JeffJo
Senior Members
Posts
44 
Joined

Last visited
Content Type
Profiles
Forums
Events
Everything posted by JeffJo

problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Mathematics does no recognize a "conventional for for a list." What you describe is just a convenient graphical representation of bijection. It has no significance. Cantor defined a string in the exact same way every other string he uses is defined. Each is a function mapping N to {'m','w'}. You are twisting it to make statements that are not true. There is no "Alist" or "Blist" except in your misrepresentations. Cantor uses another function from N to his set M, of all such strings. There is no division. He ignores nothing, because "If E0 is missing, so is D" is not true. 000000000000... 100000000000... 110000000000... etc. D is the first element here. Cantor never limits the set he uses to being a proper subset. The only restriction on what you call M1is that it is countable. You have proven nothing. You have thrown words at Cantor's proof that have nothing to do with it. 
The effect you describe is probably what is called a glory, or spectre. See https://en.wikipedia.org/wiki/Glory_(optical_phenomenon) To understand rainbows and many similar (but often unrelated) effects, start at https://atoptics.co.uk/rainbows/primrays.htm . If you move the slider, you can see how the white incoming ray separates between red and violet. But also and how that separation has nothing to do with the location of the "angle of minimum deviation" or "rainbow angle" as described in this authoritative text. But I'm getting a little tired of the people here who are less interested in a scientific approach, as much as conforming to what they learned. No matter how clearly it is contradicted by actual facts. LIke http://www.trishock.com/academic/rainbows.shtml I'm glad you agree. But this isn't a vote, it is fact. And yes, the path description holds when angles A and B apply, whether or not the function obeys Snell's Law. But another fact is that when it applies to rainbows, it always does obey Snell's Law. So such speculation is moot. The correct deviation depends on Snell's Law.

problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
That's a separator, but I think you know that. This forum does not seem to allow consecutive posts. And you don't respond to comments, you lecture on what you think must be wrong, even after admitting that you don't bother to understand what the points are. And those points are: Many people think that the outline of Cantor's second uncountability proof goes something like this: Assume that the entirety of some infinite set T (this is based on Wikipedia's naming conventions, since they name one that Cantor didn't) can be put into a bijection with the set of all natural numbers (starting with 1) N. That is, it can be listed t1, t2, t3, ... . Use any such bijection to construct a new element t0 that (A) must be in T but (B) is not in the list used in this construction. Since this contradicts the assumption that every element of T is in the list, proofbycontradiction applies and the assumption is false. THIS IS AN INCORRECT APPLICATION OF PROOFBYCONTRADICTION. You have to use all parts of the assumption to derive the contradiction. This outline does not use the part of the assumption that the entirety of T is in the list. It does, however, prove directly that the entirety is not in the list. This is close to, but not quite, what Cantor was trying to prove in the end. Nearly every "crank" objection to CDA uses this flaw in some way. The crank in question in this thread, phyti, starts by using a different bijection than what was used in the construction. Then the missing element can be found in another "orientation" of a that bijection. A more common, but less sophisticated, objection is that t0 can always be added to the bijection in the manner of Hilbert's Hotel. There are others, like https://hesperusisbosphorus.wordpress.com/2016/07/31/whatiswrongwithcantorsdiagonalargument/ . But given your claimed familiarity, I didn't think it necessary to list them. Call me silly, but I believe that if we could get more selfproclaimed experts to acknowledge this flaw in how CDA is understood, we might get fewer cranks. But these experts have to get over getting toes stepped on first. The correct outline is this: Assume that S is any infinite subset of T that can be put into such a bijection. (Cantor's lone, but minor, flaw is that he should have given an example to prove that it is possible. That way he is working with bijections that are known to exist, not supposing they might.) Construct t0 the same way. It is in T but isn't in S. If S=T, we would have the contradiction that t0 both is, and is not, in T. This is the contradiction that Cantor used, and it is different than the one that cranks think they have debunked. Specifically, it is about t0, not T, Bertrand Russell updated the terminology, in English not German, from Cantor's 1890 paper. This is something you asked to see. It says almost the exact same things, in the exact same way, with two exceptions: Russell corrected the omission of an example. For that example, he suggested one specific algorithm. He also suggested there could be "laws" defining more examples. BUT THIS WAS ONLY AN EXAMPLE FOR THE FIRST PART OF THE PROOF. Russell never suggested an algorithm or law could define the entire set. Only that it could define a listable subset. Russell never REQUIRED such an algorithm or law to define the subset S, let alone T. He said they could define S. Russell's conclusion that T can't be listed  that is, the part I call the corollary  does not suppose a definition, an algorithm, or a law. Let alone require one. Your claim that Russell committed a "howler" is based on the same misunderstanding of the difference between the two proof outlines I described. The same misunderstanding that cranks use. And that you assert does not exist, but have not defended that assertion. I do not believe that you have even considered how they are different. I feel I am justified in thinking that, based on how you responded with equal force to things you later said you didn't look at. And running away from the issue seems to be proof of this. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
No, I tried to correct you when you made false statements about details you quite clearly had not read completely. And you continue to try to make it look like I said something wrong. Google "Cantor's Diagonal Argument." If the first line of what they say is the proof contains the phrase "list of all <whatever>," that is a bad teacher. Hint: at least 99% do. No, it is not. And it is easy to demonstrate, but you have to try. You do not. To see it, look for any line in the proof that depends on the presence of that word, except the one that says "the assumption leads to a statement that contradicts 'all'." This isn't deep mathematics. Exactly. So why do you keep making arguments that are based on what you think he was thinking? What he knew, or could have known, or you think he should have known is irrelevant. Only what the proof requires. And his word was "determinate." An example is the algorithm, but no word said it must be one. Another example is a law, but again no word said it had to be one. Again, this isn't deep mathematics. As you keep saying. Yet you keep objecting as if you had. ++++++ Your disagreement is wrong, because it relies on a misinterpretation of the proof. "This analysis shows Cantor's diagonal argument published in 1891 focused on the ordered set of characters within a sequence while neglecting the orientation of the sequence, resulting in his conclusion of a missing sequence. It did not "focus on" any particular "ordered set of characters." It defined the set M of all possible, infinitelength binary strings. It said that for any ordering (not just a specific one as you use) of a subset of M (not the entire set M, as you use), that this subset necessarily omits a constructable element E0 that is in M. It doe snot utilize, or depend on, orientatton in any way. Your disagreement is wrong. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Proof by contradiction: Show that A>C for some statements A and C. Show that A>not(C) for the same statements. If A is true, then our mathematics is inconsistent. We don't want that, so we say that A is not true in this mathematics. Partial proofbycontradiction. Mention that statement B could be true. Show that A>C for some statements A and C. Show that A>not(C) for the same statements. We can say that A is not true, by contradiction. We can't say anything about B. Simply saying that it could be true does not mean we have assumed it for the purposes of proof. We need to use it in the derivation of the contradiction. Not even that B can't be true in conjunction with A being true, since that is a null statement. No statement is true in conjunction with A being true. This is my "green cheese" example, where A="there is a rational number that, when squared, equals 2" and B="the moon is made of some mix of cheese." Not a proof by contradiction: Mention that statement B could be true. Show that A>not(B). This is not a proof by contradiction, yet. This is not a contradiction, since need to use it in the derivation. It should be easy to extent this to "A and B cannot be true together." One way is that A>not(B) means B>not(A), so there can be no situation where both are true. This is a direct proof. Another is a secondary proofbycontradiction, like Cantor did. The contradiction is not "B¬(B)" as is taught for CDA. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
As I said before, anybody who doubts the result. Yes, they are cranks. Yes, I want to make this point clear to show why such cranks are wrong. I'm talking about the version that gets taught, the one that starts "Assume you have a list of all of these strings (or real numbers in [0,1])" and ends with "This contradicts the assumption that all of these strings is in the list." Take the word "all" out of it, and the steps of the proof all work the same except that no contradiction is found. This doesn't prove, yet, that it is impossible to have a complete list. But that proof isn't far. No, it really isn't a valid proofbycontradiction this way. But as Cantor did it, without that word, the corollary is a valid proof by contradiction. It is imperative for you to understand this. No, he said that an algorithm like that was an example of something that is "determinate." So was a Law. He did not require "determinate" to be either of those things. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And this is why I think you haven't tried to see what I'm saying, since I never "complained about" that proof. There were complaints associated with how others view it and you would have seen that it you even tried. Incorrect, it isn't a valid proof by contradiction. Correct, it was not Cantor's version. Incorrect, Cantor's corollary is a proof by contradiction. It isn't a proofby contradiction because it never uses the part of the assumption that says every element of M is in the list. What if I assume that the moon is made of individuallywrapped packets of bleu and green cheese, in a ratio that is equal to the square root of 2. Then I show that this leads to a contradition. Did I prove that the moon is not made of green cheese? Or anything about cheese? No, because I didn't use any of that part of the claimed assumption. Simply saying "I assume this" is meaningless if it isn't used in the derivation. It literally can be struck from the socalled proof without changing anything except the claim that there is a contradiction. But it does form the basis of a proof. You could argue that contraposition implies B>not(A), so A and B can never be true at the same time. Or, as Cantor did, that a secondary proofbycontradiction follows from it. But what people learn is invalid. And again, I never said that the two different sets of characters makes a difference. I have repeatedly said that the set using real numbers is different. "10000..." and "011111..." are different bit strings, but both are 1/2. The "properties of numbers" don't apply to the bitstrings. And the numbers do not represent the special case of Cantor's Theorem. And yet he still DID NOT REQUIRE THAT LAW, ALGORTITHM, OR ANYTHING LIKE IT. He only required "some determinant" method, and said it could be a law or algorithm. I agree that this is a fine point, but you are relying on that very fine point in your claim. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
I never said it was anything other than "fine." The point is that what is usually taught as CDA is this: Assume that a list of the entire set is possible. Prove that any list is missing an element E0. So that wasn't the entire set, as assumed. (Revisit this below) This contradiction disproves the assumption. This is invalid as a proof by contradiction, since the "assumption" that the entire set was in the list is not used to derive the supposed contradiction. So it isn't a contradiction, but it is a valid conclusion if you ignore the claim that it was assumed. But the actual proof is: For any countable subset that exists. Prove that this countable subset is missing an element E0. It follows that the entire set is uncountable, since E0 would have to both be in, and not be in, the set. And the reason this is important, is that most objections to the proof revolve around the point I asked you revisit. The "missing" string can be put into the list, so it isn't a contradiction. These doubters can't be reasoned with as long as they misinterpret the proof, and phyti is an example. He won't let go of that point. The reason I keep stressing that the "content" of CDA is not uncountability, is that one way to characterize this misinterpretation is that CDA actually depends on the argument itself being about a countable subset of M. Which I told you many times; you were also holding on to a misinterpretation of the argument, and only changed it when you read the logicmuseum translation. You were distracted from it by your objection to the Russell translation, missing the points that (A) Russell never required a Law but (B) translated every other part faithfully, including the removal of words like "manifold" that you object to. Which was my point in quoting it. The difference is: We need some gymnastics for that mapping to become a bijection, since in the trivial mapping "100000..." and "011111..." both map to "1/2." Many of the consequences that one could be tempted to put forward will deal with actual properties of numbers, which Cantor explicitly said have no part in the proof. And he is right. There is a better analogy. Let the 0's and 1's be "false" and "true." Each infinite string of 0s and 1s then defines a unique subset of the natural numbers. This way, CDA is an example, for infinite sets, of Cantor's Theorem. That the power set of any set A has greater cardinality that set A itself. The Theorem is trivial for finite sets, but it as important to show that it is true for an infinite set. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Cantor uses the set of all infinitelength binary strings, in the form he called En. He calls this set M. He also says that the proof does not utilize the properties of irrational numbers, something his 1875 proof was criticized for. This is why I do not use numbers, especially if the arguments resort to properties (like yours do). Turing has defined "Law" in such a way that the subset of M that can be generated by Turing's Laws is countable. Call this set T. Russell did not refer to T; in fact, he could not, since he did not know Turing's definition. Nor did he know that attempting a definition might make the set countable. I'm saying that this is irrelevant. RUSSELL DID NOT REQUIRE "LAWS" BY ANY DEFINITION, LET ALONE TURING"S "LAWS." He suggested that a law might be used to generate strings in what he called "some determinate manner." All he required was "some determinate manner." Call this set D. He clearly did this so that, for string En, he could say whether an,n was an "m" or a "w", rather than either an "m" or a "w". But we can make a list of T and a apply diagonalization to it to get a new string. Russell clearly thinks this process is using "some determinate manner" for generating a string. So what Russell would think, if he knew of Turing's work, and I'm not bothering to learn LATEX so "<" means a subset, is T < D <= M All I'm trying to say is that Russel's "oopsie" is in not recognizing the possible difference between D and M. Not between T and D. And that if this illdefined D is a strict subset of M, it is uncountable by the proof being discussed. I'm not saying that D is well defined by modern standards, or that Russell needed this step. I'm only saying that it doesn't invalidate the proof, and isn't a "howler." 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And my understanding is that the one at ... http://www.logicmuseum.com/cantor/diagarg.htm ... is closer to the original. This is because I have seen another translation that differed in only about five words. True, Meyer's has a more modern feel. But regardless, this is exactly what I have been summarizing the entire thread. And, there is one detail that I find important that Meyer leaves out. Where have you gotten the impression that I object to the methodology, or conclusion? All I have done is properly use the name "CDA" for the only part that, well, diagonalizes anything. The part between where Meyer's paraphrasing says "This follows from the following proposition" and "From this proposition it follows immediately." And then I call what "follows immediately" the corollary. Do you find anything less than "impeccable" about those labels? That CDA is Meyer's "this proposition" and the corollary is what "follows immediately"? I use this separation, because only the part I call CDA uses diagonalization, and conclusions that "follow immediately" from another are usually called corollaries. Or did Turing change that, too? I've even explained this, and summarized those two parts for you, more than once. I do get the impression that you don't read my objections, and only react to the fact that I objected to something in what you said. What I did object to, was where you said the content of CDA  the "proposition" Meyer described  was "uncountability." That is not mentioned in the "proposition" that I call CDA. In fact, it requires a countable set. What I did object to, in your characterization of Russell, was saying he added a "howler." He did add something that wasn't necessary, but it is not the restriction you keep saying it is. He only required that the characters au,v should be found "in some determinate manner." He did not define what that meant. Specifically, Turing's clarifications about "laws" do not apply to it. Because Russell did not require a "law." And Turing was clarifying what "law" should mean. Not defining what it did mean in 1903. Russell did give one example of "determinate manner. " It includes "the first p terms of Ep might be m’s, the rest all w’s." We can tell that this is an example, not a requirement, by reading the words "for example." He did not say that a "law" must be used. He said "Or an other law might be suggested." That's still as an example. How do you get "requirement" from "might" and "suggested"? And I said the same thing in "CDA: Any countable subset of M, the set of all infinitelength binary strings, necessarily omits a string E0 that is in M." Mine strengthens two points: I emphasized that "countable" is a necessary property of the list. Yes, it should be implied by "list." I made it explicit that the list is a subset, not the entire set. This is the part that phtyi is missing, and that is taught incorrectly. This is where Meyer departs from Cantor. Cantor said the rows of the table are "elements of M." Not "all elements of M." Russell said "any denumerable collection of E’s", not merely a table of characters. Meyer merely said "assume that there are". And the reason I keep pointing out these details, that I admit are implied in Meyer, is because they are the details that people who doubt CDA don't see. And I said the same thing in "Corollary: M is uncountable." 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
His "method" applies to any list that can be formed. The method by which is is formed is not relevant. Please, define what you think "complete" means here. Because it isn't what Cantor asked for. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And here is my (well, not so) quick response to your response. The passage is from Russell's 1903 book "The Principles of Mathematics." I only have it in PDF form where it starts on page 283. But in the document scanned, it starts on what appears to be page 532. And that document has notes in the margin indicating it was taken from page 365 of Russell's. I don't know what other versions might have been published. But it is in Chapter XLIII, "The Philosophy of the Infinite." In the introduction to the passage, Russell says "The proof [[Cantor's second on uncountability]] is interesting and important on its own account, and will be produced in outline." So not a formal rendering. But it translates, almost verbatim, what Cantor wrote. In clearer, English terminology. Which was my only point in posting it. Then, it is important to realize (and I don't think you have responded to this) that CDA and its corollary represent an informal proof only. Russell called it an outline, because he apparfently saw what Cantor wrote as an outline. It is an example of the far more formal proof that follows, that the power set of any set A has greater cardinality than that of set A. Specifically, if 'm' and 'w' are taken to mean "included" and "excluded" (and I don't know which, my German is limited), then each element of M defines a unique subset of N, the set of all natural numbers. And the reason I keep repeating this, is that I feel you are applying far to high of a standard to what Russell wrote. It updates some of the language in Cantor, almost literally. But neither is a formal proof. And Russell does not place, as restrictions, what you seem to think invalidates it as a formal proof: Russell does not say that we must use a "law" to define the strings in the list E1, E2, ..., En, ... . He merely says that the characters are to be found "in some determinate manner." He added a parenthetical comment that describes one such list, and says that "an other [sic] law might be suggested." THIS IS IN NO WAY THE REQUIREMENT YOU CLAIM IT IS. THIS IS A SUGGESTION FOR WHAT MIGHT BE USED TO MAKE A LIST. His goal for saying that appears to be twofold: To make sure that each En is different than any other. He states this explicitly, but it is actually unnecessary to the proof and Cantor did not say it. To make sure that we can "determine" each character an,n rather than just know it exists. And the reason I think you go too far, is that Russell clearly thinks that the antidiagonal is a string that is found "in some determinate manner." Turing's definition that "clarified once and for all by Turing in his famous 1936 paper" does not apply to what Russell wrote, because: Russell's concept was what Turing was trying to clarify. Russell did not require a "law," even if Turing's definition could be retroactively applied to what Russell meant. It is not part of a formal proof. This is the kind of statement that makes me think you are not reading anything that I have written, only cherrypicking for what you think I get wrong. You are assuming that I must be wrong when I contradict with you, and not looking at the context I use to contradict you. There is no part of Cantor's Second Uncountability Proof that involves adding elements to a countable list. That is the wrong context. I know that is the popular representation of it, but there is no part of the proof that says anything like it. I even outlined it before: CDA: Any countable subset of M, the set of all infinitelength binary strings, necessarily omits a string E0 that is in M. Corollary: M is uncountable. CDA (the first part above) does not mention uncountability, or add any elements to a list. It shows that any list that can exist (Russell's "in some determinate manner") is missing one. I contradicted you to point out that uncountability is not mentioned, not to say your final conclusion is wrong. CDA (the first part above) is not a proofbycontradiction. It is a direct proof, that "determines" (al a Russell) an element E0 that is not in any list that can be provided. Cantor never suggests adding E0 to the list. That is a Freshman's initial thought, when CDA (the first part above) is treated as the complete proof, and as a proofbycontradiction. That alleged contradiction is "This list that we assumed to be complete, is not in fact complete." The Freshman will have the kneejerk reaction that we can just add E0 to the list, and so remove the contradiction. The part that makes this an invalid proofbycontradiction is that saying "we assume the list is complete" is not enough  you have to use that assumption to prove the contradictory statement. And CDA does not. This "determinate manner" Russell mentions is not subject to Turing's definition. It was a vague reference to what Russell thought could exist, at the time he wrote those words, and that Turing later clarified. The corollary is not proven by adding the antidiagonal element to the list. It is a proofbycontradiction. If a listing of M could exist, then there is an E0 that both is, and is not, a member of M. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Massaging Wikipedia's informal definition (from the introduction) of a computable number, a computable string "can be computed to any desired number of characters by a finite, terminating algorithm." Russell's restriction was "[The characters comprising the strings E] are each [calculated by] some determinate manner ... which insures that the E’s of our series are all different." You are the one who translated this very nonspecific statement into "computable string." I do not disagree that the set of all computable strings, which I will call L1 (I trust you can see where I am going) is countable. So there exists a bijection between L1 and the set of all natural numbers. What CDA says is that there is a string E0 that is "determined" from this bijection, that is not in L1. Which means that it is not computable. My point is that this string E0 can satisfy Russell's intended restriction, even though it cannnot satisfy yours. It is found "in a determinate manner" as described by CDA. So there is another set L2 where every string meets Russell's restriction, but some do not meet yours. Your L1 is a strict subset of L2. L2 may, or may not, be all of Cantor's M. It may, or may not, have a useable definition. But if it does, it is uncountable. No, my argument here is that you are adding meaning to Russell's argument that he did not intend. For the obvious reason that the concepts required for this added meaning would not be developed for decades. My argument is that you can't blithely translate his vague statement into a formal definition that was beyond his time, and then turn that formal definition back on his vague statement to claim it invalidates his informal proof. He did not define what he meant by "some determinate manner" because he did not know, envision how, or care if it could be done. He was merely saying that one could "determine" every character an,n and subsequently every bn. Yes, there is a character in each randomlyaccessed position, and we don't need to "know" it to say we can invert it. But the difference between knowing it is there, and actually having it for such manipulation, is close to the subject of the Axiom of Choice. Again, not my area, but I don't know how well Choice was understood in Russell's time. My impression is that he wanted to advance the known state of the diagonal character an,n above "we know it is there and everything it can be." And that is Choice, as I understand it. +++++ But my main point is, and always has been, that the set of strings in the list used in the actual "diagonal argument" part of the uncountability proof is not the entire set M. What I consider to be "CDA" does not mention whether M is, or is not, uncountable. That's in a corollary. And to avoid drivel like what phyti spouts, we should make this clearer in all discussions of CDA. So: wtf: "How many countably infinite strings (decimal or binary makes no difference) are there? Well, there are uncountably many. That's the content of Cantor's diagonal argument." No, that's the content of the corollary to CDA. CDA: Any countable subset of M, the set of all infinitelength binary strings, necessarily omits a string E0 that is in M. Corollary: M is uncountable. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
He does if it is to be a defined object whose properties are required. Sure, if you want to call it that. Cantor doesn't, and no properties associated with "geometry" are required by any object in the proof. By this standard, he is also using vectors. Also unimportant. Cantor uses a "partial list" because that is what his proof needs at that point. You refuse to understand this, which is why you will never understand CDA. He uses it to prove only: "If E1, E2, …, Ev, … is any simply infinite [einfach unendliche] series of elements of the manifold M, then there always exists an element E0 of M, which cannot be connected with any element Ev." Please identify the words that you think support this in Cantor. Or Russell. Or Wikipedia (which, incidentally, includes the note "Cantor does not assume that every element of T is in this enumeration." So, do you want to admit that you do not understand infinite sets? The complete set of all natural numbers can be ordered, and has no last member. I'll skip your incorrect analogies based on this fallacy. I didn't ignore it. I explained why I could dismiss it. But again: Every string in Cantor's sequence of strings is identified with a unique natural number n. Cantor's string b1b2...bn... differs from the string identified with n in position, and so is different that every string in this sequence of strings. So it is not in this sequence of strings. This method explicitly rules out any possibility of a duplicate. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And I still think you do not understand my point. Which was only that "Russell's restriction of the elements of the list" does not invalidate the construction of such lists. Not that there can't be an issue with the uncountability proof, WHICH IS A COROLLARY TO CDA. And the reason for making this subtle distinction, between CDA and this corollary, is that almost every objection to CDA that I have seen is based on misunderstanding it. The sets used in the "diagonalization argument" part of the proof, are always sets that are denumerable. The full set M may or may not be. (And btw, M never contains real numbers, and CDA never mentions real numbers in any context except to say they aren't used. I know that there are relationships between the two, but I feel the discussions of CDA need to be about strings.) So a restriction that limits these lists to denumerable sets is not, by itself, a problem. But I also said that Russell went too far by stating this restriction, because we now know that there elements in his set M that cannot meet the restrictions. What I said was that his restrictions seemed necessary to him at the time he wrote it, in order to have access to all of the characters an,n. Where the potential problem lies is in the difference between L, the set of strings that can be found by his laws, and the set M that contains all strings. This isn't my area, but by my understanding what CDA (the actual argument, without the corollary) says is that applying diagonalization to any infinite list containing elements of L will produce a string that is not listed. Whether or not this string is in L is open to debate as far as I know, since it is "found in some determinate manner" but not in a finite number of steps like what your objection is based on. Maybe this is where we aren't seeing the objection the same way? If it is in L, it proves that L is uncountable, as shown in the corollary. The L that you think his restrictions define is countable. So the "finite steps" part seems like the issue. All that theory comes postRussell, and I agree he didn't know the ramifications. Or consider that there could be. I just don't think your arguments apply in the way you think they do. But I remind you, again, that this proof was never meant to be a formal proof. It is an example of Cantor's Theorem, that the power set of any set A has a greater cardinality than set A. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
No you didn't. You read looking for something you wanted to see, and you added it. My interpretation is exact. Yes he did. But a better description is that row u is the "element" Eu that is a string in the set M. (And note that he mixes u and v up sometimes). What you want to add seems to be that every element of M is associated with a natural number u, and that simply is not true. Cantor's "sequence" is the list E1, E2, …, Eu, … . It is not a string  he never uses that word, but modern version do for what Cantor calls an "element." In other words, "sequence" and "string" mean different things, so this is a very confused statement that dose not correspond to Cantor. No, the coordinate pair (u,v) references either an 'm' or a 'w'. If you want to use Wikipedia, they don't refer to coordinates, per se, but do use '0' and '1'. If you ant to claim you are using Cantor, please stick to one nomenclature, or indicate (as I try to do) which and how it fits with the other. I have no idea what this gibberish (and what follows) is supposed to mean. Why does '9' get into it? This bears no relationship to anything Cantor did. This does: For proof, let there be E1 = (a1.1, a1.2, … , a1,v, …) E2 = (a2.1, a2.2, … , a2,v, …) Eu = (au.1, au.2, … , au,v, …) …………………………. where the characters au,v are either m or w. Meaning: E1, E2, ..., Eu ... is a list of elements from M. Not a list of every element in M. I call this set of Eu the set E. It is a subset of M, but at this point it may be an improper subset (meaning all of M). Then there is a series b1, b2, … bv,…, defined so that bv is also equal to m or w ... ... if av.v = m, then bv = w And by implication, if av.v = w, then bv = m. Meaning: This is a new series of characters that fits the definition of an element in M. Cantor calls this new element E0, not s. But, while Wikipedia does call it s, it is not your s. Yours starts at "coordinates" (11,1). This really doesn't matter, except that you are not using Cantor. then one sees straight away, that the equation E0 = Eu cannot be satisfied by any positive integer u. Meaning: E0 is not in E. But, from above, it is in M. This is not a contradiction, since nowhere, so far, does Cantor say E=M. This proves the "proposition" that CDA wants (and I fixed the mixedup uses of u and v): If E1, E2, …, Eu, … is any simply infinite series of elements of the manifold M, then there always exists an element E0 of M, which cannot be connected with any element Eu." That's it. That's CDA. It does not do this: There is no "geometry" in Cantor. You can't find any words in Cantor that have that meaning. What you refer to is, indeed, the visual aid you claimed others use. Cantor never draws a diagonal. Not one starting at (1,1), nor one starting at (11,1). There is no property associated with being a "diagonal," other than it also being a string. You can't find any words in Cantor that "define it as a different thing with different properties." In fact, the closes he comes is to say it is the same kind of thing. I left that out above so I could put it here: Then consider the element E0 = (b1, b2, b3, …) of M, These b's come from the characters you can find along a diagonal (but again, Cantor never used that word). Now, if you think any of this is wrong, please point to the words in Cantor that confirm your opinion. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
I don't think I emphasized this lie enough in my pointbypoint debunk of phyti's complete misrepresentation of CDA: As a recap, Cantor first defines the set M. (I'm adding the emphasis  bold italic  to indicate an infinite set. This distinguishes the symbol for a set from those representing its elements.) It is the set of all possible binary strings. He is consistent in calling each such string an "element," not a "sequence," and he uses either E or Ev (where v is a natural number) for these elements. What Cantor does, is first say that that he is using only a subset of M, which I call E, that has the property that it can be put into an infinite list E1, E2, ... Ev, ... . Cantor never says whether this set E does, or does not, represent all of M. Only that it is countable. Phyti seems to think it does, and keeps implying that E=M without having the integrity to come out and say that this is what his claims are based on. Cantor then uses the algorithm that was, only after his proof, called his" Diagonal Argument." Or CDA. He uses it to construct the element (or string) he names E0: "Then consider the element E0 = (b1, b2, b3, …) of M ... " (you all know the rest). Please note that Cantor explicitly says that E0 is an element of M. This fact can't be missed. What phyti is lying about (I don't believe that he lacks the ability to understand this, so he must be intentionally misrepresenting it) is this: "The equation E0 = Ev cannot be satisfied by any positive integer v." In the quote at the top of this reply, phyti is claiming that in this sentence "Cantor declares E0 is not in M." Cantor is not saying that. Cantor is explicitly saying that E0 is not in E, the set listed in E1, E2, ... Ev, ... . The only way phyti's interpretation is true, is if we require E=M. That is never a requirement in the proof. If we admit that it is not in the proof, phyti's entire argument evaporates. The lie is refusing to recognize that (A) Cantor defined E as having elements from M, not being equal to M, (B) Cantor makes this crystal clear by saying that E0 is in E but not in M. and (C) then having the audacity to insist that Cantor is saying that E0 both is, and is not, a member of M when no such statement is made. in this part of the proof. It is made later, to prove that any that E meets the countability requirement cannot ever be equal to M. In other words, the very contradiction that phyti claims debunks CDA, is what it uses correctly to prove that there are uncountable sets. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
"It", meaning the diagonalization process, is about what subsets (and subsets can be "improper," meaning they equal the parent set) of Cantor's set M can be put into a surjection from the set N={1,2,3,...v, ...}. That is literally what this means: "If E1, E2, …, Ev, … is any simply infinite [sequence] of elements of the manifold M ..." What "it" is about, is the conclusion he proves, and that you doubt: "... then there always exists an element E0 of M, which cannot be connected with any element Ev." What "it" is used for, is this corollary: "From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing E0 would be both an element of M, but also not an element of M." What CDA is not about, is anything Cantor said in a paper about his philosophy. This is a different paper entirely. Drop it. A list is not the same as a set, and it is not "only a visual aid." It is the tool by which Cantor constructs the element he calls E0. A list is a surjective function from a set of consequential natural numbers {1,2,3,...v} or {1,2,3,...v...} to the set of interest. It usually is injective also, but that is not necessary here. And, while it may seem to be implied, Cantor did not require it. Russel did. Your issue is that you do not understand CDA, and you find it nonintuitive. So you look for statements that could be wrong without bothering to verify that those statements exist in CDA. They do not. These statements are part of the narrative describing what he wants to accomplish, not the proof itself. Regardless of how you choose to misinterpret them, they have no bearing on the errors you think you see. Sure. Irrelevant to CDA, but they can. The more appropriate division of sets are E, (my name for the subset he puts into a list; that is, E1, E2, E3, ..., Ev, ...) and ME, the elements in M but not in E. The important observation is that E is countable, but at this point M and ME may or may not be. By using the list structure you called "only a visual aid", Cantor constructs an element E0 that is proven to not be in E. It is not a sequence, it is an element. It is member of M but not a member of E which is put into a sequence. No error here. Get the terminology straight. Cantor calls the members of M "elements," not "sequences." Another possible name is "string." "Sequence" means the list of these elements that make the list you deny is important. There is no meaningful context for orientations of sequences, or strings. No error. Exactly. This word "diagonal" has no meaning to the proof. It is only the visual aid you misinterpret. No error. The contradiction is not a part of diagonalization. It is the "corollary" I listed above. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
You don't need to explain  I understood it from the start. You are not understanding mine. The restrictions Russell places, are placed on elements that he requires to comprise a countable set. Your objection is misplaced, because what you object to is that he cannot generate an uncountable set under those restrictions. That is irrelevant. In what way does describing how to generate a countable set invalidate the countable set that is generated? I'm not trying to "get the last word." I am asking a legitimate question about what you keep repeating. I am pointing out exactly what I think it is that you are either not seeing, or ignoring. CDA only ever makes use of a countable set. CDA proves that any such countable set cannot be the full set M (whose countability is not yet determined). Only after drawing a conclusion about countable sets, does Cantor conclude that M is uncountable. Because any subset that is countable  and as you seem to admit can be generated by the rules he described  is missing at least one element of M. And again, since you keep missing it, the rules Russell describes are never associated with the entirety of an uncountable set. No, you have discussed it for a version of that passage that is different than what Russell wrote. I don't know how I can make this any clearer: Russell's passage describes a countable set. Literally, "For let us take any denumerable collection of E’s..." The passage places restrictions on how to generate this countable set. The only place these restrictions are used is for this "collection of E's." You say it is an error because these restrictions cannot generate an uncountable set. I understand why it would be an error, if the passage was trying to generate an uncountable set. I don't understand why you think it is an error when it only needs to generate a countable set. I am suggesting that you, like phyti and what I believe is at least 99% of the people who think they understand CDA, do not fully realize that CDA is only ever applied to a countable set. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Nor will I call Russell's restrictions on how to set up the first part of that a "howler" simply because you think they don't allow for the list to be uncountable. They are supposed to do that, in part #1. I do think it wasn't necessary to do so, all he needed was to establish that (A) such lists were possible and (B) we can reference the nth character in the nth string. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
I've described it several times. The "diagonal argument" in CDA is the proof that any countable list of infinitelength binary strings will necessarily omit at least one such string. This does not mention the set of all such strings, that Cantor called M and Wikipedia called S, except to say that the constructed string is a member. It does not mention "uncountability" in any way. The content you refer to is not in the argument. It is in the followup. Yes, I am making a fine distinction. No, I don't think you misunderstand this distinction at its core. But I believe it to be the point that every "Cantor Doubter" falis to understand, because the two arguments Cantor made are usually taught as one. Like how phyti misunderstands it. And continues to demonstrate the misunderstanding. You can see it, three posts higher, in his claims about being able to add the string he called "sx" to the list he worked with. He is still thinking that the listed set is "supposed" to be M, and he is claiming that any alleged "missing" string can be put into this "supposedly complete set." So that this disproves the uncountability of M. He somehow can't, or won't, recognize that the listed set is not supposed to be M, so adding sx to the listed set has no significance. All I'm trying to make clear is that the "content" you refer to, as CDA, is the first of these two, separate arguments: Any countable set of infinitelength binary strings (Ei, for all i>=1) will necessarily omit at least one such string E0. The complete set of infinitelength binary strings, M, must be uncountable; otherwise, there would be a string E0 that both is, and is not, in M. Note that the "contradiction" usually associated with CDA, with these two arguments combined into one, is that the "assumed to be complete" set M is found to not be complete. But the contradiction Cantor actually used in the second argument only is about E0, not M. And no, I won't apologize for Cantor not making this distinction clearer. THIS IS NOT THE THEOREM HE WANTED TO PRESENT. It is an example of that theorem. It is an example of how the power set of an infinite set like N={1,2,3,4,...} cannot be put into a bijection with N. Find the rest of the same paper to see its proof, which is far more formal. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
Bzzzzt. Wrong. Read it again. The diagonalization algorithm uses elements from M, but is "within the context" of a subset that is known to be countable. But thank you for playing. Come back when you understand CDA. "If E1, E2, …, Ev, … is any simply infinite [sequence] of elements of the manifold M, then there always exists an element E0 of M, which cannot be connected with any element Ev." 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
No, I think I was quite clear that I was replying to where you attributed "content" to CDA that simply is not there. You seem to think I am arguing with you about the validity of CDA; I'm not. And you are misrepresenting what I said when you describe how I am arguing with you. The result is that I'm not sure if you understand what I said, or what Cantor (and Russell) said. But I said the sequence was denumerable. Each string of binary digits is also denumerable, but that isn't what I said. It is, hoever, how you seem to have read it. The list of such strings (or sequence of such strings) that is subjected to the process we (but not Cantor nor Russell) call diagonalization, is denumeable. Maybe Russell did add content that can only make whatever it refers to denumerable. So what? You are the one creating a side argument that has nothing to do with the thread. My point, in presenting Russell's version, was that he used language closer to modern set theory. But in a similar state of confusion, what Russell said as that each string follows a law. He wasn't as specific as you imply about what this means. We use a summation that converges to, say, the binary representation of pi as such a Law. You seem to think it is something like arbitrarily large polynomials, which form a denumnberable set. And it is still unclear which set (the string, or the sequence of strings) you think has this restriction. Or why it matters, since in the proof both are supposed to be denumerable. And I was trying to point out that Russell was adding a point that Cantor should have made, and maybe he went too far the other way. And that calling it out demonstrates that you, in addition to phyti, are not demonstrating that you understand what I am saying. And again, I don't get that you see my point when you say this. THE SET OF STRINGS USED IN DIAGONALIZATION IS ALWAYS A COUNTABLE SET. The point is that ANY COUNTABLE SET OF STRINGS IS NECESSARILY MISSING AT LEAST ONE STRING. So what it the way he defines the set of Laws, that define the sequence can only produce countable sets. They are supposed to be countable sets. I never said you did. I said you objected to what Russell added, as you confirmed here. I said Russell went too far, but that an addition was necessary. And I speculated that Russell, not I, might have wanted to avoid Axiom of Choice issues. The point of his addition was to be able to "choose" each digit in the string he constructs. But I don't know what objections he was trying to avoid, I'm not a mind reader. This one seemed like a possibility. Wikipedia provides one. That didn't seem to satisfy you, so I provided an intermediate. Again, please explain how it is invalidating to say that a denumerable sequence is defined by a denumerable algorithm. Note that I'm not agreeing, or disagreeing, with your claim about the algorithm. No, I'm trying to convince you that what most people think is CDA is flawed. But it is not what Cantor argued, so arguing about those flaws says nothing about CDA. Most "disproofs" of CDA are based on either thinking diagonalization is being applied to an uncounatble set, or to the incorrect (yes, it is) way the proof is taught as a proof by contradiction. And again, note that I am not saying CDA is invalid, or that what is taught can't be used in a valid proof. I'm saying that what is taught is invalid as the complete proof, and give the wrong impression to many. Like phyti, who can't seem to understand what is has wrong. A rough outline, with a clarification of some points that are omitted but implied, is: Assume the statement (A&B) is true, where: A = The set S contains all possible strings. B = The set S, which contains only strings, is denumerable Use Diagonalization on set S, based on statement B alone, to prove that statement A is false. So (A&B) is also false. By contradiction, (A&B) cannot be true. So when A is true, so B must be false. This is a jumble of questionable logic. But if one does not recognize that the typical opening statement that is taught as CDA ("The set R of real numbers in [0,1) is countable") is really two statements, both A and B, then most of it seems sorta kinda valid. And so it gets taught as the archetype of proof by contradiction. Cantor's actual argument is For any set S that satisfies B, Diagonalization proves that that there is a string s0 that is not in S. Assume A&B Set S satisfies B. So there is a string s0 that is not in S. Set S satisfies A. So by definition s0 is in S. Contradiction. A&B is false. The original German is translated here: With one word changed for consistency (Cantor's German used different forms of the same word), the two bits of interest say what you want, if a bit stiffly: "If E1, E2, …, Ev, … is any simply infinite [sequence] of elements of the manifold M, then there always exists an element E0 of M, which cannot be connected with any element Ev." "From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing E0 would be both an element of M, but also not an element of M." 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And from that I'm pretty sure that you are not following my point. Or Cantor's, or Russell's. By "Cantor's diagonal argument" I mean this part in Russell: This does not have the "content" you refer to. It refers to M only to say that Ep is a member. It does not refer to the countability, or uncountability, of M. This part is imbeded in a larger proof that does refer to M, and uncountability, but this actual "diagonalization" part does not. I find two shortcomings in CDA. The first is similar to your complaint, but it is not that it uses arcane language. It is informal, which leads to Cantor using the vernacular language that you find arcane. It can be informal because it isn't the theorem that the paper was presenting. That was Cantor's Power Set Theorem, and CDA is only an example of it. Llike you might show a (3,4,5) triangle before proving the Pythagorean Theorem. This is why CDA is called an argument and not a proof or theorem. The second is that Cantor did leave one thing out. The above passage applies to any list that can actually exist, not a hypothetical list "assumed for contradiction." Cantor really should have demonstrated that such lists exist, and he didn't. Examples are trivial (Ep is all m's except a w in position p) so this can be forgiven. What Russell did was to rectify this omission. He describes a more formal way to create many such lists. He may have gone too far, by limiting his lists to strings that followed a law. I suppose he was trying to get around any Axiom of Choice issues; each character in the diagonal can be found by its law. But: HE DID NOT SAY THIS REPRESENTED ALL POSSIBLE LISTS, OR THE SET M. One thing Russell included that Cantor did not, is that each string does not have to be different. They can be so limited, but the point is to show that there is no surjection from N to M. 
problem with cantor diagonal argument
JeffJo replied to phyti's topic in Linear Algebra and Group Theory
And ... the sequence he is talking about is denumerable.