## Cantor Diagonal Argument disproof

**Moderators:** mvs_staff, forum_admin, Marilyn

### Re: Cantor Diagonal Argument disproof

I have no interest in addressing anything Gofer says at the level of arcane mathematical formalisms. That he will not explain himself in plain English is merely to cloak himself in obscurantism.

It appears to me that the point he is trying to make is that ~D is not a member of the complete list of sfsc. This is irrelevant. I stated that S is preloaded with all sfsc. The Diagonal method cannot examine all of S to show that it is complete. Whereas all sfsc comprise a countable infinite, the Diagonal method cannot tell us that a countable infinite list is complete consequent to the monotonically increasing count of missing sfsc.

At an infinite limit there would remain a positive count of missing sfsc, and at this terminus ~D would be produced. However, not all of S has been examined, nor is it examinable, to find the missing sfsc which must be in the list somewhere. Being restricted to only examining a part of a list, which is intrinsic to the Diagonal method, precludes discovering whether the list is complete.

However, we know that the list of sfsc is complete because we know it is a countable infinite. That the Diagonal method fails to inform us that a countable infinite list is complete, tells us that the method is useless for determining the completeness of a list hypothetically containing the superset. Whereas the missing sfsc must be in the remainder of S which the Diagonal method cannot examine, there is no reason to believe that ~D would not also be in the remainder of a list containing both all sfsc and the hypothesized remainder of the superset.

The Diagonal method fails to confirm that the list of sfsc is complete, and that is the touchstone test which identifies the utter worthlessness of the Diagonal method.

It appears to me that the point he is trying to make is that ~D is not a member of the complete list of sfsc. This is irrelevant. I stated that S is preloaded with all sfsc. The Diagonal method cannot examine all of S to show that it is complete. Whereas all sfsc comprise a countable infinite, the Diagonal method cannot tell us that a countable infinite list is complete consequent to the monotonically increasing count of missing sfsc.

At an infinite limit there would remain a positive count of missing sfsc, and at this terminus ~D would be produced. However, not all of S has been examined, nor is it examinable, to find the missing sfsc which must be in the list somewhere. Being restricted to only examining a part of a list, which is intrinsic to the Diagonal method, precludes discovering whether the list is complete.

However, we know that the list of sfsc is complete because we know it is a countable infinite. That the Diagonal method fails to inform us that a countable infinite list is complete, tells us that the method is useless for determining the completeness of a list hypothetically containing the superset. Whereas the missing sfsc must be in the remainder of S which the Diagonal method cannot examine, there is no reason to believe that ~D would not also be in the remainder of a list containing both all sfsc and the hypothesized remainder of the superset.

The Diagonal method fails to confirm that the list of sfsc is complete, and that is the touchstone test which identifies the utter worthlessness of the Diagonal method.

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

### Re: Cantor Diagonal Argument disproof

erratum: the function I wrote in my previous post should be:

(g n m)≝{((n-1) mod 2) for m<2, 0 for n<2, (g n/2 (m-1)) for n even, (g (n+1)/2 (m-1)) for n odd}, where 'mod' is the modulus. 'a mod b' means the remainder of a after division by b.

Say we wish to calculate (g 6 3), which should equal the third bit of the string 101(0), the evaluation proceeds as follows:

(g 6 3) = (g (6/2) (3-1)) = (g 3 2) = (g (3+1)/2 (2-1)) = (g 2 1) = (2-1) mod 2 = 1. So g is recursive but will always terminate for any input n and m.

----

Robert keeps playing the same 'out of tune' song on the same broken record on the same broken player.

Btw, the anti-diagonal for the function g I just wrote down, is (1), the string of infinite 1's.

You still haven't explained how I could create the function f in my post, without "going to infinity" or "creating intermediate strings" and so on.

(g n m)≝{((n-1) mod 2) for m<2, 0 for n<2, (g n/2 (m-1)) for n even, (g (n+1)/2 (m-1)) for n odd}, where 'mod' is the modulus. 'a mod b' means the remainder of a after division by b.

Say we wish to calculate (g 6 3), which should equal the third bit of the string 101(0), the evaluation proceeds as follows:

(g 6 3) = (g (6/2) (3-1)) = (g 3 2) = (g (3+1)/2 (2-1)) = (g 2 1) = (2-1) mod 2 = 1. So g is recursive but will always terminate for any input n and m.

----

Robert keeps playing the same 'out of tune' song on the same broken record on the same broken player.

And he cries because he doesn't know how to search the internet for "list of mathematical symbols" and read about it, or can't be bothered to ask me to clarify details in my post.robert 46 wrote:I have no interest in addressing anything Gofer says at the level of arcane mathematical formalisms. That he will not explain himself in plain English is merely to cloak himself in obscurantism.

So Cantor's Lemma is irrelevant then? A question for you: notwithstanding Cantor's proof of it, is the Lemma true?It appears to me that the point he is trying to make is that ~D is not a member of the complete list of sfsc. This is irrelevant.

I assume by "preloaded", S contains all of sfsc?I stated that S is preloaded with all sfsc.

It doesn't have to! All it does is "pointing out" that the anti-diagonal can't equal any row of the list; and this has nothing to do with "intermediate strings", "the constructed string thus far" or "increasing count" or whatever, thus rebutting the rest of your post.The Diagonal method cannot examine all of S to show that it is complete.

Btw, the anti-diagonal for the function g I just wrote down, is (1), the string of infinite 1's.

You still haven't explained how I could create the function f in my post, without "going to infinity" or "creating intermediate strings" and so on.

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:I assume by "preloaded", S contains all of sfsc?robert 46 wrote:I stated that S is preloaded with all sfsc.

S could start out with anything- it could even be empty, or contain all of T. But for the point I wanted to make, assume it has all sfsc.

It doesn't have to! All it does is "pointing out" that the anti-diagonal can't equal any row of the list; and this has nothing to do with "intermediate strings", "the constructed string thus far" or "increasing count" or whatever, thus rebutting the rest of your post.The Diagonal method cannot examine all of S to show that it is complete.

Nonsense. All you are doing is making a statement that all I have said to analyze the matter is contrary to mathematical scripture, and should be rejected on the grounds of faith in the beliefs you have been indoctrinated into. You have a closed mind. Yet note that after over two and a half years of arguing this with JeffJo, he is notably silent of late. Perhaps his mind has finally opened to the correctness of what I have said, or leastwise he can't find a retort which wouldn't look foolish.

Btw, the anti-diagonal for the function g I just wrote down, is (1), the string of infinite 1's.

Assume (0) is the first element in S. The associated missing sfsc is 1(0). Assume 1(0) is the second element in S. The associated missing sfsc is 11(0). Assume 11(0) is the third element in the list. The associated missing sfsc is 111(0). Now assume that every constructed missing sfsc is the subsequent element in the list. Are there elements missing from the list? Well, where are 01(0), 001(0), 101(0), 0001(0), 0101(0), 1001(0), 1101(0), etc? These elements remain missing because the entire examinable length of S is composed of (0) and all instances of 1...1(0). Thus there are elements missing from the examinable portion of S no matter whether each constructed missing element can be found in the list or not. So what do you think the value is of a method which cannot establish that a known countable infinite list is complete, but instead tells us that it is missing elements under all circumstances?

If your answer is not "The method is worthless" then you have been taken in by an abiding belief in the infallibility of mathematical scripture.

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

### Re: Cantor Diagonal Argument disproof

Robert refuses to see that his "proof" has nothing to do with what Cantor does. My response in [] in the quote below.

Assume (0) is the first element in S [OK] .The associated missing sfsc is 1(0) [Meaningless statement!]. Assume 1(0) is the second element in S [OK]. The associated missing sfsc is 11(0) [Meaningless statement!]. Assume 11(0) is the third element in the list [OK]. The associated missing sfsc is 111(0) [Meaningless statement!]. Now assume that every constructed missing sfsc is the subsequent element in the list [Meaningless statement!]. Are there elements missing from the list? [Nope, because it looks like you are simply enumerating the set of strings having finite number of consecutive 1's followed by infinite number of zeroes.] Well, where are 01(0), 001(0), 101(0), 0001(0), 0101(0), 1001(0), 1101(0), etc? These elements remain missing because the entire examinable length of S is composed of (0) and all instances of 1...1(0) [See previous comment]. Thus there are elements missing from the examinable portion of S no matter whether each constructed missing element can be found in the list or not [Meaningless statement!]. So what do you think the value is of a method which cannot establish that a known countable infinite list is complete, but instead tells us that it is missing elements under all circumstances? [I think it's great; it does what it's suppose to do, namely construct the anti-diagonal, and not "establish" whether the list is "complete", whatever that means.]

If your answer is not "The method is worthless" then you have been taken in by an abiding belief in the infallibility of mathematical scripture. [More like you have been taken in by an abiding belief that your method has any relevancy to Cantor's proof.]

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:Robert refuses to see that his "proof" has nothing to do with what Cantor does. My response in [] in the quote below.

robert 46 wrote:Assume (0) is the first element in S [OK] .The associated missing sfsc is 1(0) [Meaningless statement!].

Take the nth character of the nth element in the list and negate it, and put it in the nth position of the developing missing element, pad with (0) on the right; this element is missing from the list so far traversed. (0) is the first element in the list. Applying the above, we get 1(0).

1(0) is the second element in the list. Apply the above, and we get 11(0) as the element currently missing from the list so far traversed. Repeat to generate 1...1(0) as the element currently missing after the examination of element n (where 1...1 is n 1s in succession).

Now assume that every constructed missing sfsc is the subsequent element in the list [Meaningless statement!].

I am hypothesizing that the list is constructed such that each intermediate missing element at row n is the next element found in the list at n+1.

Are there elements missing from the list? [Nope, because it looks like you are simply enumerating the set of strings having finite number of consecutive 1's followed by infinite number of zeroes.]

I am demonstrating that at every row n an element is produced which is missing from the list so far examined, but that the list has been previously constructed to have this missing element at row n+1. Clearly, there is no missing element at row infinity which would be found at row infinity+1 because there is no row infinity.

Well, where are 01(0), 001(0), 101(0), 0001(0), 0101(0), 1001(0), 1101(0), etc? These elements remain missing because the entire examinable length of S is composed of (0) and all instances of 1...1(0) [See previous comment].

Your previous comment shows your utter lack of understanding of the argument I am developing.

Thus there are elements missing from the examinable portion of S no matter whether each constructed missing element can be found in the list or not [Meaningless statement!].

Your comment is clueless.

So what do you think the value is of a method which cannot establish that a known countable infinite list is complete, but instead tells us that it is missing elements under all circumstances? [I think it's great; it does what it's suppose to do, namely construct the anti-diagonal, and not "establish" whether the list is "complete", whatever that means.]

It means that the elements in the set are countable if the list is complete. The Diagonal method cannot show that a known countable infinite list is complete: therefore the implication is that a known countable infinite set is an uncountable infinite set, but this is a blatant contradiction.

*****

If the list of sfsc is constructed for the purpose of preventing the monotonically increasing count of missing sfsc elements then not all of the set can be put into the list- all effort is expended in retiring the missing elements. If, however, the list of sfsc is constructed by the method I gave which produces them in order, then there is guaranteed to be a monotonically increasing count of missing sfsc elements.

If the Diagonal method does not produce a monotonically increasing count of missing sfsc then the list is incomplete because the list was not constructed by a method which makes it complete. But if the Diagonal method does produce a monotonically increasing count of missing sfsc then the Diagonal method is incapable of examining the entire list to determine its completeness.

The monotonically increasing count of missing elements does not establish that the set is an uncountable infinite: rather, it establishes that the Diagonal method is incapable of confirming that the set is a countable infinite because it inherently cannot examine all of the list.

Similarly, the hypothesized production of ~D does not establish that the set is an uncountable infinite, but also only shows that the Diagonal method cannot examine all of the list when the list is assumed to be the superset.

The failure to give the right answer for the known countable infinite set of sfsc carries over to the inability of the Diagonal method to give a definitive answer for the superset. Simply, the Diagonal method is ill-suited to tell us whether the superset is a countable or uncountable infinite.

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

### Re: Cantor Diagonal Argument disproof

Robert continues to demonstrate that he doesn't comprehend Cantor's proof and misrepresents it.

The point he seems to be trying to get across, but miserably fails at, is that if we view Cantor's algorithm as sequential, and we have constructed the fist n items (call it s) of the anti-diagonal, there will always be later strings in the list starting with s (actually infinitely many of them depending on the set being examined), and therefore the algorithm somehow fails to "examine" those strings.

However, there are at least two errors in Robert's reasoning:

1. Cantor's method isn't sequential; it constructs the anti-diagonal all at once, just like I showed by using maps/functions.

2. Even if it were sequential, once it gets to those "missing strings", the constructed anti-diagonal thus far still not equals those strings, and we therefore conclude that it will never actually be equal to any string of any row, and is hence not part of the enumeration.

The point he seems to be trying to get across, but miserably fails at, is that if we view Cantor's algorithm as sequential, and we have constructed the fist n items (call it s) of the anti-diagonal, there will always be later strings in the list starting with s (actually infinitely many of them depending on the set being examined), and therefore the algorithm somehow fails to "examine" those strings.

However, there are at least two errors in Robert's reasoning:

1. Cantor's method isn't sequential; it constructs the anti-diagonal all at once, just like I showed by using maps/functions.

2. Even if it were sequential, once it gets to those "missing strings", the constructed anti-diagonal thus far still not equals those strings, and we therefore conclude that it will never actually be equal to any string of any row, and is hence not part of the enumeration.

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:The point [Robert] seems to be trying to get across... is that if we view Cantor's algorithm as sequential, and we have constructed the first n items (call it s) of the anti-diagonal,

there will always be later strings in the list starting with s (actually infinitely many of them depending on the set being examined), and therefore the algorithm somehow fails to "examine" those strings.

However, there are at least two errors in Robert's reasoning:

1. Cantor's method isn't sequential; it constructs the anti-diagonal all at once, just like I showed by using maps/functions.

This is no more than an appeal to the miraculous. It does not in any way nullify what happens during a sequential application of the Diagonal method. Rather, it attempts to sweep-under-the- rug all the dirt which the sequential application brings to light.

2. Even if it were sequential, once it gets to those "missing strings", the constructed anti-diagonal thus far still not equals those strings, and we therefore conclude that it will never actually be equal to any string of any row, and is hence not part of the enumeration.

As the sequential application progresses, it produces strings missing from the portion of the list so far examined. Indeed some of these missing strings can be found in the continuing examination, but other missing strings are produced faster than older ones are found- so the count of them is always increasing. ~D will not be produced until the creation of these intermediate missing strings ceases. Clearly, however, the production of the intermediate missing strings never ceases because the sequential examination is endless. Thus an endless process cannot produce ~D, but will produce a monotonically increasing count of missing strings.

Clearly, if the list was preloaded with all strings of finite significant characters (sfsc) then the missing strings must be in the list somewhere. Yet the monotonically increasing count of them informs us that the sequential Diagonal method cannot find them all. It is invalid to infer from this increasing count of missing sfsc that the list is incomplete, and thereby the sfsc comprise an uncountable infinite set. We know that the sfsc are a countable infinite set because they can be produced by a sequential process. So the alternative conclusion is that the sequential Diagonal method is unable to examine the entire list to confirm that it is complete, and thereby confirm that the sfsc are a countable infinite set.

The hypothesized miraculous production of ~D is irrelevant to this prior problem of the sequential Diagonal method failing to confirm that the sfsc are a countable infinite set.

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

### Re: Cantor Diagonal Argument disproof

No, it is nothing more than logic. Just like I can create the diagonal from the function (+) of two variables, and form the function (2*), so can I, from the function g, the enumeration of all strings having finite number of 1's, construct its antidiagonal.This is no more than an appeal to the miraculous. It does not in any way nullify what happens during a sequential application of the Diagonal method.

It doesn't produce "strings", plural, but one string, the start of the anti-diagonal.As the sequential application progresses, it produces strings missing from the portion of the list so far examined.

And why is this a problem? Say we have constructed N-1 items of the anti-diagonal; do we really need to "examine" the whole string at row N? Of course not! We only need to get its N'th bit and flip it, to know the strings can't match; and that then applies to all those later strings starting with that string; so we can "retire" those strings equally fast.Indeed some of these missing strings can be found in the continuing examination, but other missing strings are produced faster than older ones are found

Name one item of the anti-diagonal that doesn't get produced?Thus an endless process cannot produce ~D,

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

### Re: Cantor Diagonal Argument disproof

Cantor never discusses supersets of infinite sets in this context. Robert is confused.robert 46 wrote:Cantor wants to prove that the superset of an infinite set is not countable.

There are two forms of Cantor Diagonalization: one uses such strings, and one uses POWER sets. Roberts doesn't seem to understand either one - the infinite strings, or power sets, so he keeps confusing them with "supersets." Please, robert, if you want to claim anything about either, stick to the strings or try to understand what a power set is.For this purpose he considers the set of all strings of infinite characters taken from a set of two characters- use {0,1}.

Interestingly, the power set version isn't restricted to infinite sets.

The set of all strings, T, is the set he proves can't be counted. It is never specifically referred to as a superset of anything (although, irrelevantly, it is proven to be a strict superset of any set of strings S that can be counted).The set of all strings of infinite characters is the superset,...

It does not. "2^infinity" is a meaningless expression, and robert's attempts to use it demonstrate his ignorance....and has 2^infinity members.

No, if a set of strings S can be counted - which is defined to mean "can be put in a numbered list" without needing justification - then that set is not T.If this superset is countable then it can be put into a list.

Again, robert demonstrates his ignorance by justifying the definition, which needs no justification. But only after he applied it to the wrong set (T, not S).Justification:

No, call the set of all strings T, and any set that can be put in a list (without any claims of it being a strict subset) S.Call the superset of infinite strings T, and S any list of elements from T.

Nobody but robert ever claims this it is attempted to put all of T in S.If all of T can be put in S then T is countable.

Using the diagonal method on the list of S, Cantor proves that the resulting string ~D, which by definition is a member of T, is not in S.Using the Diagonal method, Cantor purports to show that an element ~D can be produced which is a member of T but not in S.

To do this, all he has to assume is what he already assumed: that there is a listing of S that matches every possible natural number with a string in S.However, to do this he must go to infinity as a limit to produce all the characters of ~D.

This is possible because the Axiom of Infinity says the natural numbers exists as a set we call N.This is impossible because infinity is actually the absence of a limit.

We do not require what robert calls "being produced." Just "being defined." Every number in the set N is defined, so every character in D is defined from it.~D cannot actually be produced.

By know, robert can't help but know this, yet he still refuses to accept it.Cantor apparently knows this,

No, N is defined by induction. D is defined, because it is defined by N. Robert also continues to deliberately misunderstand how "induction" is used here.but believes that ~D can be inferred to exist consequent to induction.

robert ignores just about everything that goes on in the proof that he claims to have disproven.Cantor ignores what actually happens

Consider that, just ther eis no contradiction in the fact that every element of N is a finite number but the size of the set is infinite, every sfsc is finite even if the size of the set S that robert describes here is infinite.Consider that S is initially loaded with all the sfsc.

It is part of the same non-contradiction that allows the size of the set of even natural numbers E to be the same as the size of N, even though the #@%$! of odd numbers skipped is similarly monotonically increasing. NO INTERMEDIATE COUNT HAS ANY SIGNIFICANCE.This being the case, what do we make of the extended Diagonal method producing a monotonically increasing count of sfsc missing from S so far examined?

Whereas, by definition, ~D cannot be in SWhereas the extended Diagonal method cannot come to an infinite limit to show that S is actually missing any sfsc ...

Cantor's method doesn't need to "come to an infinite limit.", Cantor's Diagonal method cannot come to an infinite limit to show that S is actually missing ~D.

+++++

P.S. There are other things in life, than arguing with a brick wall.

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:No, it is nothing more than logic.This is no more than an appeal to the miraculous. It does not in any way nullify what happens during a sequential application of the Diagonal method.

What is logical about an invalid inductive inference? You cannot define inductive inference to work when it clearly does not.

It doesn't produce "strings", plural, but one string, the start of the anti-diagonal.As the sequential application progresses, it produces strings missing from the portion of the list so far examined.

It produces strings with an increasing number of leading characters of the anti-diagonal padded with (0) after position n for n increasing from 1. This defines a set of strings which differ from each other.

And why is this a problem?Indeed some of these missing strings can be found in the continuing examination, but other missing strings are produced faster than older ones are found

Because all the sfsc must be in a complete list of sfsc, and if the method cannot find them then it cannot tell us that the list is complete. To infer that the set of sfsc is an uncountable infinite consequent to not being able to find all elements is a false conclusion because the set of sfsc is a known countable infinite.

Say we have constructed N-1 items of the anti-diagonal; do we really need to "examine" the whole string at row N?

This is not required if we simply accept that no matter how many missing sfsc strings are retired there will still be a monotonically increasing count of them- because retiring missing sfsc strings is futile to a net reduction of the count.

Name one item of the anti-diagonal that doesn't get produced?Thus an endless process cannot produce ~D,

This statement is vague. Yet you fail to comprehend that ~D is entirely irrelevant to the problem of the Diagonal method not being able to find all elements in a known complete list of sfsc.

*****

@ JeffJo

Yes, I did confuse superset for power set.

Wikipedia, Power set wrote:Cantor's diagonal argument shows that the power set of a set (whether infinite or not) always has strictly higher cardinality than the set itself (informally the power set must be larger than the original set). In particular, Cantor's theorem shows that the power set of a countably infinite set is uncountably infinite.

To return to strings of infinite characters.

Is the set of strings of finite significant characters (sfsc) a proper subset of the set of strings of infinite characters?

Is the set of sfsc a countable infinite set?

Can the set of sfsc be put into a list in principle?

If the set of sfsc is put into a list, can Cantor's Diagonal method show that the list is not necessarily missing an sfsc element?

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?

If either method shows that the list is missing one or more sfsc elements, should this be interpreted as meaning that the set of sfsc is an uncountable infinite set?

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

### Re: Cantor Diagonal Argument disproof

Yes to all!Is the set of strings of finite significant characters (sfsc) a proper subset of the set of strings of infinite characters?

Is the set of sfsc a countable infinite set?

There exists an algorithm that does not fail to assign a unique counting number to every sfsc.Can the set of sfsc be put into a list in principle?

No to all!If the set of sfsc is put into a list, can Cantor's Diagonal method show that the list is not necessarily missing an sfsc element?

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?

If either method shows that the list is missing one or more sfsc elements, should this be interpreted as meaning that the set of sfsc is an uncountable infinite set?

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

### Re: Cantor Diagonal Argument disproof

Gofer wrote:Yes to all!Is the set of strings of finite significant characters (sfsc) a proper subset of the set of strings of infinite characters?

Is the set of sfsc a countable infinite set?There exists an algorithm that does not fail to assign a unique counting number to every sfsc.Can the set of sfsc be put into a list in principle?No to all!If the set of sfsc is put into a list, can Cantor's Diagonal method show that the list is not necessarily missing an sfsc element?

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?

If either method shows that the list is missing one or more sfsc elements, should this be interpreted as meaning that the set of sfsc is an uncountable infinite set?

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 ? In both cases it is always determined that the list is missing one or more elements.

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

### Re: Cantor Diagonal Argument disproof

And you ignored the previous times I pointed it out. And the fact that "power set" is inappropriate where you used "superset."robert 46 wrote:Yes, I did confuse superset for power set.

Wikipedia, Power set wrote:Cantor's diagonal argument shows that the power set of a set (whether infinite or not) always has strictly higher cardinality than the set itself (informally the power set must be larger than the original set). In particular, Cantor's theorem shows that the power set of a countably infinite set is uncountably infinite.

Your argument is about the one that uses strings. Mentioning power/super sets was deliberate obfuscation, a common ploy you use. And note that the section of your quote that I put in green, is the one that I almost referenced to support the green part of mine.JeffJo wrote:There are two forms of Cantor Diagonalization: one uses [infinite binary] strings, and one uses POWER sets. ... Interestingly, the power set version isn't restricted to infinite sets.

Please note this other fact you also keep ignoring: with infinite sets, showing that one set can be seen as larger than another is not enough. N can be seen as bigger then the evens E. But it can also be seen as the same size, and as smaller. Just like your "monotonically increasing" missing sets. The point of Cantor Diagonalization, is that T must always be seen as larger than N

Yes, if padded, and acknowledged many times. But irrelevant, for the reasons that I have repeated many times, and you have ignored. More obfuscation on your part.Is the set of strings of finite significant characters (sfsc) a proper subset of the set of strings of infinite characters?

You keep ignoring that these are the same question. Yet more obfuscation.Is the set of sfsc a countable infinite set?

Can the set of sfsc be put into a list in principle?

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.

- This set cannot ever include an infinite binary string that ends with infinite 1's, so it is not comparable to the set Cantor says is uncountable.

Please, try to grasp this. It isn't hard:If the set of sfsc is put into a list, can Cantor's Diagonal method show that the list is not necessarily missing an sfsc element?

- CANTOR IS NOT TRYING TO SHOW THAT YOUR LIST IS MISSING AN SFSC STRING.

The double negative is more obfuscation. Whatever you think your question was, this addresses it: Diagonalization is not trying to show that the list is missing an element that properly belongs to the listed set. It can't. Nothing can. But Diagonalization can show that there is a binary string - one that is not an sfsc, if S is the set of all sfsc - that is not listed.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 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.If either method shows that the list is missing one or more sfsc elements, should this be interpreted as meaning that the set of sfsc is an uncountable infinite set?

+++++

Why are you concerned about the properties of the set N, which cannot be "actually produced" by the definition of "actually produced" that you refuse to supply?Then why should a missing ~D, which cannot actually be produced...

Cantor's math defines every element of the infinite set N. From N, it defines the string ~D. So again, your argument is based on different criteria for proof than what is commonly deemed acceptable in the world of mathematics. You can use your limited criteria, if you want, but that does not invalidate the work of those who have a broader vision than you do. And it contains inconsistencies, like whether E is bigger, smaller, or the same size as N.

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

### Re: Cantor Diagonal Argument disproof

JeffJo wrote:Please note this other fact you also keep ignoring: with infinite sets, showing that one set can be seen as larger than another is not enough. N can be seen as bigger then the evens E. But it can also be seen as the same size, and as smaller. Just like your "monotonically increasing" missing sets. The point of Cantor Diagonalization, is that T must always be seen as larger than N

The only reason that T is seen as larger than N and uncountable is that any infinite list whatsoever will be missing ~D per Cantor's Diagonal method. However, any infinite list whatsoever will also be missing a monotonically increasing count of sfsc per the sequential Diagonal method. Yet if that list is only composed of all sfsc, there can be no missing sfsc. So what gives?

Yes, if paddedrobert 46 wrote:Is the set of strings of finite significant characters (sfsc) a proper subset of the set of strings of infinite characters?

That was stipulated.

You keep ignoring that these are the same question.Is the set of sfsc a countable infinite set?

Can the set of sfsc be put into a list in principle?

They are joined. All of a countable infinite set can be put into a list, and anything that can be put into a list is countable. Does this necessarily require a known method for putting the elements into the list?

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.

Yes, this was established long ago per the mirror Diagonal argument applying to integers.

This set cannot ever include an infinite binary string that ends with infinite 1's, so it is not comparable to the set Cantor says is uncountable.

Cantor is saying that a string of infinite significant characters is intrinsically different from an infinite string of finite significant characters. We can accept that. But it is the argument that every list must be missing an element which leads to the conclusion that the base type is an uncountable infinite. In the case of sfsc, every list is missing elements, unless some way can be found to reduce the count of missing elements to zero. This is impossible to do under the condition of a monotonically increasing count.

Please, try to grasp this. It isn't hard:If the set of sfsc is put into a list, can Cantor's Diagonal method show that the list is not necessarily missing an sfsc element?CANTOR IS NOT TRYING TO SHOW THAT YOUR LIST IS MISSING AN SFSC STRING.

Cantor shows that your set of sfsc is missing a string that belongs in T. T is different than S. S is the set that is listed. T is the set that is uncountable. ~D doesn't have to "properly belong to" S. It "properly belongs" to T.

I have no problem with this: Cantor's Diagonal method never considers the circumstance of a list of sfsc. But the sequential Diagonal method does consider this circumstance. So the same argument appears to apply: ~D missing from any infinite list of infinite strings implies the set of infinite strings is uncountably infinite; sfsc missing from any infinite list of sfsc implies the set of sfsc is uncountably infinite. Yet the latter is false. So what justification is there for the former to be true- the argument is the same?

The double negative is more obfuscation.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?

There is only one negative: "not". Rephrase:

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?

The only way this could be correct is if the count of missing sfsc gets reduced to 0. Yet that the count is monotonically increasing makes this impossible. Thus the sequential Diagonal method can never show that the list is complete: i.e not missing elements.

Whatever you think your question was, this addresses it: Diagonalization is not trying to show that the list is missing an element that properly belongs to the listed set. It can't. Nothing can. But Diagonalization can show that there is a binary string - one that is not an sfsc, if S is the set of all sfsc - that is not listed.

You say Diagonalization can't show that an element which properly belongs to the listed set is missing. Yet that is exactly what the sequential method does with the monotonically increasing count of missing sfsc elements.

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.If either method shows that the list is missing one or more sfsc elements, should this be interpreted as meaning that the set of sfsc is an uncountable infinite set?

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.

+++++

Why are you concerned about the properties of the set N, which cannot be "actually produced" by the definition of "actually produced" that you refuse to supply?Then why should a missing ~D, which cannot actually be produced...

Because I seriously doubt that inductive inference to the infinite is a valid technique.

Cantor's math defines every element of the infinite set N. From N, it defines the string ~D. So again, your argument is based on different criteria for proof than what is commonly deemed acceptable in the world of mathematics. You can use your limited criteria, if you want, but that does not invalidate the work of those who have a broader vision than you do.

And just how do they have a "broader vision" under the circumstances that they never thought about the sfsc situation?

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

### Re: Cantor Diagonal Argument disproof

Ever hear the term 'complex question' or loaded question? Look it up, and rephrase!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?

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.In both cases it is always determined that the list is missing one or more elements.

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

### Who is online

Users browsing this forum: No registered users and 4 guests