathematical notation.

Here's a translation of the above:

There exists a g in the set of functions from the natural numbers N to F such that for all functions f in F there exists an n in N such that g applied to n yields f, and where F is defined to be the set of functions from N to the set of 1 and 0.

Clearly, it's much easier to read mathematical notation.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:There exists a g in the set of functions from the natural numbers N to F such that for all functions f in F there exists an n in N such that g applied to n yields f, and where F is defined to be the set of functions from N to the set of 1 and 0.

Gofer has shown no relevancy of the above to anything in this topic.

- robert 46
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### Re: Cantor Diagonal Argument disproof

Robert has shown no relevancy in comprehending anything more advanced than 1+1=2, in this topic.

- Gofer
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