Cantor Confusion
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From: Lester Zick on 16 Dec 2006 14:25 On 15 Dec 2006 14:56:27 -0800, "Jonathan Hoyle" <jonhoyle(a)mac.com> wrote: >> In other words, Jonathan, Aristotelian syllogistic inference only >> yields truisms not truth. It can't tell us what the truth of anything >> actually is only whether or not it's true in relation to constituent >> premises of equally problematic truth. > >I cannot imagine how anyone would have expected it any different. OF >COURSE the truth of a statement is based upon the truth of its >premises. Actually the truth of a statement only rests on the falsity of its alternatives. If you regress the truth of a conclusion to the truth of premises you have achieved very little in exhaustive terms unless those constituent premises are themselves in fact true. Yet the truth of those premises remains as problematic in exhaustive terms as the truth of the conclusion itself to begin with. > I don't believe Aristotle thought his work failed to >"produce the truth he hoped it would yield". I think you have >unrealistic expectations from logic. And I disagree.Historically metaphyscists of the time were preocuppied with one primary problem, the truth of everything. And Aristotle developed his organon or instrument of syllogistic inference precisely for that purpose, to analyze and demonstrate the truth of everything. That's what drove him to pursue discovery of ultimate basic premises. He realized that in the context of syllogistic inference determination of any truth rested squarely on the truth of ultimate basic premises from which various conclusions could be derived. That's where empiricism comes from, the search for true ultimate basic premises. >Were you imagining that Aristotle was planning to come up with a system >which would determine the truth or falsity of a statement A, >irrespective of any premises? What if A = "God exists."? Or how about >A = "John is handsome."? I think Aristotle like other metaphyscists of the time was driven by exactly that. Truth of constituent premises is driven by conformance with the truth of conclusions and not vice versa.If a given conclusion is true you can only spread the truth of that conclusion in certain ways across whatever constituent premises you claim. That's where the "B" in "if A is B and B is C" comes from. Assuming "A is C" that truth can only be spread to constituent premises through some common middle term whose differences with A on the one hand and C on the other account for differences between A and C accurately and exactly. >> The truth of conclusions is not >> inherited from the truth of constituent premises as syllogistic >> inference implies and Aristotle thought. The truth of any conclusion >> must be demonstrated in its own right by demonstrating the falsity of >> alternatives by finite tautological regression to self contradictory >> alternatives. > >Ridiculous. How would you apply such a principle to the Reimann >Hypothesis? Or to something non-mathematical, like a non-reproducible >historical fact like "the U.S. went to the moon." Without *some* >assumptions (or in mathematics, axioms), you cannot arrive to any >conclusions of truth. (In the aforementioned example, there are >conspiratorialists who deny some of the more basic assumptions that you >might not realize you are even making.) So I"m not only supposed to address the demonstration of truth but actually to correct all the intellectual baggage of the last several thousand years in twenty questions? At least my assumptions regarding demonstrations of truth turn out to be true and are demonstrably so whereas your assumptions of truth turn out to be nothing but facile truisms and demonstrably nothing more. I'm actually explaining what empiricism is, where it comes from, why it is what it is, and how it came to be. You can bellyache all you want but that's the truth. Just take a look at your exclamation "ridiculous" above. That comes from a sudden realization that there is more than one way to examine and determine truth in mechanical terms and that your entire adult life you may have been looking at the problem of truth the wrong way around. I'm not in the business of making assumptions whose truth cannot be demonstrated. That's what mathematikers are for. >Perhaps I misunderstand where you are coming from, but I very much >doubt that Aristotle held the expectations you had for logic. I don't hold any expectations for Aristotelian logic and syllogistic inference which I haven't pretty much already covered. It is what it is and it isn't much more. The fact that syllogistic inference has been used in science and math uniformly since its discovery is more a testament to its exposition and formal systemization by Aristotle than any actual correctness. I don't know whether you misunderstand me or not but I assure you I'm not just out here beating my chops for amusement. Aristotle just got it exactly backwards and I'm here to correct the problem by reducing the problem of the truth of everything to its ultimate foundation in a finte tautological regression to self contradictory alternatives. ~v~~
From: mueckenh on 17 Dec 2006 04:06 Virgil schrieb: > We see it but do not believe that it does what you say it does. > > Your example assumes a bijection and constructs a "rational relation". I assume nothing but that every path gets a share of that edge through which the path runs. You cannot deny that an edge must exist if a path runs through it. I prove that there are enough shares to build 2 edges per path. > But in the absence of any original bijection there is no assurance of > any final bijection by such construction. Without any original bijection we can conclude from 1.1 ---> a 1.2 ---> b 2.1 ---> a 2.2 ---> b and the knowledge that 1.1 U 1.2 = 1 and 2.1 U 2.2 = 2 than there are as many nunmbers as letters. > And assuming what you are trying to prove is a no-no. no * no = yes Regards, WM
From: mueckenh on 17 Dec 2006 04:09 Virgil schrieb: > > > > Do you believe that paths beginn to run separated wihout any edge being > > > > inolved? > > > > > > No. And so what? But only looking at the paths that way you never get > > > the full infinite paths. > > > > If we look at least at one part of each path, then we look at all paths. > > Not so. To separate any infinite path from /all/ other infinite paths > you must look at ALL its edges, not just some of them. > The first n edges play no role. > Regards, WM
From: mueckenh on 17 Dec 2006 04:12 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing that > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > > > > > > that is required is: > > > > > > > Given any line there is an element of the diagonal not in THAT line. > > > > > > > It is not requires that: > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the diagonal > > > > > > differs form line A by element b and from line B by element a, but a is > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not in > > > > > > THAT line. It is not required that: There is an element of the diagonal > > > > > > that is not in any line." is inapplicable for linear sets. You see it > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > the diagonal is the last line. > > > > > > > > Every example with natural numbers (finite lines) is a finite example. > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > There is no last line > > > > > > Then, there is a line that comes after L_D. > > > > > > Therefore :L_D does not contain every element > > > that can be shown to exist in the diagonal. > > > > All elements that can be shown to exist in the diagonal can be shown to > > exist in one single line. > > > > Call it L_D > > L_D contains a largest element. n. > > L_D is not the last line, so there is > a line with element n+1, > > Element n+1 can be shown to exist in the diagonal. Element n+1 can be shown to exist in L_D (which is obviously a line containing n+1). All elements that can be shown to exist in the diagonal can be shown to exist in one single line. The easiest way to see that would be to name all elements of the diagonal. Every element is a natural number and as such a set containing all the smaller natural numbers. In fact it is irrelevant whether you write 1 2 3 or 123. If the diagonal contains only finite segments, then each one is in a line and vice versa each line is in the diagonal. Only if there is an infinite segment in the diagonal, then it is not in a line. But in order to have the infinite segment, you need an element which is not in any line, i.e., an infinite element. Since an infinite element is not a natural number, you have not such an element available, neither in the diagonal nor in any line. Regards, WM
From: William Hughes on 17 Dec 2006 07:25
mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing that > > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > > > > > > > that is required is: > > > > > > > > Given any line there is an element of the diagonal not in THAT line. > > > > > > > > It is not requires that: > > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the diagonal > > > > > > > differs form line A by element b and from line B by element a, but a is > > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not in > > > > > > > THAT line. It is not required that: There is an element of the diagonal > > > > > > > that is not in any line." is inapplicable for linear sets. You see it > > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > > the diagonal is the last line. > > > > > > > > > > Every example with natural numbers (finite lines) is a finite example. > > > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > > > There is no last line > > > > > > > > Then, there is a line that comes after L_D. > > > > > > > > Therefore :L_D does not contain every element > > > > that can be shown to exist in the diagonal. > > > > > > All elements that can be shown to exist in the diagonal can be shown to > > > exist in one single line. > > > > > > > Call it L_D > > > > L_D contains a largest element. n. > > > > L_D is not the last line, so there is > > a line with element n+1, > > > > Element n+1 can be shown to exist in the diagonal. > > > Element n+1 can be shown to exist in L_D (which is obviously a line > containing n+1). No. L_D is bounded. The largest element of L_D is n. L_D does not contain n+1. - William Hughes |