## Cantor Diagonal Argument disproof

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### Re: Cantor Diagonal Argument disproof

Robert, in the infinite case, there's a bijection between the "length" and the "width" of the list, meaning they have the same "size", and again making your objection to it wrong. You really can't use the normal "measurements" of the finite case.

So lets again play this simple game:

Do you agree that:

1. The function (f n):=(mod2 n) exists.

2. For every integer n, f returns a 0 or 1.

3. If g is a function from N to (N to {0,1}), I can create the function (h n):= flip (g n n).

4. There exists no integer M such that h=(g M).

5. Cantor's lemma could be stated as: for every mapping g, (4) holds.

Please state with what and why you disagree!

So lets again play this simple game:

Do you agree that:

1. The function (f n):=(mod2 n) exists.

2. For every integer n, f returns a 0 or 1.

3. If g is a function from N to (N to {0,1}), I can create the function (h n):= flip (g n n).

4. There exists no integer M such that h=(g M).

5. Cantor's lemma could be stated as: for every mapping g, (4) holds.

Please state with what and why you disagree!

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:Robert, in the infinite case, there's a bijection between the "length" and the "width" of the list, meaning they have the same "size", and again making your objection to it wrong. You really can't use the normal "measurements" of the finite case.

For all width n, length is 2^n. As n approaches infinity this relationship, length/width=2^n/n does not change. Note that the magnitude of the ratio monotonically increases with n [1]. If it was to change to length/width=2^infinity/infinity=infinity/infinity=1, there would be a discontinuity which is inexplicable. Therefore n<infinity, and there are no infinite lists or sets.

[1] 2^1/1=2; 2^2/2=2; 2^3/3=8/3=2+2/3; 2^4/4=4; 2^5/5=32/5=6+2/5; etc. Therefore, by induction 2^infinity/infinity>2. The length is greater than the width in the infinite case.

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

Robert, it is the set of all infinite strings that is being investigated, not the set of all finite strings, which indeed is countable, yet the following proposition is still true:

Proposition: for all enumerations of the set of all finite strings of a finite alphabet, Cantor's method fails to access a letter for some integer N depending on the enumeration.

which has a trivial proof; can you see it?

But you really can't make the case that Cantor's method misses an element for the infinite strings, can you? meaning, if it never misses, it must yield a letter for every integer, and therefore conforming to the requirements of being an infinite string, or map which I have been trying to tell you.

Proposition: for all enumerations of the set of all finite strings of a finite alphabet, Cantor's method fails to access a letter for some integer N depending on the enumeration.

which has a trivial proof; can you see it?

But you really can't make the case that Cantor's method misses an element for the infinite strings, can you? meaning, if it never misses, it must yield a letter for every integer, and therefore conforming to the requirements of being an infinite string, or map which I have been trying to tell you.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:Robert, it is the set of all infinite strings that is being investigated, not the set of all finite strings, which indeed is countable,

Yet the Diagonal method applied to finite strings returns a missing element: showing that it cannot examine the entire finite list for completeness.

yet the following proposition is still true:

Proposition: for all enumerations of the set of all finite strings of a finite alphabet, Cantor's method fails to access a letter for some integer N depending on the enumeration.

The method cannot examine past the number of symbols in the given width, and thereby cannot examine past the same number of elements in the list.

But you really can't make the case that Cantor's method misses an element for the infinite strings, can you?

That is what I have been doing. Implication: 2^infinity/infinity>2. If that isn't contradictory then what is???

meaning, if it never misses, it must yield a letter for every integer, and therefore conforming to the requirements of being an infinite string, or map which I have been trying to tell you.

The method never comes to completion, and therefore never creates the final anti-diagonal element. All it does is endlessly create intermediate anti-diagonal elements of finite width.

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

> Yet the Diagonal method applied to finite strings returns a missing element: showing that it cannot examine the entire finite list for completeness.

>The method cannot examine past the number of symbols in the given width, and thereby cannot examine past the same number of elements in the list.

>>But you really can't make the case that Cantor's method misses an element for the infinite strings, can you?

>That is what I have been doing. Implication: 2^infinity/infinity>2. If that isn't contradictory then what is???

answer to all above quotes:

Again, the diagonal method does not work for finite sets because it fails to access a letter for some integer. For example, what will the second letter of our constructed string be for the list [0,1]? answer: undefined. The diagonal method is only defined for lists having elements with sufficient number of letters, such as the following list: [1,10,101,1100, ...]

With "misses an element", I mean misses/fails to access a letter.

> The method never comes to completion, and therefore never creates the final anti-diagonal element. All it does is endlessly create intermediate anti-diagonal elements of finite width.

The diagonal method does not "go" anywhere. It accesses elements "all at the same time" if you will, just like a normal function would.

You should be occupying yourself with the following question: if there exists an infinite list of infinite strings, does it not have an infinite diagonal just like it has two infinite sides? If your answer is "such a lists is impossible to construct in the real world", you have failed to comprehend Cantor's proof.

>The method cannot examine past the number of symbols in the given width, and thereby cannot examine past the same number of elements in the list.

>>But you really can't make the case that Cantor's method misses an element for the infinite strings, can you?

>That is what I have been doing. Implication: 2^infinity/infinity>2. If that isn't contradictory then what is???

answer to all above quotes:

Again, the diagonal method does not work for finite sets because it fails to access a letter for some integer. For example, what will the second letter of our constructed string be for the list [0,1]? answer: undefined. The diagonal method is only defined for lists having elements with sufficient number of letters, such as the following list: [1,10,101,1100, ...]

With "misses an element", I mean misses/fails to access a letter.

> The method never comes to completion, and therefore never creates the final anti-diagonal element. All it does is endlessly create intermediate anti-diagonal elements of finite width.

The diagonal method does not "go" anywhere. It accesses elements "all at the same time" if you will, just like a normal function would.

You should be occupying yourself with the following question: if there exists an infinite list of infinite strings, does it not have an infinite diagonal just like it has two infinite sides? If your answer is "such a lists is impossible to construct in the real world", you have failed to comprehend Cantor's proof.

- Gofer
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### Re: Cantor Diagonal Argument disproof

You have a strawman argument that "creates" - pun intended - irrelevant requirements that are counter to the fundamental principles of modern mathematics. Points, lines, circles, sets, fractions, the number "1", etc., are all things that exist in the abstract only, and cannot be "created." They can be "constructed," which means only that they are, or follow logically from other, undefined (abstract) concepts and axioms. One such axiom, accepted by some (the reasonable ones) constructionists, is the axiom of induction.robert 46 wrote:I have a cogent argument that whatever is hypothesized to be "created" is not part of mathematics.

Besides, if your version of mathematics (and it is most definitely not the accepted version) denies the existence of infinite sets, why do you care that other versions, which do allow the "construction" of infinite sets, prove there are different kinds of infinite sets?

That's a complete misrepresentation of what you, Euclid, and I said. All Euclid says here is that mathematics uses logic, not that it is the basis for logic. In fact, what is says is that logic is a basis for mathematics.If mathematics is supposed to be the basis...Even Euclid would disagree: Theorems are proven deductively through a logical chain of steps leading to a single (non-contradictory) conclusion.mathematics is not supposed to be the basis for logic,

And there still is no contradiction. Oh, except that you can't name a natural number that is larger than all others.

You have been shown why your "mirror argument" is flawed.The mirror argument shows a contradiction:

The diagonal argument does not "hypothesize to create" an integer. It constructs a string comprised of an infinite combination of two characters, from a set if similar infinite strings.Whatever is hypothesized to have been created by the Diagonal method is not an integer,

If applied to the infinite representations of real numbers between 0 and 1, it creates a infinite representation of a real number between 0 and 1.

If your "mirror argument" does not do something similar, it has no bearing on Cantor's proof.

And I explained that your argument deals with finite sets, not infinite ones, so it does not mirror Cantor.I explained in my mirror argument that any intermediate missing integer must be in the list of integers at a later position.

No, this is typical robert 46 posturing. He can't deal with Cantor's proof, so he changes it to something he thinks he can deal with.robert's attempt to change Cantor's argument into something else, in order to find the phantom fault, shows that the fault lies in robert's misinterpretations.

This is typical JeffJo posturing.

I have shown how they are not extensions.[Jeffjo]doesn't actually try to refute the conclusions

- JeffJo
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### Re: Cantor Diagonal Argument disproof

By induction, Diagonalization constructs a set of Cantor Strings that can't be counted. True, it is not proven to be all of the Cantor Strings, but it is constructed.Gofer wrote:Jeff and Robert don't seem to comprehend that Cantor doesn't actually prove the EXISTENCE of uncountable sets, ...

The reason I suspect that Cantor uses proof by negation is because he says… otherwise we would have the contradiction, that a thing [Ding] E(0) would be both an element of M, but also not an element of M.

And to parrot your attempts to demonstrate that it isn't proof by contraposition, he does not say it is "proof by negation" (an unrecognized expression"), "proof of negation" (what you probably meant), "proof by contradiction" (what you argued before), or reductio ad absurdum (what he would have said if what you insist upon were indeed correct).

Nor is what you describe "proof of negation," which is the second form of proof by contradiction as outlined by Wikipedia.

And you still haven't addressed the fact that reductio ad absurdum, as well as proof of negation, works only if the contradiction FOLLOWS LOGICALLY FROM THE ASSUMPTION THAT WAS MADE. Which is not the case in Cantor's paper. No conclusions are drawn from assuming E includes all elements of M, except that it is the negation of what was proven. So not only is Cantor not claiming it is reductio ad absurdum, or proof of negation, it isn't.

what you describe is contraposition. If you want to prove that B follows from A, you can do so by proving that ~A follows from ~B. And stating what follows from ~A "contradicts B," is saying that it is ~B. Nowhere is B assumed, and no conclusion that would follow from assuming ~B are demonstrated.Now, this is proof by negation, ...

No, you didn't.because we assume something and then derive a contradiction from it.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Gofer wrote:> Yet the Diagonal method applied to finite strings returns a missing element: showing that it cannot examine the entire finite list for completeness.

>The method cannot examine past the number of symbols in the given width, and thereby cannot examine past the same number of elements in the list.

>>But you really can't make the case that Cantor's method misses an element for the infinite strings, can you?

>That is what I have been doing. Implication: 2^infinity/infinity>2. If that isn't contradictory then what is???

answer to all above quotes:

Again, the diagonal method does not work for finite sets because it fails to access a letter for some integer.

The Diagonal method does not work with any list it is given because it cannot examine the entire list, and finding a missing element from a portion of the list is irrelevant to determining whether the entire list is complete.

For example, what will the second letter of our constructed string be for the list [0,1]? answer: undefined.

The Diagonal method is necessarily restricted by what it is given. Therefore it is restricted by lists of finite width and finite length.

The diagonal method is only defined for lists having elements with sufficient number of letters, such as the following list: [1,10,101,1100, ...]

In this example you are actually assuming trailing null characters:

[1,10,101,1100, ...]=[1(0),10(0),101(0),1100(0), ...]

Whereas the trailing characters are not accessed, they are irrelevant to the result.

With "misses an element", I mean misses/fails to access a letter.

The method can only work on the data it is given.

> The method never comes to completion, and therefore never creates the final anti-diagonal element. All it does is endlessly create intermediate anti-diagonal elements of finite width.

The diagonal method does not "go" anywhere. It accesses elements "all at the same time" if you will, just like a normal function would.

Let us then consider the Sieve of Eratosthenes. If the Sieve sifts out only prime numbers, is it reasonable to say that it constructs, all-at-once, a list containing only prime numbers? If so, can that list be accessed to randomly extract prime numbers, and only prime numbers? Obviously, if the list can be accessed it will produce only prime numbers. However, it is impossible to access the list to get prime numbers in arbitrary order. Thus the Sieve of Eratosthenes does not work all-at-once on the "set of integers": it works sequentially on what it is given, which must be an integer; and it can only sequentially place primes into a list.

Thus, it is similarly unreasonable to believe that the Diagonal method works all-at-once on the list of elements. It must work sequentially on each element.

You should be occupying yourself with the following question: if there exists an infinite list of infinite strings, does it not have an infinite diagonal just like it has two infinite sides?

Only if infinity wide by infinity long is square. Yet I have proved that the length is actually 2^infinity, and that this is longer than infinity by at least a factor of 2. [1] Thus there is no diagonal of the square.

If your answer is "such a lists is impossible to construct in the real world", you have failed to comprehend Cantor's proof.

I think that you and JeffJo have failed to comprehend the absurdity of a belief in the Diagonal method working all-at-once.

[1] 2^infinity/infinity>2

2^infinity>2*infinity

Last edited by robert 46 on Fri Mar 17, 2017 1:57 pm, edited 1 time in total.

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

robert 46 wrote:I think that you and JeffJo have failed to comprehend the absurdity of a belief in the Diagonal method working all-at-once.

And I know that you refuse to even try to understand that "working all at once," or the Induction Method, is an accepted procedure in mathematics. So the only thing that is absurd here is your claim that it is not a part of mathematics. And the only contradiction is when you talk about "the integers" not being an infinite set, but can't name the largest one.

2^infinity/infinity>2

2^infinity>2*infinity

"Infinity" is not a number you can use in expressions like that.

- JeffJo
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### Re: Cantor Diagonal Argument disproof

JeffJo wrote:You have a strawman argument that "creates" - pun intended - irrelevant requirements that are counter to the fundamental principles of modern mathematics.robert 46 wrote:I have a cogent argument that whatever is hypothesized to be "created" is not part of mathematics.

The statement was in reference to an integer of infinite bits: it cannot be constructed, and in any case cannot be "created" as a concept of mathematics because it introduces a contradiction.

Points, lines, circles, sets, fractions, the number "1", etc., are all things that exist in the abstract only, and cannot be "created." They can be "constructed," which means only that they are, or follow logically from other, undefined (abstract) concepts and axioms. One such axiom, accepted by some (the reasonable ones) constructionists, is the axiom of induction.

I used induction to prove that if the expression 2^infinity has any meaning then 2^infinity>2*infinity.

Besides, if your version of mathematics (and it is most definitely not the accepted version) denies the existence of infinite sets, why do you care that other versions, which do allow the "construction" of infinite sets, prove there are different kinds of infinite sets?

Because it introduces contradictions into mathematics.

That's a complete misrepresentation of what you, Euclid, and I said. All Euclid says here is that mathematics uses logic, not that it is the basis for logic. In fact, what is says is that logic is a basis for mathematics.If mathematics is supposed to be the basis...Even Euclid would disagree: Theorems are proven deductively through a logical chain of steps leading to a single (non-contradictory) conclusion.mathematics is not supposed to be the basis for logic,

If so then mathematics with contradictions is illogical. Yet now-a-days I see formal logic as being the basis for mathematics. After all, there are theorem proving programs which work consequent to formal logic.

And there still is no contradiction.

About what in particular? There are contradictions which arise consequent to hypothesizing infinite sets.

Oh, except that you can't name a natural number that is larger than all others.

I don't need to by defining the natural numbers to be an "open-ended" set, not a "complete" set.

You have been shown why your "mirror argument" is flawed.The mirror argument shows a contradiction:

Statement of opinion not based on fact. Recapitulate your formal argument why my "mirror argument" is flawed- if you cannot find an original in this thread.

The diagonal argument does not "hypothesize to create" an integer. It constructs a string comprised of an infinite combination of two characters, from a set if similar infinite strings.Whatever is hypothesized to have been created by the Diagonal method is not an integer,

I interpret the strings as being in the domain of the integers with leading zeroes, and find that the element hypothesized to be constructed cannot be an integer.

If applied to the infinite representations of real numbers between 0 and 1, it creates a infinite representation of a real number between 0 and 1.

It does not construct it. Just because the method appears to work in the case of real numbers between 0 and 1, does not mean that the generic method is correct in consideration that it does not work in the case of integers. If the generic method was correct, it would have to work correctly in all cases- it doesn't.

If your "mirror argument" does not do something similar, it has no bearing on Cantor's proof.

It shows that the generic method does not work in all cases. This means the generic method is faulty.

And I explained that your argument deals with finite sets, not infinite ones, so it does not mirror Cantor.I explained in my mirror argument that any intermediate missing integer must be in the list of integers at a later position.

False. The integers are accepted as infinite. But this only means they are endless: i.e. open-ended. If the Diagonal method could be applied to the integers to give a conclusive result, that result would be a missing element which is not an integer. Thus the method cannot be applied to give a conclusive result in all cases.

No, this is typical robert 46 posturing. He can't deal with Cantor's proof, so he changes it to something he thinks he can deal with.robert's attempt to change Cantor's argument into something else, in order to find the phantom fault, shows that the fault lies in robert's misinterpretations.

This is typical JeffJo posturing.

By showing that the Diagonal method fails for integers, I argue that the generic method must be faulty in all cases.

I have shown how they are not extensions.[Jeffjo]doesn't actually try to refute the conclusions

Of course the mirror argument is an extension- it does everything the Diagonal method does, and then shows that the result is absurd because it is not part of mathematics.

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

JeffJo wrote:robert 46 wrote:I think that you and JeffJo have failed to comprehend the absurdity of a belief in the Diagonal method working all-at-once.

And I know that you refuse to even try to understand that "working all at once," or the Induction Method, is an accepted procedure in mathematics. So the only thing that is absurd here is your claim that it is not a part of mathematics.

Do you wish to address my "Sieve of Eratosthenes" argument? Or evade it???

And the only contradiction is when you talk about "the integers" not being an infinite set, but can't name the largest one.

Don't have to consequent to defining the integers to be an open-ended set.

2^infinity/infinity>2

2^infinity>2*infinity

"Infinity" is not a number you can use in expressions like that.

2^n>2*n, n>2. So make of it what you will by induction as n approaches infinity.

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

Then you admit that infinite sets exist?robert 46 wrote:I used induction ...

It has no meaning unless you use transcendental numbers, which you deny exist....to prove that if the expression 2^infinity has any meaning ...

Again: the only contradiction here is how you deny that infinite sets "exist," whatever you think that means, yet continue to treat them with some form of existence while evading the question "what is the largest natural number?"Besides, if your version of mathematics (and it is most definitely not the accepted version) denies the existence of infinite sets, why do you care that other versions, which do allow the "construction" of infinite sets, prove there are different kinds of infinite sets?

Because it introduces contradictions into mathematics.

Then its good that there are no contradictions.mathematics with contradictions is illogical.

And that is what Euclid said. Yet you claimed that mathematics was a formal basis for logic, the reverse of what Euclid said. And evaded it when I pointed out your contradiction.Yet now-a-days I see formal logic as being the basis for mathematics.

No, there aren't. There are contradictions arising from how you misrepresent how modern set theory deals with infinite sets.About what in particular? There are contradictions which arise consequent to hypothesizing infinite sets.

Same thing. Is it a set, or not? Call it what you will, if it is a set then Cantor's proof works.I don't need to by defining the natural numbers to be an "open-ended" set, not a "complete" set.Oh, except that you can't name a natural number that is larger than all others.

Statement of fact, that you ignore. (See * later.)You have been shown why your "mirror argument" is flawed.The mirror argument shows a contradiction:

Statement of opinion not based on fact.

Recapitulate the argument in a clear, complete manner, and I will. (If * doesn't do it already.)Recapitulate your formal argument why my "mirror argument" is flawed.

and you are wrong to do so. Every integer is represented by a finite string to digits. You can claim you make them infinite by padding them, but you still have a set of finite strings.I interpret the strings as being in the domain of the integers with leading zeroes,

Can't you understand this simple difference (call it *)? The reals between 0 and 1 ***REQUIRE*** some strings that be represented as finite strings that get padded with zeroes. The integers ***ARE ONLY*** finite strings that you pad with zeroes. The argument is not a mirror for this reason.

Nobody but you ever claimed it was. In fact, a crucial part of Cantor's proof is that the constructed string can be proven to belong to the set of all eligible strings. So this assertion not only doesn't disprove Cantor, it proves that you do not have a mirror of Cantor.and find that the element hypothesized to be constructed cannot be an integer.

It constructs it by induction.It does not construct it.

So infinite sets exist? And all we need to do to accept them is call them "open-ended" instead of "complete?" Then:The integers are accepted as infinite.

- Let C0 be a string of characters where the nth character is "m" if n is odd, and "w" if m is even. Since the integers are accepted as (open-ended) infinite, this Cantor string is also accepted as (open-ended) infinite. So infinite Cantor strings exist.
- Let C(n) the same as C0, except interchange "m" and "w" in the nth position. Now a set of (open-ended) infinite Cantor strings exists.
- Perform (open-ended) diagonalization on this set. The result is an (open-ended) infinite Cantor string that is not in that set you diagonalized.
- The same is true for any (open-ended) infinite set of Cantor Strings that can be put in 1:1 correspondence with the natural numbers. Which is all Cantor's proof requires.

Cantor diagonalization is not performed on integers. It is performed on sets that we hope will not be countable, and the integers are countable by definition. So again, I have the statement I called * that refutes your argument.But this only means they are endless: i.e. open-ended. If the Diagonal method could be applied to the integers to give a conclusive result, that result would be a missing element which is not an integer.

Gee, it can't be used to prove that a countable (by definition) set is uncountable.Thus the method cannot be applied to give a conclusive result in all cases.

Complaint: "Doctor, doctor, it hurts if I do this." Reply: "Don't do that."

Resorting to calling it "JeffJo posturing," when robert 46 runs out of ways to misrepresent the arguments, is typical robert 46 posturing.This is typical JeffJo posturing.

The diagonal method is not supposed to work on integers. I can repeat this as often as you assert this irrelevant point.By showing that the Diagonal method fails for integers,

... except operate on infinite strings.Of course the mirror argument is an extension- it does everything the Diagonal method does,...

- JeffJo
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### Re: Cantor Diagonal Argument disproof

Jeff, I suggest you read the following for the difference between 'proof of negation' (PoN) and 'proof by contradiction' (PbC): math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/. The former is constructive whereas the latter is not.

Here they are for Cantor's theorem.

M = a set of Cantor strings.

A = M is countable.

B = M equals T.

Cantor's lemma: A -> not B

Cantor's theorem: B -> not A

PoN: (A and B) -> {using the lemma} -> ((not B and B) -> contradiction) -> not A

Proof by contraposition: (A->not B)->(B->not A). Note that this is constructive

whereas (not B->not A)->(A->B) is not.

Here they are for Cantor's theorem.

M = a set of Cantor strings.

A = M is countable.

B = M equals T.

Cantor's lemma: A -> not B

Cantor's theorem: B -> not A

PoN: (A and B) -> {using the lemma} -> ((not B and B) -> contradiction) -> not A

Proof by contraposition: (A->not B)->(B->not A). Note that this is constructive

whereas (not B->not A)->(A->B) is not.

- Gofer
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### Re: Cantor Diagonal Argument disproof

Robert, Cantor's diagonal method does NOT need to access previous elements in order to calculate an element, because they are independent, just like a function. And even if the method was recursively defined, it would still produce a letter for every integer, making your objection to it once again wrong.

Btw. the Sieve of Eratosthenes, only finds primes up to a specified integer, because it is inherently imperative, using a table to cross off numbers.

And again, the method is UNdefined for sets of all finite strings of a certain length, because it fails to access a letter for some integer N, making your argument wrong that it somehow fails for finite sets.

Btw. the Sieve of Eratosthenes, only finds primes up to a specified integer, because it is inherently imperative, using a table to cross off numbers.

And again, the method is UNdefined for sets of all finite strings of a certain length, because it fails to access a letter for some integer N, making your argument wrong that it somehow fails for finite sets.

- Gofer
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**Posts:**247**Joined:**Mon May 09, 2016 8:24 am

### Re: Cantor Diagonal Argument disproof

The Sieve of Eratosthenes: The candidate set C starts as the set of all natural numbers. (Gee, is it a set? You have to accept infinite sets, defined by induction, for that to be so.) Set n=1. Then:robert 46 wrote:Let us then consider the Sieve of Eratosthenes.

- Set the new n to the lowest number in C that is greater than the current n.
- Remove all numbers in C that are a multiple of n, except n itself. (Gee, can this be done? You have to accept infinite processes, defined by induction, for that to be so.)
- Repeat endlessly. (Gee, can this be done? You have to accept infinite processes, defined by induction, for that to be so.)

The sieve "sifts out" composition numbers, and passes prime numbers.If the Sieve sifts out only prime numbers, is it reasonable to say that it constructs, all-at-once, a list containing only prime numbers?

And yes, it is reasonable, if you accept that induction works. If you don't, then you can't refer to "the natural numbers" as a set b without putting a maximum on the values in that set.

Sets do not have order.It is impossible to access the list to get prime numbers in arbitrary order.

If there is a "set of integers," then that set is defined by induction. If you accept definition by induction, that definition is "all-at-once." And the Sieve works "all-at-once."Thus the Sieve of Eratosthenes does not work all-at-once on the "set of integers"

Wrong, and internally inconsistent.Thus, it is similarly unreasonable to believe that the Diagonal method works all-at-once on the list of elements. It must work sequentially on each element.

Thus, if you accept a definition of "all natural numbers," that definition works "all-at-once" and you have accepted "all-at-once" as a way to define both processes, and sets.

Nobody but you has said anything about "square." In fact, since the set of even numbers can be put into 1:1 correspondence with the set of natural numbers, "square" is a meaningless concept when you try to apply it is infinite sets.You should be occupying yourself with the following question: if there exists an infinite list of infinite strings, does it not have an infinite diagonal just like it has two infinite sides?

Only if infinity wide by infinity long is square.

nothing.Yet I have proved...

- JeffJo
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