Why can no one in sci.math understand my simple point?
in [Logic]
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From: Ross A. Finlayson on 20 Jun 2010 04:57 On Jun 17, 10:30 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-18, Ross A. Finlayson <ross.finlay...(a)gmail.com> wrote: > > > The rationals are well known to be countable, and things aren't both > > countable and uncountable, so to have a reason to think that > > arguments about the real numbers that are used to establish that > > they are uncountable apply also to the rationals, the integer > > fractions, has for an example in Cantor's first argument, about the > > nested intervals, that the rationals are dense in the reals, so even > > though they aren't gapless or complete, they are no- where > > non-dense, they are everywhere dense on the real number line. > > As your sentence is less than coherent, I will merely point out that > it is generally poor form to use 9 commas in a single sentence except > when listing items. I will grant that parody often benefits from > abuses of ordinary sentence structure, such as, for example, and not > in any way showing that these are the only possible forms, sentences, > like this one, which are convoluted to exhibit, by way of meandering, > that they imply that mental processes, of the original writer, that > is, which may be, perhaps, less than clear, and so in some way, to > some readers, humourous. > > - Tim Some I go back and add later. Too bad no one can read it but me. No seriously still it's clear in lots of ways, I can rewrite those paragraphs as much longer. Sorry that was a bad joke. Collected, I'm very happy with the output. Each one, in its own way, has some content. You don't agree with the rationals are dense on the line? The rationals are well known to be countable, and things aren't both countable and uncountable, so to have a reason to think that arguments about the real numbers that are used to establish that they are uncountable apply also to the rationals, the integer fractions, has for an example in Cantor's first argument, about the nested intervals, that the rationals are dense in the reals, so even though they aren't gapless or complete, they are no- where non-dense, they are everywhere dense on the real number line. The rationals are well known to be countable, and things aren't both countable and uncountable, so to have a reason to think that arguments about the real numbers that are used to establish that they are uncountable apply also to the rationals, the integer fractions, has for an example in Cantor's first argument, about the nested intervals, that the rationals are dense in the reals, so even though they aren't gapless or complete, they are no- where non-dense, they are everywhere dense on the real number line. (The rationals are well known to be countable, and things aren't both countable and uncountable, so ) to have a reason to think that arguments about the real numbers that are used to establish that they are uncountable apply also to the rationals, the integer fractions, has for an example in Cantor's first argument, about the nested intervals, that the rationals are dense in the reals, so even though they aren't gapless or complete, they are no- where non-dense, they are everywhere dense on the real number line. (...) You left out the part before and after. Arguments about the uncountability of the real numbers include those derived from the numeric property of their density. For example one of them is called "Cantor's first argument for the uncountability of the reals." The constructive (computable) sets are mappable to the countable ordinals, where, the countable ordinals is the same thing as the enumerative ordinals, because, they're each countable and that's all of them. The constructive universe is complete, each in it countable, but then the results of results are results so they are infinite and their own powersets. (Sound familiar?) The existence of the constructive universe is a result. Ha what's funny is you can still read them. Basically from having an idea to write a sentence, as it's written then the parts of it are added automatically for, as you describe, contemplative pause, as well as any emplacement of comment to provide context generally. This is in the case where the writing is for a particular medium, when there's a lot of writing back in forth (in complete sentences, thank you) then to edit for readability is deferred for real intent. But, there's not really a lot of writing going on that way, rather, I much prefer the writing with the theme and the content (on the mathematics). So, I read. Warm regards, Ross Finlayson
From: Peter Webb on 20 Jun 2010 10:20 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1r04q.jrj.tim(a)soprano.little-possums.net... > On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> "Tim Little" <tim(a)little-possums.net> wrote in message >>> That is only a "finite, terminating algorithm" if the list is finite. >>> Since the only input to the algorithm is the desired precision, your >>> recipe only works if you can embed the list into the algorithm. >> >> No. You could level the same argument against Cantor's original proof. > > No you couldn't, as it makes no assumption of any finite algorithm. > Only *your* alteration of Cantor's proof assumes any sort of finite > algorithm, as it is required for the definition of "computable real". > I don't say anything about finite algorithms. I just ask for a list of computable Reals, in the same form a Cantor asks for a list of Reals. > >> For a number to be computable, it must be able to be calculated to >> any required degree of accuracy (according to the definition we have >> just both agreed on). This is obviously true of the anti-diagonal. > > Only if the list is computable, which Cantor's proof does not require. > Exactly how are my requirements for the list different from Cantor's, excepting that my list contains computable Reals and his requires any Real? > >> Cantor provides an explicit algorithm for constructing (computing) >> the anti-diagonal. > > No, the proof only provides an algorithm fragment. A full algorithm > for a computable real must take the precision as input, and nothing > else. The constructive portion of Cantor's proof takes an infinite > list of real numbers as "input", and so is not a valid algorithm for a > computable real. > No, if you want to use this definition of a computable Real (slightly different to the one we have been using, but equivalent), to determine the anti-diagonal to n places only requires the first n items on the list be examined, and hence is finite. > >>> If not, how does this differ from "suppose L is a list of real >>> numbers. Is antidiag(L) computable?" >> >> Because there is an algorithmic procedure for calculating the >> anti-diagonal to any required degree of accuracy. > > If you are given L, you say you can compute antidiag(L) regardless of > what L is. Yep. Easy. The nth digit of L is "1" if the nthe digit of the nth term is "2", otherwise it is "1". As I can easily specify every digit in its decimal expansion, it is of course computable. > If you are given x, can you not compute x+1 regardless of > what x is? By your argument, does this not mean that x+1 is always a > computable real? > As long as x is computable. And the input list by definition only contains computable Reals. > > - Tim You remind me of one of those anti-Cantor cranks who keep trying to find a flaw in Cantor's proof. My proof and his are virtually identical. Cantor proved the set of Reals is not recursively enumerable, and my proof shows the set of computable Reals is not recursively enumerable. This is not quite the same thing as being uncountable.
From: Newberry on 20 Jun 2010 10:33 On Jun 19, 6:06 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(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? That's right. There is no formula or algorithm to construct the list. It means that you would have to flip each and every digit one by one. And that is impossible. > 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. Which sequence is that? > 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. > > -- > Jesse F. Hughes > > "Of course, I don't need any more education." > -- Quincy P. Hughes (age 7)
From: Peter Webb on 20 Jun 2010 10:52 "Tim Little" <tim(a)little-possums.net> wrote in message news:slrni1r0ra.jrj.tim(a)soprano.little-possums.net... > On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> Explain to me how my requirements for the input list of all >> computable Reals is different to Cantor's requirements for the input >> list of all Reals, other than the requirement that every item on the >> list is computable? > > It is not. Your error comes later. > > Cantor's proof includes a construction taking a list L and defining an > antidiagonal real antidiag(L) from it. Your error is supposing that > this construction is a finite algorithm fitting the definition of > "computable real". It is not. By stretching the definition of > "algorithm" somewhat, it can be supposed to be an algorithm accepting > finite input n and infinite input L, and producing the n'th > antidiagonal digit. This is a stretch since algorithms are normally > not supposed to have infinite inputs. The calculation of the nth digit only requires the first n digits to n decimal places, and hence it is clearly a finite algorithm. Its simply ludicrous to suggest that the antidiagonal is not computable given that a very simple algorithm will explicitly churn it out. > > However, there is no way that you can then prove the existence of a > finite algorithm accepting only the *finite* input n and producing the > n'th antidiagonal digit. > Of course I can prove the existence of a finite algorithm. I can produce it. To produce the nth decimal of the anti-diagonal, look at the first n items on the purported list of all computable Reals. Then substitute "1" for "2" blah blah. How is that not a finite algorithm? I can produce the nth decimal place of the anti-diagonal purely by counting down to the nth Real on the list and looking at the nth digit. I can code it in Java in 3 lines. Exactly as per Cantor's proof. > >> (in more modern terminology) that there is no recursively enumerable >> mapping > > Once again, "recursively enumerable" is an introduction purely of your > own invention. As I said when I introduced it. It does not appear in Cantor's proof or mine. It is an explanation of why this somewhat paradoxical result applies, that Cantor's form of proof applied to computable Reals also produces the conclusion that any such list cannot be complete. Not that it is much of an explanation; not being recursively enumerable means pretty much the same thing as having no explicit mapping from N. As somebody else pointed out, computable Reals are enumerable but not recursively enumerable. Cantor's form of proof applied to computable Reals shows only that they are not recursively enumerable, not the stronger condition that they are uncountable. Similarly, Cantor's proof really only shows that the Reals are not recursively enumerable, not that they are also denumarable (uncountable). > It is not present in Cantor's proof, it is not present > in a modern recasting of Cantor's proof, it has no relevance at all to > Cantor's proof. Well, I think I am making a pretty good hand of my case. But yes, I know what I am saying is not accepted. I am genuinely confused by this - the fact that my opinions are unique, and I would expect therefore to be almost certainly wrong (with probability 1). On the other hand my argument seems pretty solid to me. > Cantor's proof applies to *every* mapping from N to R. > Ummm ... mine applies to every mapping from N to computable Reals. > > - Tim Really, if you have some list of computable Reals to give me, I would be happy to provide the anti-diagonal. And lets remember what a "computable Real" is, it is a Real which can be specified to any arbitrary degree of accuracy. As a practical matter, you don't have to list every one in full; I only require the first n digits.
From: WM on 20 Jun 2010 11:10
On 20 Jun., 16:33, Newberry <newberr...(a)gmail.com> wrote: > On Jun 19, 6:06 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > Newberry <newberr...(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? > > That's right. There is no formula or algorithm to construct the list. > It means that you would have to flip each and every digit one by one. > And that is impossible. Well, there are some lists defined by finite definitions like 0.1 0.11 0.111 .... But the number of these lists and the number of their diagonals is countable. > > > 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. > > Which sequence is that? O, don't be unfair. Matheolgicians "prove" their claims. They are not used to give examples. Remember Zermelo who "proved" that every set can be well ordered. Although every sober mind today knows that hisproof is wrong, set theory is built upon this lie. Otherwise the hierarchy of infinities cannot exist. Set-cranks suffer from selective damage of perception. There is no use to discuss with them. (Nevertheless we should go on as if they could understand - mainly to save those who have not yet become "mentally challenged zealots of abstract mathematics" (V.I. Arnol'd)) Regards, WM |