## Cantor Diagonal Argument disproof

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### Re: Cantor Diagonal Argument disproof

Infinite means endless, and endless means impossible. Notwithstanding that limit theory and convergence can get around the impasse in some circumstances, such is not amenable to diagonalization. The anti-diagonal, ~D, is consequent to the order of elements in the list, which is arbitrary, so one cannot project what it would be.

Analysis of the sequential Diagonal method, SDM, shows us the innate impossibility of producing ~D: it cannot be produced until all intermediate anti-diagonals, iad, have been produced and retired, but they are produced faster than they are retired. Whereas all the iad are strings of finite significant characters, sfsc, and all the sfsc can be listed in principle, it is impossible to examine the entire list of sfsc and still have non-retired iad.

Cantor gets around this contradiction by ignoring it. Assume for argument's sake that the Cantor Diagonal method, CDM, produces ~D by parallel method acting on a list of sfsc. What of the iad? Whereas each position, n, defines an iad, whereas half of them should be unique, whereas an infinite list is examined, there would be infinity/2 unique iad defined. This means that every other element in the list must be an iad to retire all those defined. But what of the other sfsc which comprise the complete list? Iad grow in number at the rate n/2, but sfsc grow in number at 2^n. The only resolution is that either:

1: not all the defined iad are retired,

or

2: not all the sfsc are considered in producing ~D.

1 implies 2 because the complete list of sfsc has all the iad.

Thus CDM would produce ~D from a portion of the complete list of sfsc. However, the complete list of sfsc is a countable infinite, so CDM should be able to access the complete list to produce ~D. This situation has an intrinsic contradiction.

Analysis of the sequential Diagonal method, SDM, shows us the innate impossibility of producing ~D: it cannot be produced until all intermediate anti-diagonals, iad, have been produced and retired, but they are produced faster than they are retired. Whereas all the iad are strings of finite significant characters, sfsc, and all the sfsc can be listed in principle, it is impossible to examine the entire list of sfsc and still have non-retired iad.

Cantor gets around this contradiction by ignoring it. Assume for argument's sake that the Cantor Diagonal method, CDM, produces ~D by parallel method acting on a list of sfsc. What of the iad? Whereas each position, n, defines an iad, whereas half of them should be unique, whereas an infinite list is examined, there would be infinity/2 unique iad defined. This means that every other element in the list must be an iad to retire all those defined. But what of the other sfsc which comprise the complete list? Iad grow in number at the rate n/2, but sfsc grow in number at 2^n. The only resolution is that either:

1: not all the defined iad are retired,

or

2: not all the sfsc are considered in producing ~D.

1 implies 2 because the complete list of sfsc has all the iad.

Thus CDM would produce ~D from a portion of the complete list of sfsc. However, the complete list of sfsc is a countable infinite, so CDM should be able to access the complete list to produce ~D. This situation has an intrinsic contradiction.

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

One can only wonder why Robert harps on and on about the "sequential diagonal method" that calculates "intermediate strings" and "retires" them, and that has absolutely nothing to do with Cantor's argument.

- Gofer
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### Re: Cantor Diagonal Argument disproof

> Infinite means endless, and endless means impossible.

Except, we don't have to calculate a single element of the anti-diagonal to PROVE that it can't equal any row; this alone rebuts everything you have said about it and the rest of your post.

> Notwithstanding that limit theory and convergence can get around the impasse in some circumstances, such is not amenable to diagonalization.

Correct, such is not amenable, because diagonalization has nothing to do with limit theory.

> The anti-diagonal, ~D, is consequent to the order of elements in the list,

Of course! Switching two rows, and the anti-diagonal might differ! So what?

> which is arbitrary, so one cannot project what it would be.

Again, so what?

Except, we don't have to calculate a single element of the anti-diagonal to PROVE that it can't equal any row; this alone rebuts everything you have said about it and the rest of your post.

> Notwithstanding that limit theory and convergence can get around the impasse in some circumstances, such is not amenable to diagonalization.

Correct, such is not amenable, because diagonalization has nothing to do with limit theory.

> The anti-diagonal, ~D, is consequent to the order of elements in the list,

Of course! Switching two rows, and the anti-diagonal might differ! So what?

> which is arbitrary, so one cannot project what it would be.

Again, so what?

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

### Re: Cantor Diagonal Argument disproof

Correct.robert 46 wrote:Infinite means endless, ...

Why?... and endless means impossible.

Consider the natural numbers. I agree that it is impossible write them all out, but AS AN ABSTRACT CONCEPT we can consider them to be a set. THAT DOES MEAN WE CAN WRITE THEM ALL OUT; it just means we have a definition that applies to every possible one. And no matter how many different ways you find to show that you can't write them all out, it is all meaningless since that is not a part of the definition of a set.

But limit theory and convergence still have absolutely nothing to do with whether of not we can consider them to be a set.Notwithstanding that limit theory and convergence

If this were true, it would be just as impossible to create a mapping between N and the strings. yet we know that is possible.The anti-diagonal, ~D, is consequent to the order of elements in the list, which is arbitrary, so one cannot project what it would be.

So something is wrong with this assertion that you keep making, but won't support except by saying over and over "is consequent to." The fact is that we don't need to know what the ordering is, or be able to "calculate" each one (which has nothing to do with limits, or convergence). If one exists, we can indeed "project" what it is for each element of the endless set, and so construct the anti-diagonal ~D.

... absolutely nothing, since it only considers finite strings (even if you pad them infinitely). It is equivalent to pointing out that the endless set of natural numbers N, and the endless set of even numbers E, can be put into a 1:1 correspondence even E is a strict subset of N. This disproves your entire thesis, which is why you ignore it.Analysis of the sequential Diagonal method, SDM, shows...

Since it doesn't demonstrate anything except the point I just made, and he did address that, this is a lie. But you will ignore that as well.Cantor gets around this contradiction by ignoring it.

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:One can only wonder why Robert harps on and on about the "sequential diagonal method" that calculates "intermediate strings" and "retires" them, and that has absolutely nothing to do with Cantor's argument.

You can wonder all you like if you are clueless about the importance of the contradictions I am uncovering. That Cantor's method leads to contradictions shows that his argument cannot be correct.

In your second post, you evade everything after the first paragraph. I assume this is because you do not understand the argument I am making.

*****

In blue is what JeffJo ignored in my post:

robert 46 wrote:Infinite means endless, and endless means impossible. Notwithstanding that limit theory and convergence can get around the impasse in some circumstances, such is not amenable to diagonalization. The anti-diagonal, ~D, is consequent to the order of elements in the list, which is arbitrary, so one cannot project what it would be.

Analysis of the sequential Diagonal method, SDM, shows us the innate impossibility of producing ~D: it cannot be produced until all intermediate anti-diagonals, iad, have been produced and retired, but they are produced faster than they are retired. Whereas all the iad are strings of finite significant characters, sfsc, and all the sfsc can be listed in principle, it is impossible to examine the entire list of sfsc and still have non-retired iad.

Cantor gets around this contradiction by ignoring it. Assume for argument's sake that the Cantor Diagonal method, CDM, produces ~D by parallel method acting on a list of sfsc. What of the iad? Whereas each position, n, defines an iad, whereas half of them should be unique, whereas an infinite list is examined, there would be infinity/2 unique iad defined. This means that every other element in the list must be an iad to retire all those defined. But what of the other sfsc which comprise the complete list? Iad grow in number at the rate n/2, but sfsc grow in number at 2^n. The only resolution is that either:

1: not all the defined iad are retired,

or

2: not all the sfsc are considered in producing ~D.

1 implies 2 because the complete list of sfsc has all the iad.

Thus CDM would produce ~D from a portion of the complete list of sfsc. However, the complete list of sfsc is a countable infinite, so CDM should be able to access the complete list to produce ~D. This situation has an intrinsic contradiction.

Note yet again his characteristic use of summary dismissal to evade the most important issues raised.

- robert 46
- Intellectual
**Posts:**2849**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

robert 46 wrote:In blue is what JeffJo ignored in my post:

And in red what robert ignores has been was shown to be false, obviating the need to respond directly from what he thinks follows from it.

robert 46 wrote:Infinite means endless, and endless means impossible. Notwithstanding that limit theory and convergence [which robert erronesouly thinks has somethign to do wiht diagonalization] can get around the impasse in some circumstances, such is not amenable to diagonalization. The anti-diagonal, ~D, is consequent to the order of elements in the list, which is arbitrary, so one cannot project what it would be.

Analysis of the sequential Diagonal method, SDM, shows [nothing about infinite lists of infinite stings] us the innate impossibility of producing ~D: it cannot be produced until all intermediate anti-diagonals, iad, have been produced and retired, but they are produced faster than they are retired. Whereas all the iad are strings of finite significant characters, sfsc, and all the sfsc can be listed in principle, it is impossible to examine the entire list of sfsc and still have non-retired iad.

Cantor [doesn't have to address it anymore than the flappuing of a butterfly's wings in China, because both are irrelevant] Assume for argument's sake that the Cantor Diagonal method, CDM, produces ~D by parallel method acting on a list of sfsc. What of the iad? Whereas each position, n, defines an iad, whereas half of them should be unique, whereas an infinite list is examined, there would be infinity/2 unique iad defined. This means that every other element in the list must be an iad to retire all those defined. But what of the other sfsc which comprise the complete list? Iad grow in number at the rate n/2, but sfsc grow in number at 2^n. The only resolution is that either:

1: not all the defined iad are retired,

or

2: not all the sfsc are considered in producing ~D.

1 implies 2 because the complete list of sfsc has all the iad.

Thus CDM would produce ~D from a portion of the complete list of sfsc. However, the complete list of sfsc is a countable infinite, so CDM should be able to access the complete list to produce ~D. This situation has an intrinsic contradiction.

Note that robert still thinks he doesn't have to respond to anything that anybody else says, yet insists that his every word must be responded to over, and over, and over, and...

And robert still has not defined what he thinks "actually produced" and/or "constructed" means; yet his entire argument is that The anti-diagonal can't be "actually produced" and/or "constructed." And he has ignored how an entire list of his sfsc can be constructed, and the ~D constructed from it.

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

### Re: Cantor Diagonal Argument disproof

Note yet again JeffJo's summary dismissal. I expect him to address each statement: to agree or disagree, and if the latter explain what is contentious.

*****

I am pleased that both JeffJo and Gofer agree that limit theory does not apply to diagonalization. This prevents them from claiming that the iad are getting closer and closer to ~D, and thus that ~D is the limit of the iad. The reason this claim cannot be made is because the iad are not algorithmic, and no projection to ~D can be calculated- unlike Sum(1/2^n, n=0...)->2. I trust this issue is now moot.

*****

I am pleased that both JeffJo and Gofer agree that limit theory does not apply to diagonalization. This prevents them from claiming that the iad are getting closer and closer to ~D, and thus that ~D is the limit of the iad. The reason this claim cannot be made is because the iad are not algorithmic, and no projection to ~D can be calculated- unlike Sum(1/2^n, n=0...)->2. I trust this issue is now moot.

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

Note again how robert won't address my comments that address his points. Just one such point, specifically, is that he won't define what "produced" means, yet insists that the diagonal isn't "produced."robert 46 wrote:Note yet again JeffJo's summary dismissal. I expect him to address each statement: to agree or disagree, and if the latter explain what is contentious.

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

### Re: Cantor Diagonal Argument disproof

I want to hear from Robert how we can prove that the functions (^2) and (1*) can't be equal, without actually having to calculate all the numbers in their range, assuming the domain is the natural numbers. Indeed, 1^2=1*1 for x=1.

analogy to Robert's argument: just because there are an infinite number of functions starting with a particular sequence of length n of 1's and 0's, doesn't mean the functions are all equal.

analogy to Robert's argument: just because there are an infinite number of functions starting with a particular sequence of length n of 1's and 0's, doesn't mean the functions are all equal.

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

### Re: Cantor Diagonal Argument disproof

JeffJo wrote:Note again how robert won't address my comments that address his points.robert 46 wrote:Note yet again JeffJo's summary dismissal. I expect him to address each statement: to agree or disagree, and if the latter explain what is contentious.

You considered the first sentence, or part of it, and summarily dismissed the rest. You were introducing a diversion, and I am not going down that garden path.

Just one such point, specifically, is that he won't define what "produced" means, yet insists that the diagonal isn't "produced."

To repeat:

MathWorld wrote: A constructive proof is a proof that directly provides a specific example, or which gives an algorithm for producing an example.

Claiming that "Poof, ~D suddenly appears" is to me not an algorithm- it is a magic trick.

What if the list is wider than long? Then it must be a finite list, and no infinitely long anti-diagonal could be produced.

What if the list is exactly as wide as long? Then it is a countable infinite in both width and length, and a missing element is defined. But how could a countable infinite list have a missing element? Only if it is not complete to start with.

Example: String n is all 0 except for character n which is 1. Missing element is (0). Append the list onto the missing element. Repeat. New missing element is (1). Append the list onto the missing element. Repeat. New missing element is 0(1). Repeat as much as you like. Does the list become longer than wide by doing this? If not then the list remains a countable infinite, notwithstanding missing elements endlessly being added to it. Therefore it is wrong to consider a countable infinite list to ever be complete. This is contradictory.

What if the list is longer than wide? Then we have the problem that examination of the width does not imply proportionate examination of the length. This means the list either has no diagonal, or a diagonal would be consequent to only a portion of the list.

P.S. I have no idea what point Gofer is trying to make.

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

Robert, my point is like I've been trying to tell you:

The anti-diagonal is a FUNCTION, and, as such, doesn't matter for how many elements it equals another function, a row for instance, just that there exists one element for which they differ. Again, If I have the function g(x,y), I can create the function g(x,x)+1 mod 2, which must necessarily be different from the function g(x,Y) for any Y.

That's about as short as you could possibly state Cantor's Lemma.

The anti-diagonal is a FUNCTION, and, as such, doesn't matter for how many elements it equals another function, a row for instance, just that there exists one element for which they differ. Again, If I have the function g(x,y), I can create the function g(x,x)+1 mod 2, which must necessarily be different from the function g(x,Y) for any Y.

That's about as short as you could possibly state Cantor's Lemma.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:The anti-diagonal is a FUNCTION, ...

A function must be shown to be valid over a specific domain. You may claim a result, but you have to prove it is valid over the domain before the result will be accepted.

- robert 46
- Intellectual
**Posts:**2849**Joined:**Mon Jun 18, 2007 9:21 am

### Re: Cantor Diagonal Argument disproof

>A function must be shown to be valid over a specific domain. You may claim a result, but you have to prove it is valid over the domain before the result will be accepted.

That's easy! Since g(x,y) is assumed to be defined for every x and y in some domain D, then surely g(x,x)+1 must be, and therefore f(x):=g(x,x)+1 mod 2. It is now trivial to show that f can't equal g(x,Y) for any Y.

And there you go, as requested!

That's easy! Since g(x,y) is assumed to be defined for every x and y in some domain D, then surely g(x,x)+1 must be, and therefore f(x):=g(x,x)+1 mod 2. It is now trivial to show that f can't equal g(x,Y) for any Y.

And there you go, as requested!

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:>A function must be shown to be valid over a specific domain. You may claim a result, but you have to prove it is valid over the domain before the result will be accepted.

That's easy! Since g(x,y) is assumed to be defined for every x and y in some domain D, then surely g(x,x)+1 must be, and therefore f(x):=g(x,x)+1 mod 2. It is now trivial to show that f can't equal g(x,Y) for any Y.

And there you go, as requested!

Carl Friedrich Gauss's views on the subject [the infinite] can be paraphrased as: 'Infinity is nothing more than a figure of speech which helps us talk about limits. The notion of a completed infinity doesn't belong in mathematics'. In other words, the only access we have to the infinite is through the notion of limits, and hence, we must not treat infinite sets as if they have an existence exactly comparable to the existence of finite sets.

https://en.wikipedia.org/wiki/Controver ... 27s_theory

However, we have established that limit theory has nothing to do with diagonalization.

I suggest you address the issues of length vs. width, and others you ignored in my prior post.

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

The rest wasn't worth consideration, since it follows from the points of yours that have been discredited.robert 46 wrote:You considered the first sentence, or part of it, and summarily dismissed the rest.

You are using it as a diversion, and an excuse to not reply to my comments.I am not going down that garden path.

To repeat: I didn't ask for the definition of a constructive proof, I asked for your definition of "producing" an object. That's not a proof.Just one such point, specifically, is that he won't define what "produced" means, yet insists that the diagonal isn't "produced."

To repeat:

MathWorld wrote: A constructive proof is a proof that directly provides a specific example, or which gives an algorithm for producing an example.

But the function CDM(S()) directly provides the anti-diagonal of the list of strings S(). Because it doesn't do it sequentially. It is your only sequential method that fails, since it is sequential.

Can't you get in through your thick skull that this is a nonsensical statement?What if the list is wider than long?

The list S() has no width; its "length" - which is the wrong word, you mean the cardinality of the list - is "countably infinite."

The strings S(n) have no length, but their "widths" - again, the wrong word, you mean the cardinality of the strings - are all "countably infinite."

--->>> ANY COUNTABLY INFINITE SET A CAN BE PUT INTO A 1:1 CORRESPONDENCE WITH ANY OTHER COUNTABLY INFINITE SET B. <<<--- Get that? Example: A is the even natural numbers, B is the natural numbers, and 2n=e is the correspondence.

--->>> BUT THE SAME A CAN ALSO BE PUT INTO 1:1 CORRESPONDENCE WITH A STRICT SUBSET OF THE SAME B. <<<--- Get that? The even numbers can be matched up with only themselves, apparently[1] "missing" half of the set.

--->>> AND B CAN ALSO BE PUT INTO 1:1 CORRESPONDENCE WITH A STRICT SUBSET OF A. <<<--- Get that? Take every fourth member of the natural numbers, and now half of the evens are unpaired.

--->>> UNMATCHED ELEMENTS ARE SIGNIFICANT ONLY IN FINITE SETS. <<<---

That's why your machinations with intermediate results have no bearing on Cantor; finite sets do not have this property. All you keep doing, is pointing out this property.

Wrong. See emphasized point above. You just found an unmatched algorithm, a known property of infinite sets. That doesn't mean there aren't matched algorithms. I've even shown you one.[2]Then it must be a finite list, and no infinitely long anti-diagonal could be produced.

Of course it's incomplete to start with. It is never assumed to be complete. --->>> YOU ARE MISREPRESENTING CDM. <<<---What if the list is exactly as wide as long? Then it is a countable infinite in both width and length, and a missing element is defined. But how could a countable infinite list have a missing element? Only if it is not complete to start with.

Wrong. See emphasized point above.What if the list is longer than wide? Then we have the problem that examination of the width does not imply proportionate examination of the length. This means the list either has no diagonal, or a diagonal would be consequent to only a portion of the list.

+++++

[1] Only if you refuse to understand that infinite sets work differently than finite ones.

[2] Oh, I forgot, it's not evasion when you completely ignore a point; but it is when I skip the points that are obviated.

- JeffJo
- Intellectual
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