Cantor Diagonal Argument disproof

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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Thu Apr 27, 2017 3:03 pm

robert 46 wrote:The only reason that T is seen as larger than N and uncountable is ...
Because if any portion of T can be counted, you can show that that portion isn't all of T.

However, any infinite list whatsoever will also be missing a monotonically increasing count of sfsc per the sequential Diagonal method.
Then don't use that method, which we keep pointing out is invalid. And you keep ignoring.

"Monotonically increasing" doesn't lead to the conclusion you want, when infinite sets are considered. Because there may be another way to count them, that doesn't have "a monotonically increasing count" of missing elements. For instance, in my last post I described how to "list" your set without missing anything.

Yet if that list is only composed of all sfsc, there can be no missing sfsc. So what gives?
You are the only one who thinks that what is missing is supposed to be an sfsc. And you are wrong that it is supposed to be an sfsc.

What you have been told many times, and choose to ignore, is that any infinite set can be put into a 1:1 correspondence with a strict subset of itself. So the fact that you found a way proves nothing, even if the missing element was supposed to be an sfsc.

Can the set of sfsc be put into a list in principle?
Didn't I just describe how to do it to you? Each natural number describes a unique sfsc, and each sfsc describes a unique natural number. So the set of all sfsc's is inherently a list.

But this has nothing to do with Cantor's proof, and your repetition of this point shows just how ignorant you are of what the proof says.

You keep ignoring that these are the same question.

They are joined.
They are the same.

Yes, this was established long ago per the mirror Diagonal argument applying to integers.
No, it was established long before that, by simple logic.

Cantor is saying that a string of infinite significant characters is intrinsically different from an infinite string of finite significant characters.
No, Cantor doesn't mention "significant characters" at all. Only you do. And you keep repeating the same mistakes about them.

We can accept that.
The royal "we" now?

But it is the argument that every list must be missing an element
No. Wrong. Invalid. Flawed. Why do you keep repeating this?

The flaw is that you keep using "element" inconsistently. But you can't help but know that by now, unless you are a complete moron. You just choose to ignore it.

The argument is that any list of infinite strings can be used to define an infinite string that is not in that list. You keep calling it an "element" without saying what it is an element of. It is an infinite string in T, but not in your sfsc's.[1]

In the case of sfsc, every list is missing elements,
No, any list of strings, including your list of sfsc's, is missing strings that are elements of T. Not elements of your set of sfsc's.[1]

This is impossible to do under the condition of a monotonically increasing count.
No, it isn't, another fact you keep ignoring.

Which are there more of, natural numbers or even natural numbers? If you ask "Which of n=1,2,3,4,5... are even?", there is a monotonically increasing count of natural numbers that are not even. If you ask "Which of e=2*n, where n=1,2,3,4,5..., are even?", there is nothing missing in either the evens e, or the natural numbers n. If you ask "Which of q=4*n, where n=1,2,3,4,5..., are even?", there is a monotonically increasing count of even numbers that you miss, but no natural numbers n that you leave out. GET IT YET?
    YOUR "MONOTONICALLY-INCREASING" ARGUMENT IS IRRELEVANT, BECAUSE WHAT YOU CALL A CONTRADICTION IS A KNOWN, AND CONSISTENT, PROPERTY OF INFINITE SETS.


I have no problem with this: Cantor's Diagonal method never considers the circumstance of a list of sfsc.
Yes, it does. It considers any infinite list of stings that can be counted. Your list of sfsc's can be counted, so it is considered.

If the set of sfsc is put into a list, can the sequential Diagonal method show that the list is not necessarily missing an sfsc element?
The double negative is more obfuscation.
There is only one negative: "not".
"Missing" is also a negative.

If the set of sfsc is put into a list, can the sequential Diagonal method show that the list is necessarily not missing an sfsc element?
Can you read? The point of the proof is not to show that there is an element of S - in your example, an sfsc - that is not in the list. It is to show that there is an element of T that is not in S, all of which is in considered to be in the list.[1]

The only way this could be correct is if the count of missing sfsc gets reduced to 0.
While still irrelevant, this is wrong. The list of missing natural numbers, when you count by two's, is never reduced to zero. Yet N and E are the same size.

You say Diagonalization can't show that an element which properly belongs to the listed set is missing.
I never said that. You said that, and I said it was the wrong point to make each time you said it.[1]

The set of all sfsc is countable. It isn't the set that Cantor proves is uncountable. And no amount of ignoring this fault in your argument will make it go away, or make your argument valid.
It is not the set that Cantor "proves" is uncountable, but it is the set which the sequential method would "prove" was uncountable if such conclusion was generically valid.
No, the proof still requires that you show that the missing element is in T, not that you show it is in S. Even if your "sequential method" were correct (you still ignore the flaws we point out in it), the listed set continues to NOT BE the set that is proven to be uncountable.[1]

+++++

And just how do they have a "broader vision" under the circumstances that they never thought about the sfsc situation?
They recognize what you ignore, and have the intelligence to see that your comparison has no relevance. So there is no need to mention it.

An example is that you can create a comparison between N and E such that:
  1. Every element of E is "found" but a monotonically-increasing number of elements of N are missing from the comparison. Just divide N into evens and odds.
  2. Every element of E is "found" and matched with every element of N, so nothing is missing. Just say e=2n.
  3. A monotonically-increasing number of elements of E are "missing" but every every element of N is used. Just say q=4n.

+++++

[1] You can ignore this all you want, but it is still the flaw in your argument. And it is the same flaw each place I use this reference.
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Re: Cantor Diagonal Argument disproof

Postby robert 46 » Thu Apr 27, 2017 9:00 pm

Gofer wrote:
robert 46 wrote:Then why should a missing ~D, which cannot actually be produced compared to the monotonically increasing count of missing sfsc which can be produced, be interpreted as meaning that the set of all strings of infinite characters is an uncountable infinite set?

Ever hear the term 'complex question' or loaded question? Look it up, and rephrase!
You have used the fallacy of quoting-out-of-context.

Whereas neither the Cantorian nor sequential Diagonal methods can show that no sfsc elements are missing from the list of sfsc, and whereas a missing element under all circumstances is an indicator that not all elements of the set can be put into the list, but whereas missing elements do not imply that the set of sfsc is an uncountable infinite, then why should a missing ~D imply that the set of infinite strings is an uncountable infinite?
In both cases it is always determined that the list is missing one or more elements.
No matter what's in the list, the anti-diagonal will never be part of it, particularly for the set of infinite strings with finite 1's (what you call sfsc), where the anti-diagonal will not have that property, and hence not be part of the list.

You are mixing contexts. I have stated that ~D is irrelevant to the sfsc context: all that is relevant is the monotonically increasing count of missing sfsc.

*****

JeffJo wrote:N is countable, trivially. You can write each element of N as a unique finite binary string, that you called an sfsc; and each sfsc as a unique, and still finite, integer. This defines a 1:1 correspondence (that is, a list) between N and your set of sfsc. You can pad each such string with infinite 0's, to make infinite strings.
Each natural number describes a unique sfsc, and each sfsc describes a unique natural number. So the set of all sfsc's is inherently a list.

Assume that the list of sfsc is contructed in order: i.e. from 0 significant bits to increasing significant bits. After examining n elements, n different missing sfsc have been created. Each of them uniquely has n significant characters, n increasing from 1. There are 2^n elements up to those elements having n significant characters. Thus, a missing element of n significant characters will be found in the list after 2^n elements have been examined. But after 2^n elements are examined there are 2^n-n net missing elements which have not been found. The count of missing elements is monotonically increasing notwithstanding that any missing element will eventually be found in the continuing examination.

When are all the missing elements found? When the count of missing elements becomes 0. However, this never happens in the sequential examination. Thus, for sequential examination, inductive inference to infinity is invalid: i.e. the examination process never ends.

But what if we do consider inductive inference to infinity as a hypothesis? This would require 2^infinity-infinity=0 (the net count of missing sfsc is reduced to 0 because all of them have been found after a "complete" examination) and ~D would then be produced. However, if 2^infinity-infinity=0 then simply 2^infinity=infinity. This means that the cardinality of T is the same as the cardinality of N because T has 2^infinity strings.

Conclusion: If inductive inference to infinity is accepted then ~D is produced but the cardinality of T is the same as N. If inductive inference to infinity is rejected then ~D is not produced. There is no circumstance under the sequential method where ~D is produced and the cardinality of T is greater than the cardinality of N.
Last edited by robert 46 on Fri Apr 28, 2017 6:05 am, edited 1 time in total.
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Re: Cantor Diagonal Argument disproof

Postby Gofer » Fri Apr 28, 2017 5:28 am

let S be the statement that Robert asks loaded questions and collected boogers as a young boy, in:

> You have used the fallacy of quoting-out-of-context.

Whereas S, there was no 'out of context' quoting.

> Whereas neither the Cantorian ....

Whereas S, S minus the boogers, yet again.

Whereas S, I find it hilarious that Robert says "I have stated that ~D is irrelevant to the sfsc context", yet just before asked a loaded question that seems to be using ~D to imply a deduction supporting his position.

> all that is relevant is the monotonically increasing count of missing sfsc.

Whereas S, there are no missing sfsc. Whereas S, every sfsc "is eventually" assigned a unique counting number by the algorithm.
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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Fri Apr 28, 2017 10:41 am

robert 46 wrote:Assume that the list of sfsc is contructed in order:
The list is still a list of finite strings, padded with 0's. And it can be defined.

After examining n elements, n different missing sfsc have been created.
Say n=5. You have "examined" - an irrelevant procedure that you won't stop referring to - the strings {(0), 1(0), 01(0), 11(0), 101(0)}. "Missing" is an ambiguous term, because you haven't said what it is missing from. And you refuse to acknowledge that it is missing from a set defined by a procedure that uses a different definition of what "belongs," than what defines what you examined. So the comparison is completely irrelevant.

And in this particular case, the only thing it is "missing from" is what you still need to "examine" because you haven't reached it yet. THIS PROVES NOTHING, because you can define how to both list, and "examine," strings in a consistent way. Every sfsc will be included in the list and in what you examine, with nothing "missing."

You can even try to follow your ill-defined suggestion[1]:

i.e. from 0 significant bits to increasing significant bits.

... which, corrected, is to examine n strings from a list made of strings that have m (not n) significant digits ENDING IN 1, for all m in N (with 0 added), like this:

{(0)}
{1(0)}
{01(0),11(0)}
{001(0),011(0),101(0),111(0)}
etc.

But you keep insisting that you have to set n=m[2], which is why you think things are missing.

But every string still has a proper place, and is not "missing" from the examined portion when you use the same definition for what to examined, as you used for what you listed. So look at n=2^m, and you get every string.

If you use different definitions for what you list, than what you “examine”, anything that is “missing” is the fault of using two definitions. I can do that to: Look at the set of natural numbers, listed by 2's. The number of "missing" odd numbers grows monotonically, but that doesn't mean they can't be found if you would take your head out of your *** and use the same method to list them, as to "examine" them.

There are 2^n elements up to those elements having n significant characters.
Then stop ignoring the fact that such a comparison is irrelevant, and what you think are “missing” elements are because you use different definitions.

When are all the missing elements found?
They don’t have to be. Tell me if you have heard this before: A property that is required for infinite sets is that the entire set can be put into a 1:1 correspondences with a strict subset of itself. All you have done, is find an example of how. I can, too:

If A=B=N, there is a listing of elements in A, as indexed by all of B, that is missing a monotonically-increasing number of elements of A. It is defined by a(b)=2*b.

And there is a listing of elements in B, as indexed by all of A, that is missing a monotonically-increasing number of elements of B. It is defined by b(a)=2*a.

The only important comparison, is that there is also a listing of elements in A, as indexed by all of B, that is not missing any elements of A. It is defined by a(b)=b.[4]

But what if we do consider inductive inference to infinity as a hypothesis?
To do so, you would have to first try to understand what inductive inference is, which you refuse to do. It defines N, but not the listing of the S made from all sfsc's. That is defined by the fact that N is defined, and a unique sfsc can be defined for every n, in a way that defines every possible sfsc.

This would require 2^infinity-infinity=0
No. It can be done is a way that appears that way to you, but only because you refuse to look at it any other way. There are other ways you ignore that don't. And how to do so has been explained many times, so you can’t have missed it unless you refuse to see it, or are a complete moron.

Conclusion:
You refuse to understand the fundamental principles required for infinite sets, so you keep seeing them as contradictions. Despite having the resolution for the contradictions explained over, and over, and over, and… (extend to infinity, it seems).

There is no circumstance under the sequential method where ~D is produced and the cardinality of T is greater than the cardinality of N.
There is a well-defined method, that you have been shown many times, that defines ~D completely.[3] It shows that if S is a set infinite binary strings that can be indexed (listed), then it is not the set of all infinite binary strings T. Since this proves that T can't be indexed by N, it is also proven that T must have greater cardinality than N.

+++++
Edited to point out where I addressed what robert claimed I ignored,
Last edited by JeffJo on Fri Apr 28, 2017 12:57 pm, edited 1 time in total.
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Re: Cantor Diagonal Argument disproof

Postby robert 46 » Fri Apr 28, 2017 12:17 pm

Here is my comment, and highlighted in blue is all that JeffJo ingnored:
robert 46 wrote:Assume that the list of sfsc is contructed in order: i.e. from 0 significant bits to increasing significant bits. After examining n elements, n different missing sfsc have been created. Each of them uniquely has n significant characters, n increasing from 1. There are 2^n elements up to those elements having n significant characters. Thus, a missing element of n significant characters will be found in the list after 2^n elements have been examined. But after 2^n elements are examined there are 2^n-n net missing elements which have not been found. The count of missing elements is monotonically increasing notwithstanding that any missing element will eventually be found in the continuing examination.

When are all the missing elements found? When the count of missing elements becomes 0. However, this never happens in the sequential examination. Thus, for sequential examination, inductive inference to infinity is invalid: i.e. the examination process never ends.

But what if we do consider inductive inference to infinity as a hypothesis? This would require 2^infinity-infinity=0 (the net count of missing sfsc is reduced to 0 because all of them have been found after a "complete" examination) and ~D would then be produced. However, if 2^infinity-infinity=0 then simply 2^infinity=infinity. This means that the cardinality of T is the same as the cardinality of N because T has 2^infinity strings.

Conclusion: If inductive inference to infinity is accepted then ~D is produced but the cardinality of T is the same as N. If inductive inference to infinity is rejected then ~D is not produced. There is no circumstance under the sequential method where ~D is produced and the cardinality of T is greater than the cardinality of N.

This summary dismissal by quoting-out-of-context and slanting which JeffJo characteristically uses is beyond the pale.

My argument was: An analysis of the sequential method shows that inductive inference to the infinite is invalid.

If inductive inference to the infinite is generically invalid then Cantor's Diagonal method is patently invalid nonsense. The mathematicians have defined inductive inference to the infinite to be a valid method without actually establishing that it is anything more than silly fantasy.
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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Fri Apr 28, 2017 1:03 pm

robert 46 wrote:Here is my comment, and highlighted in blue is all that JeffJo ingnored:
robert 46 wrote:Assume that the list of sfsc is contructed in order: i.e. from 0 significant bits to increasing significant bits.[1] After examining n elements, n different missing sfsc have been created. Each of them uniquely has n significant characters, n increasing from 1.[2] There are 2^n elements up to those elements having n significant characters. Thus, a missing element of n significant characters will be found in the list after 2^n elements have been examined[3]. But after 2^n elements are examined there are 2^n-n net missing elements which have not been found. The count of missing elements is monotonically increasing notwithstanding that any missing element will eventually be found in the continuing examination.[4]

When are all the missing elements found? When the count of missing elements becomes 0. However, this never happens in the sequential examination. Thus, for sequential examination, inductive inference to infinity is invalid: i.e. the examination process never ends.

But what if we do consider inductive inference to infinity as a hypothesis? This would require 2^infinity-infinity=0 (the net count of missing sfsc is reduced to 0 because all of them have been found after a "complete" examination) and ~D would then be produced. However, if 2^infinity-infinity=0 then simply 2^infinity=infinity. This means that the cardinality of T is the same as the cardinality of N because T has 2^infinity strings.

Conclusion: If inductive inference to infinity is accepted then ~D is produced but the cardinality of T is the same as N. If inductive inference to infinity is rejected then ~D is not produced.[5] There is no circumstance under the sequential method where ~D is produced and the cardinality of T is greater than the cardinality of N.

This summary dismissal by quoting-out-of-context and slanting which JeffJo characteristically uses is beyond the pale.

My argument was: An analysis of the sequential method shows that inductive inference to the infinite is invalid.

If inductive inference to the infinite is generically invalid then Cantor's Diagonal method is patently invalid nonsense. The mathematicians have defined inductive inference to the infinite to be a valid method without actually establishing that it is anything more than silly fantasy.


What robert ignores is how all of this - even if you ignore how I addressed what robert said I ignored, and so accept his incorrect analysis - has been shown to be irrelevant. With infinite sets, you can always make a listing that seems, as per robert's flawed analysis, to miss some. Even a monotonically increasing set of missed ones. I even gave examples, which robert ignored.

The point robert can't bring himself to address is: Infinite sets are said to have the same cardinality if, and only if, it is possible to put them in a 1:1 correspondence.

It is a known fact that you will also be able to put them in mismatched correspondences, so each time robert repeats how he thinks that he has found a mismatched one (he hasn't, he just counts differently in the sets being compared), he is just demonstrating his ignorance.
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Re: Cantor Diagonal Argument disproof

Postby Gofer » Fri Apr 28, 2017 1:18 pm

> The point robert can't bring himself to address is: Infinite sets are said to have the same cardinality if, and only if, it is possible to put them in a 1:1 correspondence.

And that is all that needs being said on the subject. If Robert can't argue from the definition, he has no business representing Cantor.

Robert's definition instead seems to be more like:

Infinite sets are said to NOT have the same cardinality if, and only if, it is possible to NOT put them in a 1:1 correspondence.
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Re: Cantor Diagonal Argument disproof

Postby robert 46 » Fri Apr 28, 2017 6:08 pm

Do either of you understand that inductive inference to the infinite cannot be applied to the sequential Diagonal method?
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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Sat Apr 29, 2017 7:00 am

robert 46 wrote:Do either of you understand that inductive inference to the infinite cannot be applied to the sequential Diagonal method?
Do you understand that inductive inference isn't used to produce the diagonal in Cantor's Diagonal Proof? Or anything having to do with Cantor's Diagonal Proof? It seems absurd that you could not, since I have been telling you so over and over. Oh, wait, you ignore those statements, don't you? And like the ostrich with its head in the sane, you think that makes the argument go away. (And this is not "posturing," it is demonstrated fact.)

I have no idea what you think "inductive inference" means. There are several meanings, in different applications, but none of them match your implied meaning. In general, it means making a broad conclusion from specific examples. But there, it is not recognized as a form of proof, just an implication.

Mathematical Induction is something very different. It starts with an example at the start of a sequence, and actually proves that a property must be carried over from any member of the sequence to the next. Even "to the infinite," despite your unfounded assertion that it can't. And, it is considered to be a form of direct proof. Even "to the infinite."

In the Axiom of Infinity, it defines a sequence of numbers that has no end, yet every possible number is defined by Mathematical Induction (not "inductive inference"). The point of the Axiom is not that every number is defined, which is already established by Mathematical Induction, but that it defines a set we call N.

Since N is defined, and you can define a function on natural numbers that produces either the character "0" or "1", infinite binary strings are also defined. Not by "inductive inference," but by direct definition based on previous definitions. If you have an infinite set of such strings in a list, Cantor's Diagonal Method is just another function that defines another infinite string.

So if infinite strings can be defined in your fantasy world, as you assume in your "sequential method," then Cantor's Diagonal Method defines one. Not by "inductive inference," regardless of what you think it means, but by applying the same kind of definition you used to define the strings in your set of sfsc's.

So, maybe you should supply your definition of "inductive inference." Or address any of the points in my posts that you ignore, like how your "monotonically increasing" sets of missing elements are irrelevant. Or admit that I haven't ignored anything you said.

Oh, wait, then you'd have to actually say what your argument is, and expose all of its flaws. And see how it is proven wrong. Silly me, thinking you'd ever be so honest. Never mind.

+++++

To repeat: A property that is required of infinite sets, is that they can be put into 1:1 correspondence with a strict subset of themselves. What that means to your argument is summarized in these examples of lists:

L1(1)=2, L1(2)=4, L1(3)=6, ..., L1(n)=2*n.
L2(1)=1/2, L2(2)=1, L2(3)=3/2, ..., L2(n)=n/2.
L3(1)=1, L3(2)=2, L3(3)=3, ..., L3(n)=n.

In L1, there is a monotonically-increasing set of natural numbers that will never appear in the list. In L2, there is a monotonically-increasing set of natural numbers are not matched with other natural numbers in the list. In L3, every natural number is matched with a unique natural number in the list.

The cardinality of an infinite set that includes an entire class of objects is not determined by 1:1 correspondences (with N, that means a list) that miss some members of the class, it is determined by whether you can make a correspondence that doesn't.

You can make a 1:1 correspondence between your sfsc's, and N, so it is a countably infinite set. And it still doesn't include a single string of infinite significant characters.

Since you can make such a list, you can define a function that makes an infinite string not in your list. It will have infinite significant characters.
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Re: Cantor Diagonal Argument disproof

Postby Gofer » Sat Apr 29, 2017 8:53 am

robert 46 wrote:Do either of you understand that inductive inference to the infinite cannot be applied to the sequential Diagonal method?
One doesn't have to go to the "infinite", but merely show that the algorithm reaches every number in the sequence.

Consider for example the Fibonacci sequence defined by f(n+1)=f(n)+f(n-1) for starting values 1 and 1; it is defined for every n, and even has an explicit formula of n. And some "functions" can only be stated as a recurrence relation, as there's no known explicit formula, but doesn't make them any less valid. Check out the "logistic map".

In this case though, Cantor's method isn't even sequential, since an element of the anti-diagonal can be calculated irrespective of the other.

----

According to the definition of countability, one only needs to find an injective map from the sfsc to N. The function g suffices:

g≝1+∑(xn*2^(n-1)) from n=1 to N

if we define a sfsc as a pair (x,N) where x is the sequence and N is the maximum position having a 1.

The anti-diagonal of the inverse of g is (1), the string of infinite 1's, which obviously is NOT a sfsc.
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Re: Cantor Diagonal Argument disproof

Postby robert 46 » Sat Apr 29, 2017 12:21 pm

Obviously, neither of you understand that inductive inference to the infinite cannot be applied to the sequential Diagonal method.

Induction means to spot and extend a trend. Inference is a method to reach a conclusion based on examples.

The inductive property of the sequential Diagonal method is that the count of missing sfsc after examining 2^n elements in the list is 2^n-n. As anyone can see, this count is monotonically increasing. However, all elements still missing after examining n elements can be found after 2^n elements have been examined. Thus the trend is that any missing element is eventually found if enough elements are examined.

Clearly, all the missing sfsc must be found before ~D can be produced because ~D has more significant characters (infinite) than any of the missing sfsc, and missing sfsc are produced in order of an increasing number of significant characters.

Not all of the missing sfsc can be found because for any number which have been found there are more yet to find.

Thus it is an invalid inference to conclude that the sequential Diagonal method can produce the missing ~D, which has infinite significant characters, because the sequential method only endlessly produces missing sfsc. One cannot use induction to infer that ~D would be produced by the sequential method. The endlessness of the sequential method precludes the production of ~D.

Whereas there is no route to the production of ~D by the sequential method, there is no reason to believe in the miraculous existence of ~D by the non-sequential Cantorian method. All the Cantorian method does is attempt to define the existence of ~D based on the belief in inductive inference to the infinite, which I have just explained is unjustified.
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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Sun Apr 30, 2017 11:01 am

robert 46 wrote:Obviously, neither of you understand that inductive inference to the infinite cannot be applied to the sequential Diagonal method.
Obviously, you think you are so privileged that you get to change the phrases others use to ones you prefer. And then supply meanings for what you changed that don't match what was said, in any way.

Induction means to spot and extend a trend.
Merriam-Webster wrote:Induction: 2a(1): inference of a generalized conclusion from particular instances.
Inference is a method to reach a conclusion based on examples.
As you should be able to see, "inference" is part of the definition of "induction." So when you add it, it is redundant. But your definition is not used - except by you - for anything having to do with Cantor Diagonalization. Even in your "sequential diagonalization," you seem to think you are using the more rigorous Math definition, when you are just making unfounded assertions.

Merriam-Webster wrote:2b: mathematical demonstration of the validity of a law concerning all the positive integers by proving that it holds for the integer 1 and that if it holds for an arbitrarily chosen positive integer k, it must hold for the integer k + 1 — called also mathematical induction
This is what "induction" should mean in any context related to Cantor Diagonalization, and it is not how you are using it.

The inductive property of the sequential Diagonal method is that the count of missing sfsc after examining 2^n elements in the list is 2^n-n.
Correction: your "sequential Diagonal method" makes two lists, using two integers m and n. It makes a list of the m=2^n sfsc's of n significant characters, and another of n<m sfsc's picked from it by unknown means. Nothing is "missing," just defined indifferent lists. Since there are other ways to list them, there is no reason to infer your "specific instance" property has any bearing on anything more general than this irrelevant comparison.

The same "inductive property" you use, after looking at the sfsc's when they are listed in an appropriate way by representing integers in binary, say there are never any missing elements of your list. Now divide the integers by some m^2 first, where m is a different function of n than what you used. Throw out the fractional portions, and guess what happens? The number of integers that produce duplicate sfsc's grows at about the same rate as your missing ones.

So obviously, there is something wrong here with the way you generalize from the specific to the general. And repeating it over, and over, and over, while ignoring the flaws doesn't help your argument.

So, go back and address all of the points you ignored, about how this is all irrelevant.

As anyone can see, this count is monotonically increasing.
And as anyone who actually looks can see, there are others ways that aren't, so this is irrelevant.

However, all elements still missing after examining n elements can be found after 2^n elements have been examined. Thus the trend is that any missing element is eventually found if enough elements are examined.
But since the list is endless, Mathematical Induction (as opposed to your invalid inferences that you claim use induction), every sfsc is eventually found.

Now, go back and define what you mean by "element" so that you can see why all this is irrelevant. "Element" of what? The sfsc's? You hit them all, even in your sequential method. As you are supposed to. Cantor's T? It contains strings that are not represented by sfsc's, and Cantor Diagonalization defines them.

Clearly, all the missing sfsc must be found before ~D can be produced because ~D has more significant characters (infinite) than any of the missing sfsc, and missing sfsc are produced in order of an increasing number of significant characters.
Please, go back and try to understand all of the properties of infinite sets that you purposely (so you can make this invalid claim) ignore.
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Re: Cantor Diagonal Argument disproof

Postby Gofer » Sun Apr 30, 2017 2:41 pm

Induction means to spot and extend a trend. Inference is a method to reach a conclusion based on examples.

The inductive property of the sequential Diagonal method [DM] is that the count of missing sfsc after examining 2^n elements in the list is 2^n-n.
There are no "missing sfsc" unless you define them to be, which you do. Besides, the DM isn't even looking at rows, but elements, making your "missing count" even larger, 2^n-1 to be exact, -1 for the formation of the anti-diagonal itself.

Clearly, all the missing sfsc must be found before ~D can be produced because ~D has more significant characters (infinite) than any of the missing sfsc, and missing sfsc are produced in order of an increasing number of significant characters.

Not all of the missing sfsc can be found because for any number which have been found there are more yet to find.

Thus it is an invalid inference to conclude that the sequential Diagonal method can produce the missing ~D, which has infinite significant characters, because the sequential method only endlessly produces missing sfsc.
How lucky for us that Cantor's method isn't sequential then, but produces ALL elements of the anti-diagonal all at once. So when those "missing sfsc" appear, they are immediately "retired", because the elements of the anti-diagonal corresponding to those entries are already calculated.

But since Robert flaunts the term "inductive inference", let's see what it really means:

Assume the "sequential" DM works for some integer n, meaning that the constructed anti-diagonal of n elements can't equal the first n elements of the first n rows. Let's add to it, element n+1, consisting of the flipped bit at position n+1 of string n+1. It is clear that the anti-diagonal of length n+1 thus far can't equal row n+1. By induction, the same property must hold for all n's.
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Re: Cantor Diagonal Argument disproof

Postby robert 46 » Sun Apr 30, 2017 7:20 pm

[Charles Sanders] Peirce began to hold that there were two utterly distinct classes of probable inferences, which he referred to as inductive inferences and abductive inferences.

Therefore the adjective "inductive" is a legitimate qualifier for a type of "inference". So much for JeffJo's characteristic nitpicking.

*****
Gofer wrote:
robert 46 wrote:The inductive property of the sequential Diagonal method [DM] is that the count of missing sfsc after examining 2^n elements in the list is 2^n-n.
There are no "missing sfsc" unless you define them to be, which you do.
The "missing sfsc" are those intermediate anti-diagonals of finite significant characters which are produced by the sequential Diagonal method. When produced they are missing from the list so far examined.
Clearly, all the missing sfsc must be found before ~D can be produced because ~D has more significant characters (infinite) than any of the missing sfsc, and missing sfsc are produced in order of an increasing number of significant characters.

Not all of the missing sfsc can be found because for any number which have been found there are more yet to find.

Thus it is an invalid inference to conclude that the sequential Diagonal method can produce the missing ~D, which has infinite significant characters, because the sequential method only endlessly produces missing sfsc.
How lucky for us that Cantor's method isn't sequential then, but produces ALL elements of the anti-diagonal all at once.

It can only be defined to do this: it is not demonstrable.
But since Robert flaunts the term "inductive inference", let's see what it really means:

Assume the "sequential" DM works for some integer n, meaning that the constructed anti-diagonal of n elements can't equal the first n elements of the first n rows. Let's add to it, element n+1, consisting of the flipped bit at position n+1 of string n+1. It is clear that the anti-diagonal of length n+1 thus far can't equal row n+1. By induction, the same property must hold for all n's.

All this says is that the intermediate anti-diagonal produced is missing from all the rows so far examined. It will be found in the remainder of the list if enough rows are examined. Yet for any missing elements ultimately found there are more to find- infinitely more because the remainder of the list is always infinite. It is impossible to logically argue that the sequential method which only produces intermediate anti-diagonals of finite significant characters can produce the single anti-diagonal, ~D, of infinite significant characters.
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Re: Cantor Diagonal Argument disproof

Postby JeffJo » Mon May 01, 2017 7:25 am

robert 46 wrote:
[Charles Sanders] Peirce began to hold that there were two utterly distinct classes of probable inferences, which he referred to as inductive inferences and abductive inferences.
Therefore the adjective "inductive" is a legitimate qualifier for a type of "inference".
Too bad poor robert is so desperate to legitimatize his usage, that he can't (A) reference where he got this quote, (B) recognize that what I said was that induction was a kind of inference, but (C) that it is different from mathematical Induction, which was my entire point, and (D) the only reason to use "inductive" as a modifier is to compare it to other kinds of inference, as this "Charles Sanders Peirce" apparently does.

So much for robert's characteristic form of evasion, by changing the subject.

+++++

Make list L1 by listing the natural numbers n, in order. Make list L2 by listing all of the natural numbers m, as a function of an increasing n, such that 2^(n-1)<=m<2^n.

For the same n, there are always 2^n-n+1 more elements in L2, than in L1. Why robert thinks this (or something very similar, which I don't care to track down) is of significance, I have no idea.

Both lists are endless/infinite, and can define everything that can be called a natural number.
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