## Cantor Diagonal Argument disproof

**Moderators:** mvs_staff, forum_admin, Marilyn

### 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
- Intellectual
**Posts:**2609**Joined:**Tue Mar 10, 2009 11:01 am

### 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
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### 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
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### 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
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### 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
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### 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
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### 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
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### 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
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### 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.

***** 2017-04-23

Anyone with an intellect would interpret my comment as a challenge to explain the relevancy of the "formula" to the topic.

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

But Gofer interprets it as an opportunity for childish parroting. My use of the expression intellectual-zombie is measured by what I see happen.

Last edited by robert 46 on Sun Apr 23, 2017 7:16 am, edited 1 time in total.

- robert 46
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

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

- Gofer
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### Re: Cantor Diagonal Argument disproof

Just for future reference, so that Robert can't claim it wasn't said, here's a complete proof on the uncountability of the set of infinite strings of 1's and 0's:

∃g∈{ℕ→F≝{ℕ→{0,1}}}(∀f∈F∃m∈ℕ((g m)=f))⇒∃n∈ℕ((g n)=f≝λx.((g x x)+1)mod2)⇒

(g n n)=(f n)⇒(g n n)=((g n n)+1)mod2⇒0=1mod2⇒↯,

where ↯ means contradiction or absurdity, and λx. is a lambda-term function abstraction.

So the existence of g, a bijective map, leads to absurdity; we therefore conclude not g.

∃g∈{ℕ→F≝{ℕ→{0,1}}}(∀f∈F∃m∈ℕ((g m)=f))⇒∃n∈ℕ((g n)=f≝λx.((g x x)+1)mod2)⇒

(g n n)=(f n)⇒(g n n)=((g n n)+1)mod2⇒0=1mod2⇒↯,

where ↯ means contradiction or absurdity, and λx. is a lambda-term function abstraction.

So the existence of g, a bijective map, leads to absurdity; we therefore conclude not g.

- Gofer
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### Re: Cantor Diagonal Argument disproof

Wikipedia, Mathematical proof wrote:As practiced, a [mathematical] proof is expressed in natural language and is a rigorous argument intended to convince the audience of the truth of a statement.

Cantor wants to prove that the superset of an infinite set is not countable. For this purpose he considers the set of all strings of infinite characters taken from a set of two characters- use {0,1}. The set of all strings of infinite characters is the superset, and has 2^infinity members. If this superset is countable then it can be put into a list. Justification: anything which can be put into a list is countable because each element of the list can be tagged with a counting number. If all elements are in the list then each element is tagged with a counting number, and there is a one-to-one correspondence between elements and counting numbers- which is the nature of countability.

Call the superset of infinite strings T, and S any list of elements from T. If all of T can be put in S then T is countable.

Using the Diagonal method, Cantor purports to show that an element ~D can be produced which is a member of T but not in S. He does this by changing character n in element n to its opposite, stringing these characters together, and then claiming that this ~D differs from every element in S.

However, to do this he must go to infinity as a limit to produce all the characters of ~D. This is impossible because infinity is actually the absence of a limit. ~D cannot actually be produced. Cantor apparently knows this, but believes that ~D can be inferred to exist consequent to induction. This is an invalid inference.

Cantor ignores what actually happens intermediate to the purported production of ~D. As one examines elements in S, intermediate strings of finite significant characters (sfsc) are produced which are not in the list so far examined. Some of these sfsc may subsequently be found in the continuing examination of S. However, as I have previously proved, the count of missing sfsc is monotonically increasing.

Consider that S is initially loaded with all the sfsc. This is reasonable because the sfsc are countable. They are countable because of the method I described which sequentially produces all the sfsc without duplication or omission. So S can in principle be preloaded with all sfsc.

This being the case, what do we make of the extended Diagonal method producing a monotonically increasing count of sfsc missing from S so far examined? Whereas S is necessarily complete with sfsc, it cannot actually be missing any sfsc. Thus if the extended Diagonal method was to come to an infinite limit, there would be missing sfsc from a list which was necessarily complete at the start. This is a contradiction which can only be resolved by recognizing that the extended Diagonal method cannot come to an infinite limit. This is necessitated by the characteristic that the infinite does not imply a limit. Thus all the extended Diagonal method can do is endlessly examine elements in S to endlessly produce a monotonically increasing count of missing sfsc, but never be able to show that S is ultimately missing any sfsc.

Whereas the extended Diagonal method cannot come to an infinite limit to show that S is actually missing any sfsc, Cantor's Diagonal method cannot come to an infinite limit to show that S is actually missing ~D. Thereby Cantor's Diagonal method dismally fails at proving what Cantor set out to establish. This failure is consequent to an invalid inductive inference.

- robert 46
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

No he doesn't! Just like I don't have to go to any limit to construct the function f in my proof, and just like you could construct the function (f n):=(g n n) from a function (g n m) of two naturals n and m without "going to infinity"; it happens all at the same time.However, to do this he must go to infinity as a limit to produce all the characters of ~D.

Remember that the definition of countability involves maps/functions, which is what I make use of in the proof, just like Cantor does in a more friendly way. It really is just lambda calculus.

Btw, mathematical notation/symbolism is a natural language, because each symbol directly identifies a particular phrase in a normal language; for instance, "there exists" is written ∃.

- Gofer
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### Re: Cantor Diagonal Argument disproof

Unfortunately, Gofer does not understand anything I say, and characteristically responds with diversions.

There is a monotonically increasing count of sfsc missing from S so far examined. So this count never goes to zero. If one did get to infinity as a limit there would be missing sfsc from list S which is actually complete with sfsc. Necessarily, this means that if the extended Diagonal method could go to infinity as a limit, there would be more of S which the method could not examine, but which would contain the missing sfsc. The reason for this is because the complete list of sfsc is longer than it is wide, and the Diagonal method cannot examine to a length greater than the width. Thus the Diagonal method is inadequate to even examine a countable infinite list (of sfsc) to determine whether it is complete. Whereas it is inadequate to examine the complete list of sfsc, the Diagonal method is necessarily inadequate to examine a complete list of infinite strings: i.e. the superset. The Diagonal method is incapable of determining whether a countable infinite list is complete, so it is useless for Cantor's intended purpose.

There is a monotonically increasing count of sfsc missing from S so far examined. So this count never goes to zero. If one did get to infinity as a limit there would be missing sfsc from list S which is actually complete with sfsc. Necessarily, this means that if the extended Diagonal method could go to infinity as a limit, there would be more of S which the method could not examine, but which would contain the missing sfsc. The reason for this is because the complete list of sfsc is longer than it is wide, and the Diagonal method cannot examine to a length greater than the width. Thus the Diagonal method is inadequate to even examine a countable infinite list (of sfsc) to determine whether it is complete. Whereas it is inadequate to examine the complete list of sfsc, the Diagonal method is necessarily inadequate to examine a complete list of infinite strings: i.e. the superset. The Diagonal method is incapable of determining whether a countable infinite list is complete, so it is useless for Cantor's intended purpose.

- robert 46
- Intellectual
**Posts:**2844**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

Unfortunately, Robert does not understand anything I say, and characteristically responds with diversions.

Let's analyze Robert's example with the set of sfsc, and try to enumerate it. The function g suffices:

(g n m)≝((n-1) mod2) for m<2, 0 for n<2, (gg n/2 (m-1)) for n even, (gg (n+1)/2 (m-1)) for n odd}.

so that (g n m) will give the m'th bit of the n'th sfsc.

For example, (g 6 m) will now be a function of m that corresponds to Robert's sfsc 101(0), and returns the correct 0 or 1 for each m. (g 6 1) returns 1; (g 6 2) returns 0, and so on.

What happens if we now construct the function (f m)≝((g m m)+1)mod2, Cantor's anti-diagonal?

Is f part of the enumeration provided by g? Suppose it is, at position nn. We get (f nn)=((g nn nn)+1)mod2, which obviously differs from (g nn nn), meaning that, contrary to our assumption, it wasn't part of g.

Notice how f, the anti-diagonal, was created as a function, without resorting to "going to infinity" or "creating intermediate strings" or various other nonsense argued by Robert!

Let's analyze Robert's example with the set of sfsc, and try to enumerate it. The function g suffices:

(g n m)≝((n-1) mod2) for m<2, 0 for n<2, (gg n/2 (m-1)) for n even, (gg (n+1)/2 (m-1)) for n odd}.

so that (g n m) will give the m'th bit of the n'th sfsc.

For example, (g 6 m) will now be a function of m that corresponds to Robert's sfsc 101(0), and returns the correct 0 or 1 for each m. (g 6 1) returns 1; (g 6 2) returns 0, and so on.

What happens if we now construct the function (f m)≝((g m m)+1)mod2, Cantor's anti-diagonal?

Is f part of the enumeration provided by g? Suppose it is, at position nn. We get (f nn)=((g nn nn)+1)mod2, which obviously differs from (g nn nn), meaning that, contrary to our assumption, it wasn't part of g.

Notice how f, the anti-diagonal, was created as a function, without resorting to "going to infinity" or "creating intermediate strings" or various other nonsense argued by Robert!

- Gofer
- Intellectual
**Posts:**283**Joined:**Mon May 09, 2016 8:24 am

### Who is online

Users browsing this forum: No registered users and 2 guests