Why can no one in sci.math understand my simple point?
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From: Tim Little on 23 Jun 2010 23:45 On 2010-06-23, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Tim Little <tim(a)little-possums.net> writes: >> You were the one applying the term "recursively enumerable" to a >> function. I was merely correcting your mistake. > > It's perfectly meaningful to say a function is or is not recursively > enumerable. I agree, certainly. The mistake in question was not in applying the term "recursively enumerable" to functions, but Peter stating that Cantor's diagonal argument proves that no function from N to R is recursively enumerable. It was just too rich to ignore that Peter was objecting to applying the term "recursively enumerable" to functions, when the only person who had done so was Peter himself. It is amusing to see a person mistake their own statement for mine and vigorously attack it on spurious grounds, and even more amusing to point out to them what they are doing. - Tim
From: Newberry on 23 Jun 2010 23:52 On Jun 23, 1:55 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 23 Jun., 15:42, Sylvia Else <syl...(a)not.here.invalid> wrote: > > > > > > > On 23/06/2010 11:10 PM, WM wrote: > > > > On 23 Jun., 14:51, Sylvia Else<syl...(a)not.here.invalid> wrote: > > > >> Cantor doesn't rely on being able to identify a last digit. He's just > > >> saying that no matter how far down the list you look, you'll find that > > >> the element at that point doesn't match the anti-diagonal. > > > > That is wrong. Cantor uses the alleged "fact", that infinity can be > > > completed, i.e., that that the infinite list can be finished. > > > If he would only assume what you say, then the anti-diagonal could > > > remain in the unknown part of the list. You know and appreciate that > > > after *any* line number n there are infinitely many more lines? > > > Yet it's clear that if you look at any later line m, you will still find > > that it doesn't match the anti-diagonal. > > But that does not help. It is as clear that at any later line you have > yet seen less lines than you will have to see. Therefore, without > finished infinity, you cannot see any line. Most of them, nearly all > of them (infinitely many compared to finitely many) will never be > seen. > > > > > > > > > >> But you can't > > >> even begin to formulate his proof if you can't identify the first > > >> element of the list (and hence first digit of the anti-diagonal) either. > > > > Isn't it enough, also in my case, to know that every antidiagonal has > > > a first digit? In fact it has. Every antidiagonal is constructed from > > > a list with a first line (that is the previous antidiagonal) and the > > > remaining list. Every line has a finite number n. > > > I agree that every anti-diagonal you added is well defined. But for your > > proof to work you also have to look at the anti-diagonal of the list > > after you've added all of the (infinitely many) constructed > > anti-diagonals. > > How should that be possible? How many there ever may have been > constructed. The majority of infinitely many will remain to be > constructed. > > And even if infinitely many diagonals would have been added, then the > resulting list would yield another diagonal. Why do you think that > would be a complete list withot its own diagonal? > > > To construct the first digit of that anti-diagonal you > > have to look at the first element in the list. But it has no first > > element > > It is very naive to believe that there would result a list without a > first line. > In my construction every list has a first line and every diagonal will > be placed in front of it. How should a list without first line occur? > > - any element you might claim is the first is in fact preceded > > > by infinitely many other elements. > > That is ridiculous! How should a list without first element come to > existence? > > > > > Your proof falls apart if you cannot construct the anti-diagonal which > > you claim should have been in the list. > > If it is in the list, I can construct it and place it in front of the > list. Why should that situation change? > > > > > As I observed earlier, this problem can be obviated > > What problem? There is no problem. There is an infinite sequence of > lists and of antidiagonals. If the sequence is interrupted by a list > without containing its antidiagonal, then Cantor's argument fails. If > the sequence is not interrupted by a list containing its antidiagonal, > then the countable set of antidiagonals is not listable. This is interesting. So where is the error in the argument? > > >> First and last are interchangeable, of course, but with your > > >> construction above, you can't specify either the first or the last. > > > > As I told you, my notation is only an abbreviation for the following > > > definition: > > > 1) Take a list L0 of all rational numbers. > > > 2) Construct its antidiagonal A0. > > > 3) Add it at position 0 to get (A0,L0) > > > 4) Construct the antidiagonal A1. > > > 5) and so on. > > > With a resulting 'list' which is infinite at both ends. > > No, you are completely in error. How should a missing first line come > into existence? If it occured in the process of construction, I would > stop and boast that the preceding list has no antidiagonal. Why should > I continue? And, first of all, *how* should I continue? To create > another blank line from a list without first line??? > > Can you tell me that? Can you suppose that you could or would like to > produce such nonsense-list? > > Regards, WM > > > > - Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Virgil on 24 Jun 2010 00:09 In article <b0258a2f-1679-492d-99e9-3397093bb054(a)t10g2000yqg.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > On Jun 23, 1:49�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <8cd5db0d-46a5-4deb-8fe5-9a28acad5...(a)k39g2000yqb.googlegroups.com>, > > > > �Newberry <newberr...(a)gmail.com> wrote: > > > Cantor's proof starts with the assumption that a bijection EXISTS, not > > > that it is effective. > > > > Actually, a careful reading shows �Cantor's proof merely assumes an > > arbitrary INJECTION from N to R (originally from N to the set of all > > binary sequences, B) which is NOT presumed initially to be surjective, > > and then directly proves it not to be surjective by constructing > > OK, so construct it assuming injection. It is simplest to do for an injection from N to B. A member of B, a binary sequence, is itself a function from N to an arbitrary two element set which, without loss of generalization, we may take to be S = {0,1}. A list of such functions is then equivalent to a single function from NxN, the Cartesian product to {0,1}, say F(-,-), so that for each m in N, the function F(m,-): N -> {0,1}: m |--> F(n,m) is one of the binary functions in the list. For any such list of binary functions, define a new function g(-):N -> {0,1}:n |--> 1 - F(n,n), or more briefly, g(n) = 1-F(n,n). This new function is the desired "antidiagonal" for the list of lists F(-,-) and g(-) is not in the original list since g(-) differers from each F(n,-) at n. > > > the > > "antidiagonal"as a member of the codomain not in the image. > > > > Thus proving that ANY injection from N to R (or B) fails to be > > surjective. > > > > For some unknown reason, the DIRECT "anti-diagonal" proofs given by > > Cantor, and his followers, are often misrepresented as being proofs by > > contradiction, but they never were.
From: WM on 24 Jun 2010 03:05 On 24 Jun., 01:02, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > a) Does this list contain the anti-diagonal of > > (..., An, ... A2, A1, A0, L0)? > > This is not a list of numbers. L0 is not a number, it is a list. > > Therefore (..., An, ... A2, A1, A0, L0) does not have an anti-diagonal. You are wrong. The symbols above abbreviate the sequence of lists An .... A0 L0 Each of them has an antidiagonal that either is not in the list (then the set of all of them is unlistable) or is in a list. Then Cantors argument is wrong. Regards, WM
From: WM on 24 Jun 2010 03:13
On 24 Jun., 02:39, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 8:44 PM, WM wrote: > > > > > > > On 23 Jun., 06:47, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >> Rather than argue that VMs proposition fails on that point, I wanted to > >> address the flaw, in order to find a more substantial objection. > > > My initials ar WM. > > > There is no flaw in the argument: Your bijection either contains all > > constructed antidiagonals. Then a list contains also its antidiagonal, > > because every list has an antidiagonal and your bijection (that is > > only a permutation of my list) contains all lines and antidiagonals. > > I.e., there is no missing antidiagonal of a "limit" list outside of > > the bijection. > > > Or there is a last diagonal of the limit list that does not belong to > > your bijection (and to my list). Then there is a countable set (the > > set that includes this last diagonal) that is not listable. > > > Think over that only a little bit. It is not difficult to understand. > > > Regrads, WM > > Thing is, I just cannot see that you've constructed a countable set that > should contain its own anti-diagonal when expressed as a list. When it does not contain it, then the antidiagonal is not listed. That is all. > All > you've done is construct a countable set of countable sets, each of > which contains the anti-diagonals of its predecessors, but none of which > by construction can be expected to contain its own anti-diagonal. That is why I did it. The set of elements of L0 and antidiagonals A0, A1, ... is countable, but cannot be listed. > > You've then claimed, but without any kind of formal argument, that if > you take this to the limit, Why should there be a limit? Why should I agrre to take this to a limit? There is no limit in n. Every one is followed by another one. > the result is a countable set that should > contain its own anti-diagonal when expressed as a list. Then Cantor's argument would fail. There are no other escapes: Either there is some list that contains its antidiagonal or there is no list that contains it. Regards, WM |