## Cantor Diagonal Argument disproof

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### Re: Cantor Diagonal Argument disproof

Consider:

n is a natural number

D[] is a global string

S[] is a global list of strings

S[p][q] is the character at position q extracted from the string at position p

Function Produce_D(n)

D[n]=S[n][n]

Produce_D(n+1)

End Function

Load S with any non-duplicated strings of constant width of characters taken from a 2-character set.

Produce_D(1)

This function will either crash the system, or not return a result consequent to not having an exit test. If n becomes greater than the width of strings, or n become greater than the length of S then the system crashes consequent to an out-of-range error.

If n never becomes out-of-range consequent to endless strings and an endless list, it does not return a value.

If an exit test is put in:

if (n<(width of S) and n<(length of S) and n<(width of D)) then Produce_D(n+1)

...then the system does not crash (assuming idealized resources). If it terminates then it returns a D with finite significant characters.

However, if the width of S is endless, and the length of S is endless, and the width of D is endless then the exit condition is never satisfied and the function never returns a value: i.e. it never returns a D with endless significant characters.

This is what we are up against with endlessness: if it uses the exit condition then it doesn't do what Cantorians want it to do, and if it doesn't use the exit condition then it still doesn't do what they want it to do. Attempting to define a parallel method with endless agents does not solve the problem of endlessness. It merely sweeps one endlessness_problem under the rug by invoking another endlessness_problem to cover it.

n is a natural number

D[] is a global string

S[] is a global list of strings

S[p][q] is the character at position q extracted from the string at position p

Function Produce_D(n)

D[n]=S[n][n]

Produce_D(n+1)

End Function

Load S with any non-duplicated strings of constant width of characters taken from a 2-character set.

Produce_D(1)

This function will either crash the system, or not return a result consequent to not having an exit test. If n becomes greater than the width of strings, or n become greater than the length of S then the system crashes consequent to an out-of-range error.

If n never becomes out-of-range consequent to endless strings and an endless list, it does not return a value.

If an exit test is put in:

if (n<(width of S) and n<(length of S) and n<(width of D)) then Produce_D(n+1)

...then the system does not crash (assuming idealized resources). If it terminates then it returns a D with finite significant characters.

However, if the width of S is endless, and the length of S is endless, and the width of D is endless then the exit condition is never satisfied and the function never returns a value: i.e. it never returns a D with endless significant characters.

This is what we are up against with endlessness: if it uses the exit condition then it doesn't do what Cantorians want it to do, and if it doesn't use the exit condition then it still doesn't do what they want it to do. Attempting to define a parallel method with endless agents does not solve the problem of endlessness. It merely sweeps one endlessness_problem under the rug by invoking another endlessness_problem to cover it.

Last edited by robert 46 on Fri May 19, 2017 7:14 pm, edited 1 time in total.

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

What is a "global string?" You do realize that computer terminology is not the same as math terminology, don't you? And that computers don't have a way to represent the infinite, so any analogy using them is completely irrelevant unless you assume a hypothetical computer that can?robert 46 wrote:Consider:

n is a natural number

D[] is a global string

Yep, it looks like you are trying to foist off another invalid analogy, this time using computer language. So let's just assume what you are probably not, that we can have infinite lists, and infinite strings.S[] is a global list of strings

S[p][q] is the character at position q extracted from the string at positiion p

Function Produce_D(n)

D[n]=S[n][n]

Produce_D(n+1)

End Function

No. At this point, your analogy becomes invalid. To be valid, you have to "load" S with strings of infinite length. Which the hypothetical computer we need to assume can do.Load S with any non-duplicated strings of constant width of characters taken from a 2-character set.

But you knew that, didn't you? Either that, or you have ignored every comment made in response to your invalid arguments. Which makes one wonder how you found the parts you think were evaded, but were not.

Oh, so you are trying to define "produce?" Since we assume infinite sets exist, your definition has to allow for them, and you know as well as I that you are choosing this method because it doesn't.Produce_D(1)

So it is this system (i.e., your methodology) that is at fault, not Cantor's.This function will either crash the system, ...

Nope. You do understand what "random access" means, don't you?or not return a result consequent to not having an exit test.[1]

And again, it is the way you handle endlessness that is at fault. But, again, you knew that.This what we are up against with endlessness.

+++++

Let's look at the Axiom of Infinity:

- 1 is a natural number
- If 1 is a natural number, then so is n+1.
- The entirety of such natural numbers constitutes a set we call N.

Now, go back to:

- Function Produce_D(n)

D[n]=S[n][n]

Produce_D(n+1)

End Function

+++++

[1] What robert means here, is that a non-square, finite array cannot allow access to a diagonal element for every possible row and/or column. He even thinks this is a ground-breaking revelation, that none before him thought of.

What robert ignores, and so is evading the truth, is that in the array is endless in both dimensions, no matter how it was constructed aspect-ratio-wise, it is always possible to identify a diagonal element. Because there is no row (or column) which lacks a corresponding column (or row).

But robert will ignore this point. That Cantor didn't overlook this point - as robert has claimed repeatedly. It just doesn't take much intellect to see that it is an invalid point; that every row has a corresponding column, no matter how you define them.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

JeffJo wrote:Function Produce_D(n)

D[n]=S[n][n]

Produce_D(n+1)

End Function

Given this hypothetical computer, it has the capability to run this program without crashing. And to cut off your obvious retort, this is what Wikipedia means when it said that the diagonal was constructed.

Unfortunately, the function is always producing without ever having produced a final result- consequent, of course, to endlessness because there is no final result. "Final" is a contradiction in the context of "endlessness".

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

As always, robert's response is "If I set it up as a finite process, it can only be a finite process in finite time. Therefore, it must be a finite process." Here, he ignores the hypothetical nature that is require, that makes it possible to create the infinite string.robert 46 wrote:Unfortunately, the function is always producing without ever having produced a final result- consequent, of course, to endlessness because there is no final result. "Final" is a contradiction in the context of "endlessness".

He also ignores, AS I PREDICTED HE WOULD:

JeffJo wrote:[1] What robert means here, is that a non-square, finite array cannot allow access to a diagonal element for every possible row and/or column. He even thinks this is a ground-breaking revelation, that none before him thought of.

What robert ignores, and so is evading the truth, is that in the array is endless in both dimensions, no matter how it was constructed aspect-ratio-wise, it is always possible to identify a diagonal element. Because there is no row (or column) which lacks a corresponding column (or row).

But robert will ignore this point. That Cantor didn't overlook this point - as robert has claimed repeatedly. It just doesn't take much intellect to see that it is an invalid point; that every row has a corresponding column, no matter how you define them.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

If we consider a list of all strings of n characters taken from a set of two characters, the length of the list is 2^n and the width is n. The ratio of length to width is 2^n/n, and we can see that this is a monotonically increasing number as n increases. This means that the length becomes a greater multiple of the width as n increases. If we extend this toward endless, we see that the multiple increases without limit. So the multiple approaches 2^infinity/infinity=infinity. This means that the part examined by diagonalization of the endless list approaches 0% of the entire list.

Now, JeffJo would have us believe that the diagonal examination accesses the entire list. This would be correct if 2^infinity=infinity, for then the ratio of length to width is 2^infinity/infinity=infinity/infinity=1, which is square. But what accounts for the discontinuity from a monotonically increasing ratio of length to width which approaches infinite to abruptly become equality? Only the contradiction which arises between 2^infinity/infinity=infinity and 2^infinity/infinity=1. How can this be resolved? Ans: By realizing that we are actually always dealing with finite n because endlessness does not imply an obtainable end. Thus the correct conclusion is that diagonalization is only capable of processing a negligible amount of the entire list as it approaches endless.

Now, JeffJo would have us believe that the diagonal examination accesses the entire list. This would be correct if 2^infinity=infinity, for then the ratio of length to width is 2^infinity/infinity=infinity/infinity=1, which is square. But what accounts for the discontinuity from a monotonically increasing ratio of length to width which approaches infinite to abruptly become equality? Only the contradiction which arises between 2^infinity/infinity=infinity and 2^infinity/infinity=1. How can this be resolved? Ans: By realizing that we are actually always dealing with finite n because endlessness does not imply an obtainable end. Thus the correct conclusion is that diagonalization is only capable of processing a negligible amount of the entire list as it approaches endless.

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

... then we are treating the set as a set that has an end; that is, a finite set. And we will never be able to extend the properties we observe to infinite sets.robert 46 wrote:If we consider a list of all strings of n characters taken from a set of two characters, ...

For example, if we label our rows with 1, 2, 3,...2^n, and our columns with 1,2,3...,n, then the length of the list is 2^n and the width is n. The ratio of length to width is 2^n/n, and we can see that this is a monotonically increasing number as n increases.

BUT WE WILL ALWAYS BE ABLE TO FIND A DIAGONAL ELEMENT IN ANY ROW IF WE CAN EXTEND THE VALUE OF n ENDLESSLY.

Since we accept, by axiom, the concept of an infinite set, including an array that can be extended endlessly in either dimension, the it is obvious that any aspect ratio we (artificially) contrive is irrelevant. As is done in this next quote, which is rendered irrelevant by what I just said, but Robert refuses to see how it was so I'm including it:

And once again, this is meaningless since endless diagonalization can still "approach" (wrong word) any diagonal element in the entire list.This means that the length becomes a greater multiple of the width as n increases. If we extend this toward endless, we see that the multiple increases without limit. So the multiple approaches 2^infinity/infinity=infinity. This means that the part examined by diagonalization of the endless list approaches 0% of the entire list.

And if Robert wants to contest that, he should try to identify a diagonal element that can't be reached.

Robert still tries to use "infinity" like a number, and it isn't. The cardinality of the sets Robert refers to are all the same, because the sets can be put into a 1:1 correspondence, despite the fact that they are constructed in differently-sized chunks. His concept of finding a discontinuity requires identifying an end to the lists.Now, JeffJo would have us believe that the diagonal examination accesses the entire list. This would be correct if 2^infinity=infinity, for then the ratio of length to width is 2^infinity/infinity=infinity/infinity=1, which is square. But what accounts for the discontinuity from a monotonically increasing ratio of length to width which approaches infinite to abruptly become equality? Only the contradiction which arises between 2^infinity/infinity=infinity and 2^infinity/infinity=1. How can this be resolved? Ans: By realizing that we are actually always dealing with finite n because endlessness does not imply an obtainable end. Thus the correct conclusion is that diagonalization is only capable of processing a negligible amount of the entire list as it approaches endless.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

JeffJo wrote:Robert still tries to use "infinity" like a number, and it isn't. The cardinality of the sets Robert refers to are all the same, because the sets can be put into a 1:1 correspondence, despite the fact that they are constructed in differently-sized chunks. His concept of finding a discontinuity requires identifying an end to the lists.robert 46 wrote:Now, JeffJo would have us believe that the diagonal examination accesses the entire list. This would be correct if 2^infinity=infinity, for then the ratio of length to width is 2^infinity/infinity=infinity/infinity=1, which is square. But what accounts for the discontinuity from a monotonically increasing ratio of length to width which approaches infinite to abruptly become equality? Only the contradiction which arises between 2^infinity/infinity=infinity and 2^infinity/infinity=1. How can this be resolved? Ans: By realizing that we are actually always dealing with finite n because endlessness does not imply an obtainable end. Thus the correct conclusion is that diagonalization is only capable of processing a negligible amount of the entire list as it approaches endless.

This situation has the exact same problem as the monotonically increasing count of IADs. In the one: Every IAD will appear somewhere later in the list, but for every IAD found there are more and more to find. In the other: For every column found there are more and more rows to find [1]. In both cases it is inconceivable that an "end" can be reached where all the IADs are found and all the rows have been processed because this would introduce a fundamental contradiction. This means that diagonalization is always in the realm of producing finite IADs and never reaches the realm of a produced ultimate anti-diagonal ~D. Thus whatever function Produce_D(n) is defined, the function never exits, so no statements after the function call will ever be processed. If we believe diagonalization works we wind up in the wonderful world of Oz, but clicking our shoes together three times won't get us back to reality.

[1] If n columns are processed, there are 2^n-n more rows to process. If 2^n columns are processed there are 2^2^n-2^n more rows to process. If 2^2^n columns are processed there are.... Well, you should get the idea: there are 2^...2^n columns to process and n can be any natural number, but this is less than the number of columns to process to get ~D. This should not be surprising: it is always less than the number of columns to process to get ~D because there is no number of columns to process to get ~D. ~D is outside the realm of natural numbers [2]. Therefore it cannot be said that ~D is a string of characters in any sense which we have familiarity with. This makes ~D a mythological beast, so how is it meaningful to argue about a mythological beast???

[2] Every natural number has a finite value. The set of natural numbers is said to be infinite, but infinity is not a natural number. ~D is said to have infinite characters, but this is to treat infinity as if it was a natural number.

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

And if just 1 column is processed, there are infinity-1=infinity rows left to process; but so what? Your arguments reveal that you don't really comprehend the problem. It has nothing to do with whether an algoritm actually produces all elements, considered as a set, of the anti-diagonal, which it can't, but whether each one eventually gets calculated, which it will. That's the algorithmic viewpoint.If n columns are processed, there are 2^n-n more rows to process. If 2^n columns are processed there are 2^2^n-2^n more rows to process.

Here's a better one, functional, which you have neither addressed nor rebutted:

1. The function f(x,y) exists.

2. I can create the function d(x) := f(x,x)+1 mod 2.

3. The function d is not equal to λx.f(x,Y) for any Y. [3]

[3] "λx" reads "a function of x", and is a lambda abstraction.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Almost like that was the point. Oh, wait, thatWAS the point.robert 46 wrote:This situation has the exact same problem as the monotonically increasing count of IADs.

What Robert ignored:

JeffJo wrote:For example, if we label our rows with 1, 2, 3,...2^n, and our columns with 1,2,3...,n, then the length of the list is 2^n and the width is n. The ratio of length to width is 2^n/n, and we can see that this is a monotonically increasing number as n increases.

BUT WE WILL ALWAYS BE ABLE TO FIND A DIAGONAL ELEMENT IN ANY ROW IF WE CAN EXTEND THE VALUE OF n ENDLESSLY.

Since we accept, by axiom, the concept of an infinite set, including an array that can be extended endlessly in either dimension, the it is obvious that any aspect ratio we (artificially) contrive is irrelevant.

... that is shorter than those examined, but none that are longer so this is irrelevant, ...In the one: Every IAD...

will appear somewhere later in the list,

Until Robert can produce a definition of what he wants "produced" to mean, the rest of what he wrote is meaningless, and deserves to be ignored.

Yet it is only Robert who keeps trying to use infinity as a natural number, and the extension of the set.Every natural number has a finite value. The set of natural numbers is said to be infinite, but infinity is not a natural number. ~D is said to have infinite characters, but this is to treat infinity as if it was a natural number.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:And if just 1 column is processed, there are infinity-1=infinity rows left to process; but so what?robert 46 wrote:If n columns are processed, there are 2^n-n more rows to process. If 2^n columns are processed there are 2^2^n-2^n more rows to process.

One can project from the finite to the infinite and get a meaningful answer if there is a convergence which can be calculated. Diagonalization does not have an algorithmic convergence because it is consequent to the order of elements in the list which is ad hoc. ~D in theory could be any string, but the probability it would have a trailing repeating sequence is effectively zero. The point is not that there are more rows to process, but that there are contradictions which arise if one thinks ~D exists.

It has nothing to do with whether an algoritm actually produces all elements, considered as a set, of the anti-diagonal, which it can't, but whether each one eventually gets calculated, which it will. That's the algorithmic viewpoint.

The algorithmic viewpoint is that all the IADs are produced, but that it is contradictory to think that ~D can be considered to exist because IAD production is endless. There is no transition from IADs to ~D.

The only way to have ~D is to define it to exist because one cannot get ~D by projecting from the finite. It is all a matter of believing in the fantasy or not. People have beliefs for a wide variety of reasons which greatly outnumber rational thought.

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

And if you apply the same kind of "algorithmic viewpoint" to the production of the fodder used to create IADs, SFSC padded with infinite zeros, you can't "produce" a single one of them either. Proving that the fault is in the "algorithmic viewpoint".robert 46 wrote:The algorithmic viewpoint is that all the IADs are produced, but that it is contradictory to think that ~D can be considered to exist because IAD production is endless. There is no transition from IADs to ~D.

Of course, a definition of "actually produce" a justification for the "algorithmic viewpoint" would show why this if true, and why you won't provide either.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

In the Wizard of Oz movie, some of the Kansas people were caricatured in the Land of Oz:

Farmhands

Hunk -> Scarecrow

Hickory -> Tin Man

Zeke -> Cowardly Lion

Professor Marvel, the fortune-teller -> The Wizard

Almira Gulch -> Wicked Witch of the West

The Wizard rules over the Munchkins (the "little people") through fast-talk and deception.

Set math has similar exaggerations:

finite sets become infinite sets;

finite strings become infinite strings;

finite lists become infinite lists.

Cantor rules over those of small mind similarly.

Many people have no difficulty recognizing the Land of Oz as a fantasy. Some people have great difficulty recognizing set mathematics as a fantasy.

Farmhands

Hunk -> Scarecrow

Hickory -> Tin Man

Zeke -> Cowardly Lion

Professor Marvel, the fortune-teller -> The Wizard

Almira Gulch -> Wicked Witch of the West

The Wizard rules over the Munchkins (the "little people") through fast-talk and deception.

Set math has similar exaggerations:

finite sets become infinite sets;

finite strings become infinite strings;

finite lists become infinite lists.

Cantor rules over those of small mind similarly.

Many people have no difficulty recognizing the Land of Oz as a fantasy. Some people have great difficulty recognizing set mathematics as a fantasy.

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

Unfortunately for Robert, his being so fond of algorithms and programming, it wasn't "fast-talk", because Cantor's theorem has application to computable functions (CF), which are the algorithms/computer programs that provably halt in finite time for every input.

Let (f n) be a CF that, for every input n, returns CF's, who in turn return a computable number for every input m. Then the function (d n) := ((f n) n + 1) clearly is CF, but not equal to (f N) for any N.

A consequence of the above is that there exists NO computer algorithm/code that can enumerate [1] all the CF's. This is related to the halting problem.

As Robert can clearly see, there's no mentioning of infinite sets anywhere.

[1] Enumerate, as in the program will eventually produce any CF in finite time - and NOT that it will ever end after having produced all CF's, because all of them would be infinite.

Let (f n) be a CF that, for every input n, returns CF's, who in turn return a computable number for every input m. Then the function (d n) := ((f n) n + 1) clearly is CF, but not equal to (f N) for any N.

A consequence of the above is that there exists NO computer algorithm/code that can enumerate [1] all the CF's. This is related to the halting problem.

As Robert can clearly see, there's no mentioning of infinite sets anywhere.

[1] Enumerate, as in the program will eventually produce any CF in finite time - and NOT that it will ever end after having produced all CF's, because all of them would be infinite.

Last edited by Gofer on Tue May 23, 2017 1:18 pm, edited 1 time in total.

- Gofer
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### Re: Cantor Diagonal Argument disproof

So is Cantors theory a fancy way of telling us we cannot count to infinity?

- Edward Marcus
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### Re: Cantor Diagonal Argument disproof

No, that would be what Robert is really trying to say but can't seem to comprehend that that is all he's saying; and that is not what Cantor's theory says.Edward Marcus wrote:So is Cantors theory a fancy way of telling us we cannot count to infinity?

See my post above for an application!

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