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Re: Peer-reviewed arguments against Cantor Diagonalization
Posted:
Oct 29, 2012 11:44 PM
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On Monday, October 29, 2012 9:38:59 PM UTC-5, Peter Webb wrote:
> Arturo Magidin wrote:
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> > On Monday, October 29, 2012 9:37:20 AM UTC-5, JRStern wrote:
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> > > On Sun, 28 Oct 2012 19:20:51 -0700 (PDT), Arturo Magidin
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> > >
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> > > <magidin@member.ams.org> wrote:
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> > >
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> > >
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> > >
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> > > >> How can this be unclear?
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> > >
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> > > >
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> > >
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> > > > Because I did not simply state the conclusion.
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> > >
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> > > > I gave you the "diagonal argument".
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> > >
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> > >
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> > >
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> > > I did not recognize the argument you gave as a/the "diagonal"
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> > >
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> > > argument.
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> >
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> > That suggests you are not very familiar with the argument.
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> >
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> > > I saw a separate argument for the same conclusion.
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> >
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> > It was not.
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> >
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> > > The diagonal argument as I've seen it starts with enumerations of
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> > >
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> > > numbers, selects out digits, and constructs from those digits, and
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> > >
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> > > then makes an assertion about the result.
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> >
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> > Like I said: if you want to discuss specific steps in a specific
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> > argument, then you must provide that argument explicitly and in
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> > detail. Otherwise, we may be talking about different things.
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> >
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> > I will allow that there are many expositions of the argument that are
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> > not very precise, or very clear, or that are overly complicated (it
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> > is very common to present the argument as an argument by
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> > contradiction, although it really is not such). But that only makes
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> > it all the more necessary that you be very explicit and very precise
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> > about exactly what argument you are talking about.
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> >
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> > > > If you are questioning whether it is "coherent", then you should
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> > >
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> > > > point to whatever point you find incoherent, rather than simply
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> > >
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> > > > quote and then give a sentence fragment.
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> > >
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> > >
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> > >
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> > > This is (perhaps too obviously) not my area of expertise, and rather
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> > >
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> > > than try to present my own proof or even formal objection at this
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> > >
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> > > point - I came looking for further background.
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> >
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> > A common presentation of the corollary (which in reality can be
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> > obtained from the presentation I gave), is the following:
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> >
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> > THEOREM. If N is the natural numbers, (0,1) are the real numbers
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> > between 0 and 1, not inclusively, and f:N-->(0,1) is any function,
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> > then f is not onto. That is, there exists a real number r(f) (which
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> > depends on f) such that f(n)=/=r(f) for every n in N.
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> >
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> > Proof. Suppose that f:N-->(0,1) if any function from the natural
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> > numbers to the numbers in (0,1) (this is equivalent to a function
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> > from N to the reals, using the bijection between (0,1) and the reals
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> > afforded by the arctan function and a simple linear map, like I did).
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> >
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> > I claim that f is not onto.
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> >
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> > To that end, for each n in N, consider the decimal expansion of f(n)
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> > (if f(n) has two expansions, select the one with only finitely many
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> > nonzero entries). That is,
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> >
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> > f(n) = Sum_{i=1}^infty (a_{n,i}/10^n) 0<= a_{n,i} <= 9
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> >
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> > (this is equivalent to saying f(n) = 0.a_{1,1}a_{1,2}a_{1,3}.... is
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> > the decimal expansion)
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> >
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> > We construct a sequence b_n of digits as follows:
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> >
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> > b_n = 5 if a_{nn}=/=5
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> > b_n = 6 if a_{nn} = 5.
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> >
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> > The existence of (b_n) is guaranteed by the Principle of Induction,
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> > which is a theorem of ZF (follows from the construction of N, which
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> > is well-defined by the Axiom of Infinity).
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> >
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> > Let r(f) = Sum_{i=1}^{infty} b_n/10^n.
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> >
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> > This is a real number, since the sequence is Cauchy. It is less than
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> > 1, because it is term by term strictly smaller than Sum (9/10^n) = 1;
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> > it is greater than 1 because it is term by term strictly larger than
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> > Sum (1/10^n) = 0.11111.... Thus, r(f) is a real number in (0,1).
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> >
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> > I claim that r(f)=/=f(n) for all n. First, note that r(f) has a
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> > unique decimal expansion; thus, if f(n) has two decimal expansions,
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> > then it cannot equal r(f). Hence, in order to ensure that r(f)=/=f(n)
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> > it suffices to show that there is a k such that b_k=/=a_{n,k}
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> > (because either f(n) has a unique decimal expansion, in which case
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> > the fact that b_k=/=a_{n,k} suffices; or else f(n) has two decimal
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> > expansions, in which case it cannot equal r(f)).
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> >
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> > By construction, b_k=/=a_{k,k}; hence r(f)=/=f(n). Thus, r(f) is not
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> > in the range of f, so f is not onto. QED
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> >
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> > As to its validity: we are assuming that f is given; the fact that
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> > f(n) has a decimal expansion follows by construction of the real
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> > numbers as equivalence classes of Cauchy sequences. Our ability to
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> > define b_n is guaranteed by the fact that we know f and we know the
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> > expansion of f(n) (because we know f); and the fact that we have
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> > defined a full sequence follows by induction. That r(f) is a real
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> > number follows because the sequence of partial sums is Cauchy, hence
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> > it represents a unique real number.
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> >
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> >
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> You have made the proof slightly more rigorous by pointing out the
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> anti-diagonal is Real using Cachy sequences.
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> Two problems with this:
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> 1. The cranks don't dispute the anti-diagonal is a Real, so you are not
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> addressing their issue (whatever it is, but its not that the
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> anti-diagonal is not Real).
This is only a "problem" if you think that my presentation was an attempt at convincing or arguing with a crank. It was not. I was addressing the Original Poster of this thread, which seems to suffer not from crankiness or "anti-Cantorianism", but rather from simple and pure lack of knowledge about the subject (and a lack of desire to acquire that knowledge, prefering instead to ask and talk in vague terms). It was an attempt at presenting a specific instance of the argument so that he could provide specific questions about specific steps... something, alas, he seem sunwilling to do.
> 2. Your average Cantor crank wouldn't recognise a decimal expansion is
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> a Cauchy sequence, or that Cauchy sequences specify Real numbers.
Again, only a problem if you think I believe I have provided an "answer" to cranks. I have neither the intention nor the desire to do so.
> You may have been better off proving |P(S)| > |S| (where you have
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> access to the axioms of ZF)
Amusingly enough, I did do so earlier.
> and then showing |c| > |N| is a special
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> case where S = N. As of course others have done in this thread.
Where "me" happens to be one of the value of "others [...] in this thread".
--
Arturo Magidin
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