Why can no one in sci.math understand my simple point?
in [Logic]
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From: Tim Little on 28 Jun 2010 04:36 On 2010-06-27, Jesse F. Hughes <jesse(a)phiwumbda.org> wrote: > I haven't seen his definition, but he seems to suggest that the > machine that repeatedly overwrites a single square with 0 or 1 > computes a real number. Is that consistent with his definition? Yes, apparently it represents the real number 0, as all machines do that never produce a tape state having at least two "2" symbols. As I recall, the mapping goes something like this: the successive states in which there exist at least two "2" symbols on the tape define a sequence of binary strings between the leftmost pair of "2"s. Then there is a mapping from sequences of binary strings to reals. I don't think the latter mapping was explicitly given, but there are plenty of suitable options. - Tim
From: Tim Little on 28 Jun 2010 04:38 On 2010-06-27, Newberry <newberryxy(a)gmail.com> wrote: > I was replying to this: > "It implies a listing must exist, but does not provide such a > listing." > > "It", in this context is the statement "all computable reals are > countable." > > If an antidiagonal existed it would prove that there was no such > list. No, it merely implies that every such listing has an uncomputable antidiagonal, which further implies that the listing itself is an uncomputable function. - Tim
From: Tim Little on 28 Jun 2010 04:59 On 2010-06-28, Owen Jacobson <angrybaldguy(a)gmail.com> wrote: > So, here is an informal presentation of Cantor's diagonal argument > that avoids the word "list" (as well as a few other common verbal > shortcuts): > > 1. Let S be the set {0, 1}^N. > 2. For any function L from N to S, we can identify an element of S not > in the image of L. Peter is incapable of separating the usual informal phrase "we can..." from his fixed idea of "there exists a finite algorithm that can...". So, for example: 2. For any function L from N to S, there exists an element of S not in the image of L. No mention of "we can identify it" or even "given L we can find it". Those are irrelevant distractions that he will be unable to see past to the actual matter of the proof. - Tim
From: Tim Little on 28 Jun 2010 05:07 On 2010-06-28, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > This occurs in step 4. You state we can "identify an element d of S > that is on the diagonal". Unless the list is explicitly defined, you > can't "identify" the digit in position x. See Owen? I told you he wouldn't be able to ignore the informal fluff phrase "we can identify", and so fails to interpret it correctly. Mathematically, it means nothing more than "there exists". - Tim
From: WM on 28 Jun 2010 05:29
On 28 Jun., 11:07, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-28, Peter Webb <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > > > This occurs in step 4. You state we can "identify an element d of S > > that is on the diagonal". Unless the list is explicitly defined, you > > can't "identify" the digit in position x. > > See Owen? I told you he wouldn't be able to ignore the informal fluff > phrase "we can identify", and so fails to interpret it correctly. > > Mathematically, it means nothing more than "there exists". And mathematically "there exists" means nothing. Regards, WM |