KJW Posted January 8 Share Posted January 8 5 hours ago, phyti said: https://drive.google.com/file/d/17Zz39YzAEEvD5pWB0f4NihEucQfZ3RHU/view?usp=sharing This asked me to sign in to Google, so please present it on the forum. Link to comment Share on other sites More sharing options...

phyti Posted January 8 Author Share Posted January 8 Another frustration of the internet system! They offer 'file sharing' which should work without being a subscriber! Here is the pdf (many viewers don't have MS os). Cantor's illusion.pdf Link to comment Share on other sites More sharing options...

KJW Posted January 8 Share Posted January 8 6 hours ago, phyti said: Cantor's illusion.pdf 96.59 kB · 1 download Thank you. The flaw in your argument is that Cantor was dealing with infinite lists of infinite sequences, whereas you chose to reject the notion of infinite lists of infinite sequences and instead deal with finite lists of finite sequences, and in doing so, misapplied the diagonal argument. In fact, your argument against the diagonal argument actually proved Cantor's theorem in the case of finite lists of finite sequences. In the case of finite sequences, one can always create a finite list of all of them. Therefore, if one applies the diagonal argument to a finite list of all the possible finite sequences, then naturally the negation of any diagonal will form a sequence that is part of the list not covered by the diagonal. That the total number of sequences is greater than the length of the sequences is actually what Cantor's theorem is saying in the case of finite sequences. But when one considers infinite sequences, the total number of possible sequences cannot be listed, as proven by the diagonal argument. Link to comment Share on other sites More sharing options...

phyti Posted January 9 Author Share Posted January 9 (edited) Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence! The 'infinite list' is a contradiction of terms. If it has no last element, it would always exist in an incomplete state. You ignore the fact that the rule of formation determines the geometric form of the list, not the word 'infinite'. If Cantor' diagonal argument is false, his followers would miss their 'feel good moments', believing they can comprehend deep things. He would have had better credibility by descending from a high place with a stone tablet with aleph0 etched on it. Edited January 9 by phyti Link to comment Share on other sites More sharing options...

KJW Posted January 9 Share Posted January 9 (edited) But you have not shown that Cantor's diagonal argument is false. All you have done is denied the existence of infinite lists and infinite sequences. And you are incorrectly applying Cantor's diagonal argument to finite lists of finite sequences. Edited January 9 by KJW Link to comment Share on other sites More sharing options...

phyti Posted January 11 Author Share Posted January 11 KJW; But you have not shown that Cantor's diagonal argument is false. The flaw is his definition of a diagonal sequence/string/s. It depends on all s listed, and differs from those. He defines the initial list as consisting of horizontal s. That's why the random properties of the s and the list which is itself a random s are included. If the s are parallel there can't be any influence of one on any other. I.e. they are independent. All you have done is denied the existence of infinite lists and infinite sequences. False. I use the infinite set N. I haven't seen an answer to 'what is the magic n when the split occurs' that he describes in excerpt 2 below. And you are incorrectly applying Cantor's diagonal argument to finite lists of finite sequences. Those are the only ones we can form. No one has or can produce an infinite list. It's not an abstraction but a fantasy. No one will ever see a list for the set N. excerpts from 'Cantor on Set Theory': 1. "This correlation, if it is possible at all, is, as one easily sees, always completely determinate; and since in the widened number sequence there is always one and only one number alpha such that the numbers preceding it (from 1 on) on the natural succession have the same Anzahl, one must set the "Anzahl" of both these "well-ordered" sets directly to alpha, if alpha is an infinitely large number, and to the number alpha-1 which immediately precedes alpha, if alpha is a finite integer." 2. "When I conceive the infinite … there follows for me a genuine pleasure … in seeing how the concept of integer [der ganze Zahlbegriff], which in the finite has only the background of number [Anzahl], as it were splits into *two* concepts when we ascend to the infinite" He rambles on about determinate order/well ordered. How can there be an ordered set/list for an infinite set? Using {0, 1} in that succession: 1. (000000...), 2. (000000... where/when do we put the first 1? If you have two identical s that only differ in the last symbol, you can't sort them, since there is no last symbol. Thus most if not all mathematical operations that are defined for finite (things with boundaries) can't be applied to things without boundaries, including measurement (counting). But that is what Cantor is trying to do. The meaningless phrase in red is more of the nonsense cited by Wittgenstein. Link to comment Share on other sites More sharing options...

KJW Posted January 11 Share Posted January 11 (edited) 13 hours ago, phyti said: Quote But you have not shown that Cantor's diagonal argument is false. The flaw is his definition of a diagonal sequence/string/s. It depends on all s listed, and differs from those. He defines the initial list as consisting of horizontal s. That's why the random properties of the s and the list which is itself a random s are included. If the s are parallel there can't be any influence of one on any other. I.e. they are independent. Where did you get the idea that the sequences are random? The more correct term is "arbitrary", which is not the same as "random". In fact, Cantor attempts to construct an infinite list of all possible infinite sequences. But regardless of how that is attempted (represented by a list of arbitrary sequences), the list will always be missing the sequence constructed from the negation of the diagonal. In the case of finite sequences, which Cantor does not consider, one can form a complete list of all possible sequences. But then the diagonal argument becomes establishing that the list of sequences is longer than the sequences themselves. It seems to me that you are placing too much importance on the structure of the list of sequences while not placing enough importance on what the diagonal argument is saying. 13 hours ago, phyti said: Quote All you have done is denied the existence of infinite lists and infinite sequences. False. I use the infinite set N. False. Where do you use the infinite set [math]\mathbb {N}[/math]? In an earlier post you said: "Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence!" If that's not denying the existence of infinite lists and infinite sequences, then what is? 13 hours ago, phyti said: I haven't seen an answer to 'what is the magic n when the split occurs' that he describes in excerpt 2 below. I see nothing wrong with the notion of going from the finite to the infinite. It's a qualitative transition that does not imply a "magic [math]n[/math]". 13 hours ago, phyti said: Quote And you are incorrectly applying Cantor's diagonal argument to finite lists of finite sequences. Those are the only ones we can form. No one has or can produce an infinite list. It's not an abstraction but a fantasy. No one will ever see a list for the set N. I realised after had I posted that what I said was ambiguous. And you chose the unintended interpretation. I didn't mean that it was incorrect to apply Cantor's diagonal argument to finite lists of finite sequences. I meant that your application of Cantor's diagonal argument was incorrect. Specifically, you drew the wrong conclusion from the length of the list being longer than the length of the sequences. Also, because Cantor was dealing with infinite lists of infinite sequences, your dealing with finite lists of finite sequences is largely irrelevant. I should remark that although Cantor's diagonal argument applied to infinite lists of infinite sequences only produces one missing sequence (the negation of the diagonal), in fact the cardinality of missing sequences is the same as the cardinality of all possible sequences. Any list of sequences is a proverbial drop in the ocean compared to the set of all possible sequences. Edited January 11 by KJW Link to comment Share on other sites More sharing options...

phyti Posted January 13 Author Share Posted January 13 KJW; [/quote] Where did you get the idea that the sequences are random? The more correct term is "arbitrary", which is not the same as "random". In fact, Cantor attempts to construct an infinite list of all possible infinite sequences. But regardless of how that is attempted (represented by a list of arbitrary sequences), the list will always be missing the sequence constructed from the negation of the diagonal.[/quote] 'Arbitrary' and 'random' are synonyms for each other. 1. Before he defines b, all s(equences) are horizontal, there are no diagonal s. 2. He defines D from the list as all the elements with coordinates (u, u). 3. If he forms the negation of D as E0 and states it's missing from the list, he contradicts himself, since he has assumed D is a member of the list. The building set is binary {m, w}, they occur in pairs! If D then E0. Link to comment Share on other sites More sharing options...

KJW Posted January 14 Share Posted January 14 2 hours ago, phyti said: 'Arbitrary' and 'random' are synonyms for each other. No. This is what Wikipedia says about the use of "arbitrary" in mathematics: Quote In mathematics, arbitrary corresponds to the term "any" and the universal quantifier ∀ (for all), as in an arbitrary division of a set or an arbitrary permutation of a sequence. Its use implies generality and that a statement does not only apply to special cases, but that one may select any available choice and the statement will still hold. For example, one might say that: "Given an arbitrary integer, multiplying it by two will result in an even number." Even further, the implication of the use of "arbitrary" is that generality will hold - even if an opponent were to choose the item in question. In which case, arbitrary can be regarded as synonymous to worst-case. This is different to "random", which is associated with chance events and probability, neither of which apply to "arbitrary". 2 hours ago, phyti said: 1. Before he defines b, all s(equences) are horizontal, there are no diagonal s. 2. He defines D from the list as all the elements with coordinates (u, u). 3. If he forms the negation of D as E0 and states it's missing from the list, he contradicts himself, since he has assumed D is a member of the list. The building set is binary {m, w}, they occur in pairs! If D then E0. Are you talking about finite lists of finite sequences, or are you talking about what Cantor is talking about... infinite lists of infinite sequences? It seems to me that you are talking about finite lists of finite sequences, in which case, you can't argue against Cantor, who is talking about infinite lists of infinite sequences. If you are talking about infinite lists of infinite sequences, then you are not taking into account the infinite nature of the sequences and lists. Link to comment Share on other sites More sharing options...

phyti Posted January 14 Author Share Posted January 14 KJW; "Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence!" Anyone can define an infinite list. No one can produce one. Do you understand the difference? It seems to me that you are placing too much importance on the structure of the list of sequences while not placing enough importance on what the diagonal argument is saying. If that's what you think, then here is another version that doesn't consider the form of an infinite list. Cantor's illusion.pdf KJW; Something went wrong! "Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence!" Anyone can define an infinite list. No one can produce one. Do you understand the difference? It seems to me that you are placing too much importance on the structure of the list of sequences while not placing enough importance on what the diagonal argument is saying. If that's what you think, then here is another version that doesn't consider the form of an infinite list. Cantor's illusion.pdf Link to comment Share on other sites More sharing options...

KJW Posted January 15 Share Posted January 15 Quote "Per the constructivist view, no one can nor have formed an infinite list. No one can even form an infinite sequence!" Anyone can define an infinite list. No one can produce one. Do you understand the difference? In terms of what this thread is about... no. But if you invoke physical reality, I will call foul. Quote This paper shows E0 must be a member of L. In your paper, you said that L is a binary tree graph that represents the Cantor set M. Then yes, E_{0} must be a member of L. That is actually a part of Cantor's proof. But it is not what the diagonal argument is about. It seems to me that you don't really understand what it is that Cantor is proving. The sequence formed by negation of the diagonal is an element of the set of all the possible sequences (it is represented in the binary tree graph). This is actually important to Cantor's proof and is why the negation of the diagonal is made of the same symbols as the sequences. When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences. Thus, the list of sequences is incomplete, proving that the set of sequences has a higher transfinite cardinality than the list of sequences, which has a transfinite cardinality of [math]\aleph_0[/math]. You appear to fail to distinguish between the set of sequences and the list of sequences, which is what Cantor's proof is about. Link to comment Share on other sites More sharing options...

phyti Posted January 16 Author Share Posted January 16 Look again at the binary tree. There are no diagonal sequences. All are horizontal, as originally shown by Cantor. The diagonal is his misdirection. Link to comment Share on other sites More sharing options...

phyti Posted January 17 Author Share Posted January 17 (edited) KJW; Quote In your paper, you said that L is a binary tree graph that represents the Cantor set M. Then yes, E0 must be a member of L. That is actually a part of Cantor's proof. But it is not what the diagonal argument is about. It seems to me that you don't really understand what it is that Cantor is proving. The sequence formed by negation of the diagonal is an element of the set of all the possible sequences (it is represented in the binary tree graph). [Here you agree E0 is a member of L, which represents M.] This is actually important to Cantor's proof and is why the negation of the diagonal is made of the same symbols as the sequences. When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences. Thus, the list of sequences is incomplete, proving that the set of sequences has a higher transfinite cardinality than the list of sequences, which has a transfinite cardinality of ℵ0. You appear to fail to distinguish between the set of sequences and the list of sequences, which is what Cantor's proof is about. [Here you say E0 is not a member of L, which represents M. ?] [Cantor doesn't have to prove the list is incomplete. Since the set M has no last member by definition of 'infinite', neither does the list L.] [The binary tree shows both D as a member of Mo and E0 as a member of M1, thus they can't coexist in the same subset, but are still members of M and L. There is no other place for them. Why did Cantor the mathematician ignore this? (because it served his purpose?) The diagonal D is similar to the original sequences, being a string of 0's and 1's, but 1. it is redundant since it is already in the binary tree as a horizontal sequence, 2. its definition required the list to be complete, 3. there are no diagonal sequences.] KJW; In your paper, you said that L is a binary tree graph that represents the Cantor set M. Then yes, E0 must be a member of L. That is actually a part of Cantor's proof. But it is not what the diagonal argument is about. It seems to me that you don't really understand what it is that Cantor is proving. The sequence formed by negation of the diagonal is an element of the set of all the possible sequences (it is represented in the binary tree graph). [Here you agree E0 is a member of L, which represents M.] This is actually important to Cantor's proof and is why the negation of the diagonal is made of the same symbols as the sequences. When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences. Thus, the list of sequences is incomplete, proving that the set of sequences has a higher transfinite cardinality than the list of sequences, which has a transfinite cardinality of ℵ0. You appear to fail to distinguish between the set of sequences and the list of sequences, which is what Cantor's proof is about. [Here you say E0 is not a member of L, which represents M. ?] [Cantor doesn't have to prove the list is incomplete. Since the set M has no last member by definition of 'infinite', neither does the list L.] [The binary tree shows both D as a member of Mo and E0 as a member of M1, thus they can't coexist in the same subset, but are still members of M and L. There is no other place for them. Why did Cantor the mathematician ignore this? (because it served his purpose?) The diagonal D is similar to the original sequences, being a string of 0's and 1's, but 1. it is redundant since it is already in the binary tree as a horizontal sequence, 2. its definition required the list to be complete, 3. there are no diagonal sequences.] Edited January 17 by phyti Link to comment Share on other sites More sharing options...

KJW Posted January 17 Share Posted January 17 (edited) Quote Quote In your paper, you said that L is a binary tree graph that represents the Cantor set M. Then yes, E0 must be a member of L. That is actually a part of Cantor's proof. But it is not what the diagonal argument is about. It seems to me that you don't really understand what it is that Cantor is proving. The sequence formed by negation of the diagonal is an element of the set of all the possible sequences (it is represented in the binary tree graph). [Here you agree E0 is a member of L, which represents M.] Quote This is actually important to Cantor's proof and is why the negation of the diagonal is made of the same symbols as the sequences. When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences. Thus, the list of sequences is incomplete, proving that the set of sequences has a higher transfinite cardinality than the list of sequences, which has a transfinite cardinality of ℵ0. You appear to fail to distinguish between the set of sequences and the list of sequences, which is what Cantor's proof is about. [Here you say E0 is not a member of L, which represents M. ?] You think I contradicted myself, don't you? That's because you fail to distinguish between the set of sequences and the list of sequences. You fail to recognise that there are sequences in the set of sequences that are not in the list of sequences. By failing to recognise this, you are unable to understand Cantor's proof because this is precisely what Cantor is proving. By assuming that the list of sequences has all the sequences that are in the set of sequences and using this assumption in an attempt to disprove Cantor's theorem, you are begging the question. Quote [Cantor doesn't have to prove the list is incomplete. Since the set M has no last member by definition of 'infinite', neither does the list L.] By "the list is incomplete", I mean that there are sequences in the set that are not in the list, and that is precisely what Cantor is proving. Having "no last member" doesn't come into it. But by assuming that there is only one infinite, you are assuming that Cantor's theorem is false in your attempt to disprove Cantor's theorem, again begging the question. Edited January 17 by KJW Link to comment Share on other sites More sharing options...

phyti Posted January 18 Author Share Posted January 18 KJW; 1. You think I contradicted myself, don't you? 2. That's because you fail to distinguish between the set of sequences and the list of sequences. You fail to recognise that there are sequences in the set of sequences that are not in the list of sequences. By failing to recognise this, you are unable to understand Cantor's proof because this is precisely what Cantor is proving. By assuming that the list of sequences has all the sequences that are in the set of sequences and using this assumption in an attempt to disprove Cantor's theorem, you are begging the question. 3. By "the list is incomplete", I mean that there are sequences in the set that are not in the list, and that is precisely what Cantor is proving. Having "no last member" doesn't come into it. But by assuming that there is only one infinite, you are assuming that Cantor's theorem is false in your attempt to disprove Cantor's theorem, again begging the question. 1. More like, make up your mind, which is it. 2. As a model the binary tree contains all possible sequences of 0 and 1. Cantor knows, by defining a diagonal sequence D, and forming its negation E0 as a horizontal sequence, the E0 is excluded from the set. It's another flaw in his manipulation of the list. The horizontal sequence has coordinates (u, v) where u is a constant, and v varies from 1 upward through N. The diagonal sequence has coordinates (u, v) where both u and v vary from 1 upward through N. The first is one dimensional, the second is two dimensional. He uses two different definitions of a sequence. If his math was as rigorous as you think, E0 would have inherited the properties of D and also be a diagonal. 3. The definition of 'infinite' literally means 'without end', thus it IS a factor. Two different members of the set N can be compared to determine the greater of the two. No member of the set N can be compared to determine the greatest of the set. No matter how many members added to a collection, N has a corresponding n that expresses 'how many'. N is inexhaustible. There are no superlative degree of infinity. When the container is empty, it can't get emptier. If you think Cantor was infallible, he admits to occasionally making mistakes. You also claimed 'the man', or 'motive' is irrelevant. A person is what they think. The more you know about the person, helps to understand their ideas, and helps to dispel false claims about them or their work. Link to comment Share on other sites More sharing options...

KJW Posted January 18 Share Posted January 18 (edited) Quote Quote You think I contradicted myself, don't you? More like, make up your mind, which is it. I actually said: "E_{0} must be a member of L". I didn't say that E_{0} is not a member of L. You incorrectly inferred that from: "When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences". E_{0} is a member of the set of sequences. E_{0} is not a member of the list of sequences. This is not a contradiction. The list of sequences is not the set of sequences. Cantor proved that the list of sequences is not the set of sequences. You are not going to convince me otherwise. Quote As a model the binary tree contains all possible sequences of 0 and 1. The infinite binary tree is actually ambiguous. It can represent all possible sequences of 0 and 1, or it can represent all possible finite sequences of 0 and 1, which is still infinite. The latter can actually be made into a list. The former cannot be made into a list. Your argument becomes invalid if you assume the binary tree represents all possible sequences of 0 and 1 but treat it like it represents all possible finite sequences of 0 and 1. Due to the subtlety of the distinction between these two cases, I will not accept an argument from you based on a binary tree and will assert that a binary tree is not a list. Quote Cantor knows, by defining a diagonal sequence D, and forming its negation E_{0} as a horizontal sequence, the E_{0} is excluded from the set list. It's another flaw in his manipulation of the list. That Cantor can define an E_{0} that is excluded from the list is what proves his theorem. I suspect that you think he has forced a result. It is true that regardless of how a list is constructed, even if it is formed from a set that can be listed, the E_{0} derived from this list will be excluded from the list. But E_{0} isn't just excluded from the list. E_{0} is also a member of the set of all possible sequences of 0 and 1. Therefore, the exclusion of E_{0} from the list proves that the set of all possible sequences of 0 and 1 cannot be listed. But earlier in this thread, I showed that the set of all finite sequences can be listed. I then applied the diagonal argument to this list to obtain an E_{0} that is excluded from the list. However, in this case the E_{0} obtained was also not a finite sequence, and therefore a proof using the diagonal argument that the set of all finite sequences cannot be listed fails. Quote The horizontal sequence has coordinates (u, v) where u is a constant, and v varies from 1 upward through N. The diagonal sequence has coordinates (u, v) where both u and v vary from 1 upward through N. The first is one dimensional, the second is two dimensional. He uses two different definitions of a sequence. If his math was as rigorous as you think, E0 would have inherited the properties of D and also be a diagonal. This is just plain silly and brings to mind the term "cargo cult mathematics". One point that you've overlooked is that one doesn't need to consider sequences to prove Cantor's theorem. One can prove Cantor's theorem using sets, and the mappings between their elements and their subsets. So, if you're focusing specifically on sequences, then that won't invalidate proofs based on alternative notions. Quote The definition of 'infinite' literally means 'without end', thus it IS a factor. But one can still make statements that are true for every element of an infinite set. Quote If you think Cantor was infallible, he admits to occasionally making mistakes. I never suggested that Cantor was infallible. And everyone makes mistakes every now and then. So what? If there was an error in Cantor's proof, then the error isn't just Cantor's, it's all the people who have accepted Cantor's proof as valid, including myself. So, you have to do better than mention Cantor's fallibility. What about the fallibility of everyone else who accepted Cantor's proof? Quote You also claimed 'the man', or 'motive' is irrelevant. A person is what they think. The more you know about the person, helps to understand their ideas, and helps to dispel false claims about them or their work. Of course it's irrelevant. We're talking about a mathematical theorem, not Mein Kampf. The validity of a mathematical theorem is self-evident, not based on the author. Edited January 18 by KJW Link to comment Share on other sites More sharing options...

phyti Posted January 19 Author Share Posted January 19 (edited) KJW; Quote 1. The infinite binary tree is actually ambiguous. It can represent all possible sequences of 0 and 1, or it can represent all possible finite sequences of 0 and 1, which is still infinite. The latter can actually be made into a list. The former cannot be made into a list. Your argument becomes invalid if you assume the binary tree represents all possible sequences of 0 and 1 but treat it like it represents all possible finite sequences of 0 and 1. 1. You are reading things into my post that aren't there! This follows fig.2. "Each s has no last v and L has no last row." The ellipsis (...) means continue forever. It's an 'infinite' list. Any continuous path in the tree represents a sequence. Fig.3 shows the property of symmetry which means both D and E0 are present in L which represents M. There is nothing to prove as Cantor presents his list, before he defines b in his paper. 2. For every Cantor (v), there are 2v sequences in a finite list. Since v is an integer from the set N, there is no largest v. Thus the set of all finite sequences cannot be listed. The list has no end. It's simple. 3. I am NOT questioning Cantor's theorem. The subject is the 'diagonal argument'. 4. The general population depends on the knowledge and experience of a few who specialize in various fields, doctors, lawyers, politicians, science, etc. for their 'expert' guidance. Those who accept a theory will typically know less about it than the author. The patient trusts the treatment prescribed by their doctor. The viewer doesn't have to know how tv communication works to get the benefits, or internal combustion for the ability to travel. Before 1670 people believed light moved instantaneously since that's what science believed. Then theory was revised to define light motion with a finite speed. 5. Are these people in error because they disagree with Cantor? Brouwer, Leopold Kronecker, Poincaré, Gauss, Wittgenstein If Cantor's theory is in error, it's his error, it's his idea/mental construct. Bertrand Russell, Mayberry (2000, p.10) has noted that "The set-theoretical axioms that sustain modern mathematics are self-evident in differing degrees. One of them – indeed, the most important of them, namely Cantor's axiom, the so-called axiom of infinity – has scarcely any claim to self-evidence at all". Weyl classical logic was abstracted from the mathematics of finite sets and their subsets …. Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and Quote 2. But earlier in this thread, I showed that the set of all finite sequences can be listed. I then applied the diagonal argument to this list to obtain an E0 that is excluded from the list. However, in this case the E0 obtained was also not a finite sequence, and therefore a proof using the diagonal argument that the set of all finite sequences cannot be listed fails. 3. One can prove Cantor's theorem using sets, and the mappings between their elements and their subsets. So, if you're focusing specifically on sequences, then that won't invalidate proofs based on alternative notions. 4. If there was an error in Cantor's proof, then the error isn't just Cantor's, it's all the people who have accepted Cantor's proof as valid, including myself. So, you have to do better than mention Cantor's fallibility. What about the fallibility of everyone else who accepted Cantor's proof? 5. The validity of a mathematical theorem is self-evident, not based on the author. finally applied it, without justification, to the mathematics of infinite sets. This is the Fall and original sin of [Cantor's] set theory …." (Weyl, 1946) _________________ "You fail to recognise that there are sequences in the set of sequences that are not in the list of sequences. By failing to recognise this, you are unable to understand Cantor's proof because this is precisely what Cantor is proving." Do you think repeating this multiple times will make it true? Cantor fails to prove his idea via the diagonal argument with badly executed manipulation of the list. You don't like the conclusion which is; humans can't comprehend 'infinity'. Edited January 20 by phyti The post is not uploaded as typed!?? Link to comment Share on other sites More sharing options...

KJW Posted January 20 Share Posted January 20 (edited) Quote Quote The infinite binary tree is actually ambiguous. It can represent all possible sequences of 0 and 1, or it can represent all possible finite sequences of 0 and 1, which is still infinite. The latter can actually be made into a list. The former cannot be made into a list. Your argument becomes invalid if you assume the binary tree represents all possible sequences of 0 and 1 but treat it like it represents all possible finite sequences of 0 and 1. You are reading things into my post that aren't there! I'm not reading anything into your post other than that we are on different pages. You introduced the binary tree and I'm saying that I'm not going to accept any arguments from you involving the binary tree, and why. Quote The ellipsis (...) means continue forever. It's an 'infinite' list. An ellipsis is convenient but is hardly a rigorous way to represent the infinite. And given that this topic is about hierarchies of the infinite, an ellipsis must not be a crucial part of a proof about the infinite. Quote Quote But earlier in this thread, I showed that the set of all finite sequences can be listed. I then applied the diagonal argument to this list to obtain an E_{0} that is excluded from the list. However, in this case the E_{0} obtained was also not a finite sequence, and therefore a proof using the diagonal argument that the set of all finite sequences cannot be listed fails. For every Cantor (v), there are 2^{v} sequences in a finite list. Since v is an integer from the set N, there is no largest v. Thus the set of all finite sequences cannot be listed. The list has no end. It's simple. It's not simple. We wouldn't be on page 7 of this thread if it was simple. But I feel I need to clarify what I mean by an "infinite list". An infinite list is a one-to-one mapping between a given set and the set of natural numbers. That is, for every natural number there is an element of the given set to which it maps, and for every element of the given set, there is a natural number to which it maps. The set of sequences can't be listed because regardless of the mapping from the set of natural numbers to the set of sequences, there are sequences for which there is no corresponding natural number. But this is not the case for the set of finite sequences which can be mapped one-to-one with the set of natural numbers. Quote Quote One can prove Cantor's theorem using sets, and the mappings between their elements and their subsets. So, if you're focusing specifically on sequences, then that won't invalidate proofs based on alternative notions. I am NOT questioning Cantor's theorem. The subject is the 'diagonal argument'. You have rejected Cantor's theorem from the outset. Any discussion of the diagonal argument is ultimately about Cantor's theorem. Otherwise, what is the point of discussing the diagonal argument? Quote Those who accept a theory will typically know less about it than the author. That's actually not true. For example, relativity is much better understood by modern day physicists than it was by Einstein. That's because in the time since Einstein, physicists have had the opportunity to learn more about the theory than Einstein ever could. Quote Are these people in error because they disagree with Cantor? Brouwer, Leopold Kronecker, Poincaré, Gauss, Wittgenstein Sure, why not? I did say that everyone makes mistakes every now and then. By the way, why is Gauss in this list? Quote If Cantor's theory is in error, it's his error, it's his idea/mental construct. Mathematics is not an idea/mental construct. Quote Quote The validity of a mathematical theorem is self-evident, not based on the author. "The set-theoretical axioms that sustain modern mathematics are self-evident in differing degrees. One of them – indeed, the most important of them, namely Cantor's axiom, the so-called axiom of infinity – has scarcely any claim to self-evidence at all" I said it was the validity of a mathematical theorem that is self-evident. I wasn't talking about axioms, which are often said to be self-evident truths. Thus, I am able to verify Cantor's proof to my own satisfaction and without reference to Cantor. Quote Quote "You fail to recognise that there are sequences in the set of sequences that are not in the list of sequences. By failing to recognise this, you are unable to understand Cantor's proof because this is precisely what Cantor is proving." Do you think repeating this multiple times will make it true? Do you think ignoring it multiple times will make it false? By assuming that the set of sequences is the same as the list of sequences, your argument against Cantor's diagonal argument is invalid as begging the question. Although Cantor proved that there are sequences in the set of sequences that are not in the list of sequences, even without such a proof, you need to recognise that this is a possibility that needs a proof one way or the other. So, by assuming that the set of sequences is the same as the list of sequences, you have made an error regardless of the validity of Cantor's proof. By contrast, I have not assumed that there are sequences in the set of sequences that are not in the list of sequences, that has been proven, and prior to a proof, it was something to be proven. Quote humans can't comprehend 'infinity' Don't project your inability to comprehend infinity onto others. In particular, don't project your inability to comprehend infinity onto me. Edited January 20 by KJW Link to comment Share on other sites More sharing options...

phyti Posted January 20 Author Share Posted January 20 KJW; You introduced the binary tree and I'm saying that I'm not going to accept any arguments from you involving the binary tree, and why. Show me the beginning of a sequence that is not in the binary tree. Link to comment Share on other sites More sharing options...

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