Why can no one in sci.math understand my simple point?
in [Logic]
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From: |-|ercules on 19 Jun 2010 16:03 Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote ... > Newberry <newberryxy(a)gmail.com> writes: > >>> Because every infinite sequence of digits represents a real number? And >>> the antidiagonal is one such sequence? >> >> If it does not exist then it does not represent anything let alone a >> number. >> >> Now it is clear that it does not exist. Since all the reals are on the >> list and the anti-diagonal would differ from any of them. This >> violates the assumption. Hence the anti-diagonal does not exist. > > Wow. Are you saying that the *sequence of digits* specified by the > anti-diagonal does not exist? > > Anyway, in a sense, you're right. If we assume that every real is > represented by a sequence on the list, then we can prove that every > sequence occurs on the list (ignoring the issue of multiple > representations). And yet, we can also show that a particular sequence > is not on the list. This is a contradiction and hence *our assumption* > that every real is represented by a sequence on the list must be false. > > You've never seen proof by contradiction in your life? How remarkable. > You see the words but you don't read. Cantor's proof should work if you give a definition for ALL antidiagonals instead of a SPECIFIC antidiagonal. CONSTRUCT A REAL An AD(n) =/= L(n,n) PROOF THAT IT'S NEW An AD(n) =/= L(n,n) THEREFORE [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Higher Infinity Do you agree with the above proof? Herc
From: |-|ercules on 19 Jun 2010 16:10 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 19/06/2010 7:07 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... >>> On 19/06/2010 4:11 PM, |-|ercules wrote: >>> >>>> To support your argument you should at least show that you've formed a >>>> new sequence of digits. >>> >>> I'll explain it simply then. The first digit of the created number >>> differs from the first digit of the first number in the list. The >>> second digit differs from the second digit of the second number in the >>> list. >>> >>> In general, digit n differs from digit n of the nth number in the list. >>> >>> So for all n, the created number differs from number n. Therefore the >>> created number is not in the list - it is a new sequence of digits. >> >> No I've told you all 20 times that does not create any new sequence at all. >> >> All you've done is >> CONSTRUCT a digit sequence like so >> An AD(n) =/= L(n,n) >> >> And then you say, it's different to each number like so >> >> PROOF >> An AD(n) =/= L(n,n) >> >> But you have not demonstrated a NEW SEQUENCE OF DIGITS. > > How can it not be a new sequence of digits if it's not in the list? > >> >> All you've done is this >> >> [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Superinfinity >> >> Your actual 'proof' is a specific example of the above 'proof'! >> >> [ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Superinfinity >> >> Do you agree with the above version of Cantor's proof? > > That is not a statement of Cantor's proof. For a start, it leaves out > the assumption that the list of numbers is a list of all the reals. > >>>> >>>> If you actually read my derivation of herc_cant_3 instead of blindly >>>> dismissing it, >>>> you'll see it holds, just like all digits of PI appear in order below >>>> this line, if interpreted >>>> correctly. >>>> >>>> Herc >>>> >>>> ___________________ >>>> >>>> 3 >>>> 31 >>>> 314 >>>> 3141 >>>> ... >>>> >>>> >>> >>> herc-cant-3 is not a derivation. It's a wild leap of faith. Nothing is >>> proved therein. >>> >>> Sylvia. >> >> >> Then which step do you disagree with? >> >> >> defn(herc_cant_3) >> The list of computable reals contains every digit (in order) of all >> possible infinite sequences. >> >> Derivation >> >> Given the increasing finite prefixes of pi >> >> 3 >> 31 >> 314 >> .. >> >> This list contains every digit (in order) of the infinite expansion of pi. >> >> Given the increasing finite prefixes of e >> >> 2 >> 27 >> 271 >> .. >> >> This list contains every digit (in order) of the infinite expansion of e. >> > > This one: > >> Given the increasing finite prefixes of ALL infinite expansions, >> that list contains every digit (in order) of every infinite expansion. > > You provide no justification for that statement. It doesn't follow from > what came previously. You just assert it. > > Sylvia. If all digits of a single infinite expansion can be contained with increasing finite prefixes, and the computable set of reals has EVERY finite prefix, then all digits of EVERY infinite expansion are contained. Herc
From: WM on 19 Jun 2010 16:42 On 19 Jun., 16:00, Sylvia Else <syl...(a)not.here.invalid> wrote: > ZFC makes claims in the context of ZFC. You can't take it down using a > different set of axioms, because ZFC doesn't make statements under those > other axioms. If you want to attack ZFC, as distinct from inventing > competing sets of axioms, your only viable course is to seek to show > that it is inconsistent. There are things more elementary than ZFC. Induction for instance: If a list is constructed as follows: 0.0 0.1 0.11 0.111 .... then the anti-diagonal p = 0.111... does not exist or it is in one and the same line of the list. Proof: All digits of p are in the list within their appropriate columns. Otherwise they could not constitute p. So either all digits are within one line (containg p) or not. If not, then there must be at least two lines (or more) containing digits of p but not all ofd them. Now we prove by induction: If n lines contain digits of p then one of these lines contain all digits that are contained by those n lines. Therefore n lines can be substituted by one line. This can be proven for all finite n. But there are no other than finite line nunbers. Hence the proof is complete. The usual "refutation" of this argument is: There cannot be shown two lines with different digits of p because an infinity of lines is required. But it is obviously not sensible to claim an infinitude of lines while not even being able to show two lines which are required. No, for a set in linear order it is not possible to get the digits of p distributed over more than one line. So, if all are there, then all are within one line. This proves that "not all are there". Finished infinity is nonsense. Therefore aleph_0 und uncountability should be recognized as what they are: pure selfcontradictions. In particular because it is clear that set theory has not any single application in science including mathematics. (if it had, one could easily see it fail. But it cannot fail because it does yield any concrete result apart from the "hierarchy of infinities" that has the same relevance as Russell's tea pot). Regards, WM
From: Virgil on 19 Jun 2010 18:13 In article <e21e5c3a-7a2a-4722-987b-7f46e33cf3dc(a)5g2000yqz.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > Cantor's "proof" shows either that a countabke set is uncountable or > that a number that can be distinguished from every other number cannot > be distinguished from every other number. Only in WM's world. In the real world things are quite different.
From: Virgil on 19 Jun 2010 18:17
In article <9473bde7-4b41-4e83-9d5a-b8a02892e592(a)z10g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > All finite definitions of numbers map on infinite sequences of digits. > But infinite sequences of digits do not map on finite definitions. Sure they do, at least for many meanings of "map". |