## Cantor Diagonal Argument disproof

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### Re: Cantor Diagonal Argument disproof

Note that what Robert call sfcs, strings of finite significant characters, I call Tf, the set of all strings having a 1 at some maximum position. For example, the string 01011(0), a string 01011 padded with infinite zeroes, has a 1 at a maximum position of 5.

Statement without proof! Off the top of my head, I can think of one that might work.1(01) is a string of infinite significant characters. It is not in the starting list for good reason: there is no method to include in the list all the strings with infinite repeating sequences.

Kinda hard when you write a lot of nonsense hard to follow.but that you need to address my posts statement by statement to show that you can follow the argument I am developing.

No, Tf isn't defined by a method, but by the definition I just gave.Tf is defined as being produced by a method which is expected to put all the strings of finite significant characters into Tf and concurrently into a list S.

The Diagonal Method doesn't produce "strings", plural, but ONE string that is not listed, namely the diagonal, hence its name. And the string it produces for a listing of Tf, DOES NOT have the property of those in Tf, which you need to prove if you are claiming that.However, if the extended Diagonal method produces ~D by examining S, there would also be strings of finite significant characters which were not found in S.

Irrelevant what your "production method" does! The fact that Tf fits Cantor's Lemma, means that its listing will always produce an anti-diagonal not in Tf; you still haven't rebutted this fact!That such strings would be missing from S (which recall is countable) then they would also be missing from Tf. This would imply that the production method did not actually create the set of all strings of finite significant characters, Tf, nor the list of same, S.

As I have been telling you many times, Cantor's Theorem is not about the "existence", whatever that means, of infinite sets, but if the set of all infinite strings does "exist", it can't be countable.Thus it is wrong to believe there exists a set Tf containing all strings of finite significant characters. And by extension wrong to believe in the existence of any countable infinite set.

It's not a "circular statement"; it's just the way things are. We're not talking "existence", but logic and deduction. Come to grips with that!Tf is defined according to its definition

A circular statement devoid of meaningful content. Essence does not imply existence. Come to grips with that.

- Gofer
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### Re: Cantor Diagonal Argument disproof

The obvious one, where I described an abstract proof of property that belongs to only odd numbers, or could not belong to even ones.Gofer wrote:What example?Constructive math was not appropriate to the example.

Only if you believe in constructionist math, and are trying to construct something. I don’t believe in constructionist math, and didn’t try to construct something, so you are wrong on two counts.Constructive math is always relevant, always.

The abstract proof was that if the something exists, if has certain properties. This is an example of how you try to assert the rules that everybody else must play by, and then claim they are wrong when they play by other perfectly valid rules.

Which was the (again, obvious) point of using “odd” vs. “non-even.” One includes a word you just implied makes it a negative statement, and one does not so it is a positive statement. Yet the statements are equivalent, proving that this one source that claims “Proof of Negation” (which you still haven’t admitted was wrong to call “Proof by Negation”) is different than “Proof by Contradiction” is either wrong, or has left out a critical detail.Which is usually done by searching the statement for the word "not".The point was that to distinguish how a proof outline applies differently to ϕ and ¬ϕ, you have to identify which is the inherently positive proposition, and which is inherently negative.

Why can’t you just come to terms with the fact that you are the one who claims others are wrong based solely on the semantics you chose to accept?

No, because you don’t want to accept the explanation.Because you haven't explained it very well.

And what you keep ignoring, is that for the lemma to prove the proposition by a form proof-by-contradiction (which includes proof-of-negation), it has to show that the absurdity (in this case, ~B) follows from B. It doesn’t. It follows from A. So, BY CONTRAPOSIITON, and only BY CONTRAPOSIITON, B->~A follows directly from the lemma A->~B.I never said the lemma follows from assuming B - see previous comment.

Which is exactly what Cantor said.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Not believing in constructive math is like saying you don't believe in algorithms, a statement devoid of logic. And obviously we ARE trying to construct something, namely a proof of B->not A.Only if you believe in constructionist math, and are trying to construct something. I don’t believe in constructionist math, and didn’t try to construct something, so you are wrong on two counts.

This is error in logic. odd and non-even are not the same thing. Proving something to be odd means there's direct proof of that fact. Proving something non-even, there's direct proof of assuming it even leads to absurdity/contradiction.Which was the (again, obvious) point of using “odd” vs. “non-even.” One includes a word you just implied makes it a negative statement, and one does not so it is a positive statement. Yet the statements are equivalent

'proof of negation' exists in both classical and constructive math, whereas 'proof by contradiction', which implies 'the excluded middle', only exists in the former.proving that this one source that claims “Proof of Negation” (which you still haven’t admitted was wrong to call “Proof by Negation”) is different than “Proof by Contradiction” is either wrong, or has left out a critical detail.

Why can't you just come to terms with the fact that the two are different proof techniques, REGARDLESS what they're called.Why can’t you just come to terms with the fact that you are the one who claims others are wrong based solely on the semantics you chose to accept?

This is error! We are tasked with proving B->not A, which per definition equalsAnd what you keep ignoring, is that for the lemma to prove the proposition by a form proof-by-contradiction (which includes proof-of-negation), it has to show that the absurdity (in this case, ~B) follows from B.

B->(A->absurdity), which means if we assume (A->not B), B and A, it leads to a contradiction. Since (A->not B) is theory, we must conclude that B-> not A. The logical formula of this is

(A->~B)->(B->(A->false)), which of course is contraposition.

No, I just proved it using 'proof of negation'. See wikipedia.org/wiki/Contraposition, "simple proof by contradiction"; note the first part which is really 'proof of negation', whereas the second one is 'proof by contradiction'.So, BY CONTRAPOSIITON, and only BY CONTRAPOSIITON, B->~A follows directly from the lemma A->~B.

Obvious projection on your part, as Cantor never explicitly stated as much. And there is evidence to the contrary, namely Cantor's statement "otherwise we would have the contradiction that the anti-diagonal would both a part of T and not part of T", paraphrased. You still haven't explained this phrase or rebutted that it could be 'proof of negation'.Which is exactly what Cantor said.

- Gofer
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### Re: Cantor Diagonal Argument disproof

[edit: make stronger arguments]

Here, I don't accept that you have to construct the entire class, EVEN BY LESS RESTRICTIVE STANDARDS THAN YOU IMPLY, before you can prove properties of that class. Proving the proposition "All even numbers >2 are non-prime" does not require me to construct the entire set. If a number n is even, and n>2, then by definition n=2*m where m>1.

If you want to debate that, construct for me a non-empty set of odd numbers that are not non-even, or non-even numbers which are not odd. Or do you not believe you have to meet the same standards that you insist I meet?

And I keep bringing up the names, because it shows how you cannot not admit to having made any mistake at all. Even after having changed your argument twice; from "oh, yes, that was the correct name" to "maybe it was better phrased the other way," and then to "the other way is right."

B="S=T."

A="S is countable."

We are "tasked with" proving B->¬A.

We do so by proving A->¬B, which is Cantor's lemma.

By saying "Contrapostion!" at this point, B->¬A is proven. By saying "Note that B is the negation of what we proved - that is, ¬B - so the statement is proven" implies "Contraposition!" By saying "the assumption that S=T contradictions what we proved, that S<T" implies "B is the negation of what we proved."

What is so hard to see here?

(A) Note that it is a proof of Contraposition, by contradiction. There are other ways to prove it - one is given immediately prior to yours, and one immediately after - so your implication (note how Gofer never states his points? Only implies them, and then if you guess wrong he jumps on you?) that contradiction is necessarily a precursor to contrapoisition is fallacious.

(B) Note that your passage accepts the Principle of The Excluded Middle (where it says "assuming that we are dealing with concrete statements that are either true or not true"). So if you don't, this isn't a proof.

(C) Look at the introduction to that page, where it says that the only difference between Proof by Contradiciton, and Proof by Contraposition, is that Contraposition applies only to propositions fo the form A->B? What form does our proposition take?

Now, look at wikipedia.org/Proof_by_contrapositive. In the introduction, see where it says "In mathematics, proof by contraposition is a rule of inference used in proofs" ?

Did he, or did he not, explicitly say "this is a proof by contradiction?" No? Gee, maybe you should apply the same standards to yourself, that you do to me.

Did he, or did he not, say that the contradiction of "S is all of T" follows from "S is countable" (that is, identify the contrapositive)? I agree he used different terminology, but that terminology is all you are arguing about.

Have I, or have I not, said that Cantor never explicitly identified the form of proof that he used?

Now, look at wikipedia.org/Proof_by_contradiction, under "Principle." See where it says, under both forms of the proof, that the form is to assume ¬P and then imply the contradiction (whichever form it takes) from that assumption? Does the contradiction Cantor explicitly identifies follow from the assumption you say he makes, that S=T? Or is it merely the implication that B="S=T" is the negation of what was proven in the lemma, A->¬B?

Then, look at your source. See where he says "assume ϕ and derive absurdity" ? Does he need to add that this means you must derive absurdity from the assumption? Are you really that obtuse?

And as I keep explaining to you, for it to be proof by contradiction, the contradiction has to FOLLOW FROM the assumption. And it does not. So even if Cantor intended this to be Proof by Contradiction - for which there is no explicit evidence of the kind you insist I provide - it fails to be.

The rebuttal is the fact you keep ignoring, that the absurdity has to follow from the assumption. As even Andrej Bauer said.

I believe you can use it where it is appropriate, and I explained what the difference is. You ignored it.Gofer wrote:Not believing in constructive math ...

Here, I don't accept that you have to construct the entire class, EVEN BY LESS RESTRICTIVE STANDARDS THAN YOU IMPLY, before you can prove properties of that class. Proving the proposition "All even numbers >2 are non-prime" does not require me to construct the entire set. If a number n is even, and n>2, then by definition n=2*m where m>1.

Yes, they are. You don't need to assume the Principle of the Excluded Middle, if you can prove it.This is error in logic. Odd and non-even are not the same thing.

If you want to debate that, construct for me a non-empty set of odd numbers that are not non-even, or non-even numbers which are not odd. Or do you not believe you have to meet the same standards that you insist I meet?

Come back when you understand what the Principle of the Excluded Middle is, and can use the phrase properly so we can understand what you mean.'proof of negation' exists in both classical and constructive math, whereas 'proof by contradiction', which implies 'the excluded middle', only exists in the former.

Because (A) I find no evidence that anybody but you and your one source believe there is a difference, and (B) neither of you has identified how to tell a positive proposition from a negative one. So if you replace ¬Φ with θ, and Φ with ¬θ (recognizing the Principle of the Excluded Middle), you get the same proofs fitting the opposite names. To make them different, you have to reject the Principle of the Excluded Middle, and identify how that middle fits in these negations.Why can't you just come to terms with the fact that the two are different proof techniques, REGARDLESS what they're called.

And I keep bringing up the names, because it shows how you cannot not admit to having made any mistake at all. Even after having changed your argument twice; from "oh, yes, that was the correct name" to "maybe it was better phrased the other way," and then to "the other way is right."

What error?This is error! We are tasked with proving B->not A

B="S=T."

A="S is countable."

We are "tasked with" proving B->¬A.

We do so by proving A->¬B, which is Cantor's lemma.

By saying "Contrapostion!" at this point, B->¬A is proven. By saying "Note that B is the negation of what we proved - that is, ¬B - so the statement is proven" implies "Contraposition!" By saying "the assumption that S=T contradictions what we proved, that S<T" implies "B is the negation of what we proved."

What is so hard to see here?

- Cantor never explicitly identified a form of proof.
- He identified a contradiction.
- But the only thing the contradiction established was the statements that we call B and ¬B.

- Defining B and ¬B satisfies the requirements for proof by contraposition.
- Defining this contradiction does not satisfy the requirements for proof by contradiction.

See wikipedia.org/wiki/Contraposition, "simple proof by contradiction";

(A) Note that it is a proof of Contraposition, by contradiction. There are other ways to prove it - one is given immediately prior to yours, and one immediately after - so your implication (note how Gofer never states his points? Only implies them, and then if you guess wrong he jumps on you?) that contradiction is necessarily a precursor to contrapoisition is fallacious.

(B) Note that your passage accepts the Principle of The Excluded Middle (where it says "assuming that we are dealing with concrete statements that are either true or not true"). So if you don't, this isn't a proof.

(C) Look at the introduction to that page, where it says that the only difference between Proof by Contradiciton, and Proof by Contraposition, is that Contraposition applies only to propositions fo the form A->B? What form does our proposition take?

Now, look at wikipedia.org/Proof_by_contrapositive. In the introduction, see where it says "In mathematics, proof by contraposition is a rule of inference used in proofs" ?

???Obvious projection on your part, as Cantor never explicitly stated as much.

Did he, or did he not, explicitly say "this is a proof by contradiction?" No? Gee, maybe you should apply the same standards to yourself, that you do to me.

Did he, or did he not, say that the contradiction of "S is all of T" follows from "S is countable" (that is, identify the contrapositive)? I agree he used different terminology, but that terminology is all you are arguing about.

Have I, or have I not, said that Cantor never explicitly identified the form of proof that he used?

Now, look at wikipedia.org/Proof_by_contradiction, under "Principle." See where it says, under both forms of the proof, that the form is to assume ¬P and then imply the contradiction (whichever form it takes) from that assumption? Does the contradiction Cantor explicitly identifies follow from the assumption you say he makes, that S=T? Or is it merely the implication that B="S=T" is the negation of what was proven in the lemma, A->¬B?

Then, look at your source. See where he says "assume ϕ and derive absurdity" ? Does he need to add that this means you must derive absurdity from the assumption? Are you really that obtuse?

... and what evidence is there that he meant this was a proof by contradiction? C'mon, you insist I need an explicit statement. Where is yours? Simply using the word "contradiction" does not mean the same thing.And there is evidence to the contrary, namely Cantor's statement "otherwise we would have the contradiction

And as I keep explaining to you, for it to be proof by contradiction, the contradiction has to FOLLOW FROM the assumption. And it does not. So even if Cantor intended this to be Proof by Contradiction - for which there is no explicit evidence of the kind you insist I provide - it fails to be.

I've done almost nothing but explain, and re-explain, that phrase. And why it can't be a form of Proof by Contradiction.You still haven't explained this phrase or rebutted that it could be 'proof of negation'.

The rebuttal is the fact you keep ignoring, that the absurdity has to follow from the assumption. As even Andrej Bauer said.

Last edited by JeffJo on Wed Apr 12, 2017 9:00 am, edited 1 time in total.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:Note that what Robert call sfcs [sfsc], strings of finite significant characters, I call Tf, the set of all strings having a 1 at some maximum position.

Sfsc is a generic descriptor. To Gofer, Tf is a set of sfsc. To me, S is a list of sfsc.

Statement without proof! Off the top of my head, I can think of one that might work.robert 46 wrote:1(01) is a string of infinite significant characters. It is not in the starting list for good reason: there is no method to include in the list all the strings with infinite repeating sequences.

Not unless it has infinite nested loops. But I invite you to describe what you think may work.

but that you need to address my posts statement by statement to show that you can follow the argument I am developing.

Kinda hard when you write a lot of nonsense hard to follow.

So, you don't have a rudimentary understanding and the intellect to ask questions to clarify the matter??? That you can't even interpret "Note that the 1st and 3rd entries are duplicates" and "Note that the 1st, 3rd, 5th, 7th entries are duplicates" to find meaning is not auspicious.

No, Tf isn't defined by a method, but by the definition I just gave.Tf is defined as being produced by a method which is expected to put all the strings of finite significant characters into Tf and concurrently into a list S.

Your definition is not rigorous. In any case, I can define a unicorn to be a horse-like animal with a spiral horn in the center of its forehead and a tail like that of a lion; yet this essence does not establish existence. I assure you it would be futile to argue about unicorns.

The Diagonal Method doesn't produce "strings", plural, but ONE string that is not listed, namely the diagonal, hence its name.However, if the extended Diagonal method produces ~D by examining S, there would also be strings of finite significant characters which were not found in S.

I have extended the method beyond what Cantor had the intellect to understand. Otherwise he would have mentioned it, unless he recognized that it refutes his thesis, in which case he may have suppressed it. We will likely never know for sure, but I think his religious training provided blinders which prevented him from seeing the extended method.

And the string it produces for a listing of Tf, DOES NOT have the property of those in Tf, which you need to prove if you are claiming that.

Whereas ~D is never produced, that imaginary unicorn is irrelevant.

Irrelevant what your "production method" does! The fact that Tf fits Cantor's Lemma, means that its listing will always produce an anti-diagonal not in Tf; you still haven't rebutted this fact!That such strings would be missing from S (which recall is countable) then they would also be missing from Tf. This would imply that the production method did not actually create the set of all strings of finite significant characters, Tf, nor the list of same, S.

You can't follow and understand what I say, so you always revert back to your programming.

Thus it is wrong to believe there exists a set Tf containing all strings of finite significant characters. And by extension wrong to believe in the existence of any countable infinite set.

As I have been telling you many times, Cantor's Theorem is not about the "existence", whatever that means, of infinite sets, but if the set of all infinite strings does "exist", it can't be countable.

Do you agree that the method:

"1. Put 0(0) into list S.

2. For all strings of n characters, padded with 0 after position n, if character n is 1 then put it into list S.

3. n=n+1; loop to 2."

... will put all sfsc into S? What is the ultimate count of elements in S? Is it 2^infinity? If you agree that it is 2^infinity, then how does this differ from the count of all infinite strings- which is also 2^infinity???

Tf is defined according to its definition

A circular statement devoid of meaningful content. Essence does not imply existence. Come to grips with that.

It's not a "circular statement"; it's just the way things are.

And what if I define Tf as:

1. Put 0(0) into set Tf.

2. For all strings of n characters, padded with 0 after position n, if character n is 1 then put it into set Tf.

3. n=n+1; loop to 2.

... Why isn't my constructive definition better than your non-constructive definition???

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

I've ignored most of your argument with Gofer, but there is a difference between the set T in Cantor's method (the set of all possible binary strings), and what you call sfsc. The string "111..." is in T but is not a sfsc. Even an extended one, constructed by your methods.robert 46 wrote:Sfsc is a generic descriptor. To Gofer, Tf is a set of sfsc. To me, S is a list of sfsc.

All of your arguments, in one way or another, misrepresent an infinite set as a finite one in some way. That's why they are wrong.

No method that you accept, but there is a method. It's called inference. If you don't accept it, then there is no set N of all natural numbers. And so the question of whether another set T, that you also don't accept, can be put into a bijection with N is moot to you. But not to those with the intellect to understand that rejecting inference implies both that there is a largest natural number, and that there can't be.robert 46 wrote:1(01) is a string of infinite significant characters. It is not in the starting list for good reason: there is no method to include in the list all the strings with infinite repeating sequences.

You have misrepresented what Cantor did, so it is not an extension at all. And you seem to lack the intellect to even try to understand why.I have extended the method beyond what Cantor had the intellect to understand.

It is produced by induction, and that is what you refuse to try to understand.Whereas ~D is never produced, that imaginary unicorn is irrelevant.

Do you agree that the method:

"1. Put 0(0) into list S.

2. For all strings of n characters, padded with 0 after position n, if character n is 1 then put it into list S.

3. n=n+1; loop to 2."

... will put all sfsc into S? What is the ultimate count of elements in S? Is it 2^infinity? If you agree that it is 2^infinity, then how does this differ from the count of all infinite strings- which is also 2^infinity???

I agree that this method is incomprehensible. For one thing, you never initialized n. For another, it fails to recognize that an infinite set can be put into a bijection with a subset of itself (example: the natural numbers and the even natural numbers). For another, it tries to deny the existence of transfinite numbers - which is the only realm where "2^infinity" makes sense - and then uses them to disprove themselves. For another, it is wrong. Every string you add to S contains a finite prefix that contains 1's and 0's, and an infinite suffix that contains only 0's

- JeffJo
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### Re: Cantor Diagonal Argument disproof

JeffJo wrote:...there is a difference between the set T in Cantor's method (the set of all possible binary strings), and what you call sfsc. The string "111..." is in T but is not a sfsc. Even an extended one, constructed by your methods.robert 46 wrote:Sfsc is a generic descriptor. To Gofer, Tf is a set of sfsc. To me, S is a list of sfsc.

Will you admit that the number of "1" characters in sequence is unlimited in the totality of strings of finite significant characters?

No method that you accept, but there is a method. It's called inference.robert 46 wrote:1(01) is a string of infinite significant characters. It is not in the starting list for good reason: there is no method to include in the list all the strings with infinite repeating sequences.

I was referring to a constructive method. Gofer challenged that he knew of one, but has not followed up on that claim.

If you don't accept it, then there is no set N of all natural numbers. And so the question of whether another set T, that you also don't accept, can be put into a bijection with N is moot to you. But not to those with the intellect to understand that rejecting inference implies both that there is a largest natural number, and that there can't be.

Not when one takes the middle road that some sets are extendable, although never complete.

You have misrepresented what Cantor did, so it is not an extension at all. And you seem to lack the intellect to even try to understand why.I have extended the method beyond what Cantor had the intellect to understand.

I think what Cantor did was fast-talk a lot of credulous people into believing him in the manner of any snake-oil salesman who selectively leaves out important details. The claim is that the snake-oil cures what-ails-you, but the main ingredient is grain alcohol which just causes the consumer to become drunk and forget about what ails him. Why haven't you and Gofer been lulled into a stupor?

It is produced by induction...Whereas ~D is never produced, that imaginary unicorn is irrelevant.

What if induction does not apply to the infinite, and you have only been taken in by a snake-oil salesman?

Do you agree that the method:

"1. Put 0(0) into list S.

2. For all strings of n characters, padded with 0 after position n, if character n is 1 then put it into list S.

3. n=n+1; loop to 2."

... will put all sfsc into S? What is the ultimate count of elements in S? Is it 2^infinity? If you agree that it is 2^infinity, then how does this differ from the count of all infinite strings- which is also 2^infinity???

I agree that this method is incomprehensible.

That is your problem.

For one thing, you never initialized n.

Do you think you can infer what the initial value of n is from the context??? You and Gofer appear incapable of putting dots on "i"s, and crosses on "t"s. Does everything have to be spelled out for you in excruciating detail?

For another, it fails to recognize that an infinite set can be put into a bijection with a subset of itself (example: the natural numbers and the even natural numbers).

The question is: what is the count of all sfsc in Tf; and how does this compare to the count of all infinite strings in T.

For another, it tries to deny the existence of transfinite numbers - which is the only realm where "2^infinity" makes sense - and then uses them to disprove themselves.

Is the cardinality of Tf the same as the cardinality of T?

Every string you add to S contains a finite prefix that contains 1's and 0's, and an infinite suffix that contains only 0's

But all of these strings are also in T.

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

The number of "1" characters that can be in an unconstrained, infinite binary string is unlimited. The number of "1" characters in any sequence you have described is finite.robert 46 wrote:Will you admit that the number of "1" characters in sequence is unlimited in the totality of strings of finite significant characters?

Get that? Any such string has a finite number of "1"s; but what that finite number is has no limit.

This is not a contradiction, which I know will be your reply. It is the same as saying that every natural number is finite, but the set itself is infinite. This is the distinction you still refuse to consider.

And I was referring to construction by inference. Like most mathematicians, I reject the idea that you have to be able to write done the end result for it to "exist" in mathematics. Defining an algorithm that describes it is sufficient construction.I was referring to a constructive method.

"Doctor, Doctor, it hurts when I do this!" "Don't do that."Not when one takes the middle road that some sets are extendable, although never complete.If you don't accept it, then there is no set N of all natural numbers. And so the question of whether another set T, that you also don't accept, can be put into a bijection with N is moot to you. But not to those with the intellect to understand that rejecting inference implies both that there is a largest natural number, and that there can't be.

You can't selectively choose which sets are extendable by inference, and which aren't. That's what I meant when I said you treat the infinite as finite. "Taking the middle road," as you call it, it choosing the parts of whichever theory helps you demonstrates your lies. And yes, I do mean to call them the lies that they are.

If the natural numbers N can be called a set, than so can T.

The only fast-talking being done here, is by you. You even admit to "taking the middle road," which is fast-talking your way out of the contradictions your approach causes.I think what Cantor did was fast-talk...

Since the point of induction is to"apply to the infinite," this is another (blatant) attempt at fast-talk.What if induction does not apply to the infinite, ...

No, it's yours, since the flaws I pointed out are real.That is your problem.

Yep. Do you think you can write a comprehensible algorithm, without requiring such inference? If you want to demonstrate a point, that onus is on you.Do you think you can infer what the initial value of n is from the context???

It is a countably infinite set, since it is equivalent to a set of finite strings.The question is: what is the count of all sfsc in Tf; ...

Can you read? It does not compare to T, which includes infinite strings that cannot be found by your algorithm.... and how does this compare to the count of all infinite strings in T.

And any time you find a set of such strings that is countable, from it Cantor and I can find a string that is is also in T, but is not in your set.But all of these strings are also in T.

- JeffJo
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**Posts:**2605**Joined:**Tue Mar 10, 2009 11:01 am

### Re: Cantor Diagonal Argument disproof

(JeffJo loves to parrot)JeffJo wrote:Since the point of induction is to"apply to the infinite," this is another (blatant) attempt at fast-talk.robert 46 wrote:What if induction does not apply to the infinite, ...

This is what the url says: "Principle of Mathematical Induction

The truth of an infinite sequence of propositions P_i for i=1, ..., infinity is established if (1) P_1 is true, and (2) P_k implies P_(k+1) for all k. This principle is sometimes also known as the method of induction."

The problem is in "i=1, ..., infinity". What this means is:

i=1, ..., i=infinity

But as JeffJo might point out, whereas infinity is not a number i=infinity is not a valid assignment. It should be:

i=1, ...

From this we get P_k implies P_(k+1) for all k+1<infinity. That is what induction is actually about. Extending it to "i=1, ..., i=infinity" is to introduce non-mathematical fantasy into the method.

Yep. Do you think you can write a comprehensible algorithm, without requiring such inference? If you want to demonstrate a point, that onus is on you.Do you think you can infer what the initial value of n is from the context???

What JeffJo is too timid to say is that n=1 is the obvious starting value, and all he has been doing is make himself look like a time-wasting, nitpicking nuisance- which, however, is in character.

It is a countably infinite set, since it is equivalent to a set of finite strings.The question is: what is the count of all sfsc in Tf; ...

This is interesting because what JeffJo is saying is that the (count of sfsc in Tf)=2^n, but he IS NOT allowing induction to let n=infinity, but only n->infinity (approaches).

It does not compare to T, which includes infinite strings that cannot be found by your algorithm.... and how does this compare to the count of all infinite strings in T.

This is also interesting because JeffJo is saying that the (count of infinite strings in T)=2^n, where he IS allowing induction to let n=infinity.

He is being inconsistent. Is n=infinity a valid assignment or not???

And any time you find a set of such strings that is countable, from it Cantor and I can find a string that is is also in T, but is not in your set.But all of these strings are also in T.

JeffJo must now agree that the constructive method I described does in fact produce all the sfsc for Tf without duplication. T contains all of Tf. Consider that we try to construct T by starting with the construction of Tf. Needless to say, the construction of Tf goes on endlessly. If JeffJo does not allow induction to imply that Tf is a complete set then T cannot be complete without all of Tf. However, if JeffJo does allow induction to imply Tf is a complete set then T can be complete since nothing is missing from Tf. The count of elements in an incomplete Tf is 2^n, but the count of elements in a complete Tf is 2^infinity. Yet, the count of elements in T is the powerset which is 2^infinity. Thus if Tf is complete, it provides all the elements in T, and there is nothing more to add; but if Tf is incomplete T is also incomplete.

We get the contradiction that a complete Tf is a countable infinite with 2^infinity elements, but T is an uncountable infinite with 2^infinity elements. If Tf is a countable infinite and is complete then T must also be a countable infinite and complete because they have the same number of elements. But if Tf is not complete then T must also be incomplete which implies that there is no uncountable infinite set which is complete.

- robert 46
- Intellectual
**Posts:**2832**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

(robert loves to pick only the arguments he can misrepresent, to reply to)

He repeats the same invalid comparison in several ways in the rest of his diatribe, none of which are worth responding to. The flaw is the same in each.

+++++

The integers are defined by induction: the set N contains 1, and if it contains n, it also contains n+1. It is countably infinite because it can be put into a 1:1 correspondence with itself.

The set P, of all pairs natural numbers like (3,4), appears to "go up in size" by n^2. Just look at partial sets sequentially, using all pairs of integers less than or equal to n. But it can be proven to be countable - a formula exists that (A) for any n, produces a unique element of P and (B) by induction produces every element of P. This is not a contradiction, since an infinite set can be a proper subset of itself.

The set RT (for "robert's T"), of all finite binary strings padded with infinite 0's, appears to "go up in size" by 2^n. Just look at partial sets sequentially, using only binary strings of n characters. But it can be proven to be countable - a formula exists that (A) for any n, produces a unique element of RT and (B) by induction produces every element of RT. This is not a contradiction, since an infinite set can be a proper subset of itself.

Here, robert deliberately ignores the fact that Mathworld is using the standard shorthand for the "potential infinity," which means the sequence is unbounded. That the index i increases without bound, not that it is ever assigned the value "infinity." This is an obvious - and universally accepted - implication. Something robert should accept, he thinks I am required to accept his not-so-obvious, and the opposite of what is universally accepted, implications.robert 46 wrote:This is what the url says: "Principle of Mathematical Induction

The truth of an infinite sequence of propositions P_i for i=1, ..., infinity is established if (1) P_1 is true, and (2) P_k implies P_(k+1) for all k. This principle is sometimes also known as the method of induction."

The problem is in "i=1, ..., infinity". What this means is:

i=1, ..., i=infinity

But as JeffJo might point out, whereas infinity is not a number i=infinity is not a valid assignment.

He repeats the same invalid comparison in several ways in the rest of his diatribe, none of which are worth responding to. The flaw is the same in each.

+++++

The integers are defined by induction: the set N contains 1, and if it contains n, it also contains n+1. It is countably infinite because it can be put into a 1:1 correspondence with itself.

The set P, of all pairs natural numbers like (3,4), appears to "go up in size" by n^2. Just look at partial sets sequentially, using all pairs of integers less than or equal to n. But it can be proven to be countable - a formula exists that (A) for any n, produces a unique element of P and (B) by induction produces every element of P. This is not a contradiction, since an infinite set can be a proper subset of itself.

The set RT (for "robert's T"), of all finite binary strings padded with infinite 0's, appears to "go up in size" by 2^n. Just look at partial sets sequentially, using only binary strings of n characters. But it can be proven to be countable - a formula exists that (A) for any n, produces a unique element of RT and (B) by induction produces every element of RT. This is not a contradiction, since an infinite set can be a proper subset of itself.

- JeffJo
- Intellectual
**Posts:**2605**Joined:**Tue Mar 10, 2009 11:01 am

### Re: Cantor Diagonal Argument disproof

JeffJo wrote:The set RT (for "robert's T"), of all finite binary strings padded with infinite 0's, appears to "go up in size" by 2^n. Just look at partial sets sequentially, using only binary strings of n characters. But it can be proven to be countable - a formula exists that (A) for any n, produces a unique element of RT and (B) by induction produces every element of RT. This is not a contradiction, since an infinite set can be a proper subset of itself.

I provided that method to sequentially produce unique elements, but the set still has 2^infinity elements. So address what I said (there is no point in changing Tf, which has prior precedence, to RT):

robert 46 wrote:1. T contains all of Tf.

2. Consider that we try to construct T by starting with the construction of Tf.

3. Needless to say, the construction of Tf goes on endlessly.

4. If JeffJo does not allow induction to imply that Tf is a complete set then T cannot be complete without all of Tf.

5. However, if JeffJo does allow induction to imply Tf is a complete set then T can be complete since nothing is missing from Tf.

6. The count of elements in an incomplete Tf is 2^n, but the count of elements in a complete Tf is 2^infinity.

7. Yet, the count of elements in T is the powerset which is 2^infinity.

8. Thus if Tf is complete, it provides all the elements in T, and there is nothing more to add; but if Tf is incomplete T is also incomplete.

9. We get the contradiction that a complete Tf is a countable infinite with 2^infinity elements, but T is an uncountable infinite with 2^infinity elements.

10. If Tf is a countable infinite and is complete then T must also be a countable infinite and complete because they have the same number of elements.

11. But if Tf is not complete then T must also be incomplete which implies that there is no uncountable infinite set which is complete.

...statement by statement to try to refute it.

- robert 46
- Intellectual
**Posts:**2832**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

Cantor first produced a similar method (actually, closer to the one about pairs), to show that the rational numbers are countable.robert 46 wrote:I provided that method to sequentially produce unique elements, but the set still has 2^infinity elements.

What robert can't bring himself to consider - or to accept that Cantor not only understood it, he understood it better than robert does - is that this doesn't make the count infinity^2, or 2^infinity. All countable infinities can be put into 1:1 correspondence, even robert constructs them in chunks of n^2, or 2^n. The size of the chunks you construct is completely irrelevant, if they can still be put into 1:1 correspondence with the natural numnbers.

Yes, it does.robert 46 wrote:1. T contains all of Tf.

Okay.2. Consider that we try to construct T by starting with the construction of Tf.

Yes, it does. That does not mean it finds every string that is in T. "Goes on endlessly" does not mean "finds every possibility robert wants it to find." And induction still says that every element it finds is equivalent to a unique natural number.3. Needless to say, the construction of Tf goes on endlessly.

And in fact, it can never produce 0(1), 1(1), 00(1), 01(1), 10(1), etc.; all of which are in T.

Sure, its a complete set. A complete set of all strings of finite 1's and 0's, followed by infinite 0's. It still cannot produce 1(1).4. If JeffJo does not allow induction to imply that Tf is a complete set

Every natural number is finite. The size of the set of all natural numbers is infinite. This size is called aleph0. Aleph0 cannot be considered to be a member of the set, but it is the size of the set. It is a transfinite number, and cannot be used in math as you try to do.

Wrong. The chunks you construct have size 2^n, but the size of the set is aleph0. There is no contradiction in constructing different countable infinities in differently-sized chunks, one going up as n, one going up as n^2, and another going up as 2^n.6. The count of elements in an incomplete Tf is 2^n, but the count of elements in a complete Tf is 2^infinity.

If you use transfinite numbers, some may think they can write aleph1=2^aleph0. But this means something different than 2^infinity in your point #6.7. Yet, the count of elements in T is the powerset which is 2^infinity.

No, it doesn't. It doesn't contain 1(1). See the rebuttal to point #3.8. Thus if Tf is complete, it provides all the elements in T, and there is nothing more to add; but if Tf is incomplete T is also incomplete.

No, we get robert using the same symbols in contradictory ways, and claiming the contradiction isn't in those different ways.9. We get the contradiction that a complete Tf is a countable infinite with 2^infinity elements, but T is an uncountable infinite with 2^infinity elements.

Invalid deduction. See #3.10. If Tf is a countable infinite and is complete then T must also be a countable infinite and complete because they have the same number of elements.

Invalid deduction. See #3.11. But if Tf is not complete then T must also be incomplete which implies that there is no uncountable infinite set which is complete.

+++++

Challenge for robert: write out your construction of Tf, that you think makes it look like its size is 2^infinity. But each time you add a string to Tf, put a number in set N1 that is calculated by bit1+2*bit2+4*bit3+... . This full set set can be constructed by induction. What is its size?

- JeffJo
- Intellectual
**Posts:**2605**Joined:**Tue Mar 10, 2009 11:01 am

### Re: Cantor Diagonal Argument disproof

JeffJo wrote:Yes, it does.robert 46 wrote:1. T contains all of Tf.Okay.2. Consider that we try to construct T by starting with the construction of Tf.Yes, it does. That does not mean it finds every string that is in T. And induction still says that every element it finds is equivalent to a unique natural number.3. Needless to say, the construction of Tf goes on endlessly.

And in fact, it can never produce 0(1), 1(1), 00(1), 01(1), 10(1), etc.; all of which are in T.

Then how can Tf be complete without going to infinity?

Sure, its a complete set. A complete set of all strings of finite 1's and 0's, followed by infinite 0's. It still cannot produce 1(1).4. If JeffJo does not allow induction to imply that Tf is a complete set

If we have all of the strings of n significant characters, there are 2^n of them. Assume the string 1011(0): how many characters are in this string? Is it 4+infinity or just infinity? It has to be the latter. So how many characters are in a string of n significant characters? It is not n+infinity, but just infinity.

Now if we let n increase, the total number of strings of n significant characters is 2^n, and the superset of all strings of n characters has 2^n elements; so the mere fact that one set has the strings padded with (0) and the other doesn't does not change the number of elements- the count is the same.

So what is the length of a string with infinite significant characters? Is it infinity+infinity or just infinity? For the same reason it cannot be n+infinity, it cannot be infinity+infinity either. A string of infinity significant characters is identical to an infinite string. This is to say that there is no padding with (0). Thus if Tf is complete, which is to say n=infinity, then it morphs to become all of T. This must be the case because if n is allowed to be infinity to define T by induction then n must also be allowed to be infinity to define Tf by induction. Both have the same number of elements [1] 2^infinity; and whereas T contains all of Tf, if Tf has the same number of elements as T then Tf contains all of T. If T contains all of Tf, and Tf contains all of T then T=Tf of logical necesssity. There are no elements of Tf which are not in T, and vice versa.

The problem is whether the method actually produces all of Tf. The method is sequential, and any sequential process can load a list. Any list which can be loaded is necessarily countable. If an infinite list can be loaded then that list is a countable infinite. Whereas completely loading Tf by the method is concurrent with loading a list, Tf must be a countable infinite. But completely loading Tf appears to be the same as completely loading T (they wind up with the same number of elements, and the same elements). So why isn't T a countable infinite also?

The only argument against this appears to be that Tf cannot be completely loaded; i.e. induction does not apply to Tf. However, if Tf cannot be completely loaded then it is missing elements. Whereas T contains all of Tf, if Tf is missing elements then T must also be missing elements. Therefore T cannot be complete if Tf is not complete. The completeness of T is contingent on the completeness of Tf. Thereby we cannot prohibit induction applying to Tf and still save the completeness of T.

[1] None of these elements is longer than infinity, nor shorter than infinity.

- robert 46
- Intellectual
**Posts:**2832**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

Robert, your problem is that you are not working with the definition of countability, which is "A is said to be countable if there exists an injective [1] map from A to a subset of N, the natural numbers".

So there is really no need, or is error, to write down an algorithm that "creates" A or "counts" objects in A.

In our case, all we need to do is find one injective map for Tf, which Jeff has already given you a hint how that might be accomplished. Go ahead and try to prove the proposition below.

Proposition: the map g proves that Tf is countable, where (g s):= sum (s n)*2^(n-1) over all n, and s is a map representing a sequence of 1's and 0's.

But even though g proves that Tf is countable, so do the maps 2*g, or 3*g, ..., all which actually fail to produce all of N; there's room over, so to speak.

[1] a map f is said to be injective if (f a)=(f b) implies a=b for all a and b in A.

So there is really no need, or is error, to write down an algorithm that "creates" A or "counts" objects in A.

In our case, all we need to do is find one injective map for Tf, which Jeff has already given you a hint how that might be accomplished. Go ahead and try to prove the proposition below.

Proposition: the map g proves that Tf is countable, where (g s):= sum (s n)*2^(n-1) over all n, and s is a map representing a sequence of 1's and 0's.

But even though g proves that Tf is countable, so do the maps 2*g, or 3*g, ..., all which actually fail to produce all of N; there's room over, so to speak.

[1] a map f is said to be injective if (f a)=(f b) implies a=b for all a and b in A.

- Gofer
- Intellectual
**Posts:**282**Joined:**Mon May 09, 2016 8:24 am

### Re: Cantor Diagonal Argument disproof

Yes, there are. If we have every natural number less than or equal to 2^n, there are 2^n of them, also. Neither statement has any relevance. We can build up either set in this progression that increases as 2^n, and what we get is a countably infinite set.robert 46 wrote:If we have all of the strings of n significant characters, there are 2^n of them.

It has countably-infinite many. The number is not "infinity," or "4+infinity."Assume the string 1011(0): how many characters are in this string? Is it 4+infinity or just infinity?

But if you insist on using those expressions, EACH IS JUST AS CORRECT AS THE OTHER. You can start with the first character, and put the string in a 1:1 correspondence with N. You skip four characters, and put it the rest in a 1:1 correspondence with N.

But it still has what you call a finite number of significant characters.

It has to be either. And you are again treating the infinite as if it were finite, when it pleases you to do so.It has to be the latter.

It can be either. But they still have a a finite number of significant characters, and the entire set can still be put in 1:1 correspondence with N.So how many characters are in a string of n significant characters? It is not n+infinity, but just infinity.

... and amazingly, is still countably infinite.Now if we let n increase, the total number of strings of n significant characters is 2^n, and the superset of all strings of n characters has 2^n elements; ...

... and is still countably infinite....so the mere fact that one set has the strings padded with (0) and the other doesn't does not change the number of elements- the count is the same.

I know not what a "string of infinity significant characters" is. Infinity is not a number, and can't be used this way.A string of infinity significant characters is identical to an infinite string.

I know not what "is identical to" is supposed to mean. (1), a string with infinitely-many significant characters, is not identical to (0), an infinite string.

You are also implying that this somehow crosses the line from finite significant characters, to infinite significant characters. And it doesn't.

Saying "n=infinity" is like saying "the three balls are both the same." It's jibberish.Thus if Tf is complete, which is to say n=infinity, then it morphs to become all of T.

But no, it doesn't "morph" to anything. You seem to be saying that since N is infinite, a member of it morphs to become aleph0. This is wrong, and it is the point you continue to refuse to address.

Your problem is that you want it to so badly, you are blind to why it can't.The problem is whether the method actually produces all of Tf.

Please, define what you want Tf to be, and what "fully loaded" means.The only argument against this appears to be that Tf cannot be completely loaded;

You seem to have defined Tf to be the set of all possible binary strings of finite significant characters, padded with (0). Induction defines every possible member of that set, which is what I take "fully loaded" to mean. All of them have finite significant characters, by definition. So (1) is not part of that set, and cannot ever be "loaded." But it is in T.

+++++

If "longer" and "shorter" have any meaning here, it is not the one you are using.[1] None of these elements is longer than infinity, nor shorter than infinity.

A subset of any of them can be put in a 1:1 correspondence with all of N, which is a possible interpretation of what "longer" means. Any of them can be put in a 1:1 correspondence with a subset of N, which is a possible interpretation of what "shorter" means.

Your problem is still trying to treat the infinite as if it were finite. You keep denying the properties that an infinite set must have.

- JeffJo
- Intellectual
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