Re: Idiocy of Muckenheim was Re: countability of reals
From: Piotr Sawuk (piotr5_at_unet.univie.ac.at)
Date: 01/26/05
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Date: 26 Jan 2005 03:45:15 GMT
In article <353l3qF4hliarU1@individual.net>,
"|-|erc" <h@r.c> writes:
> "Piotr Sawuk" <piotr5@unet.univie.ac.at> wrote in
>> In article <34oocqF4dne5aU1@individual.net>,
>> "|-|erc" <h@r.c> writes:
>> > -------------------------------------s-o-s------------------------------------
>> > <bryant_j_j@yahoo.com> wrote in message
>> >> |-|erc wrote:
>> >> >
>> >> -------------------------------------s-o-s------------------------------------
>> >>
>> >> >
>> >> > I only did calculus to 1st year, and stats to 2nd year. programming
>> >> degree.
>> >> >
>> >> > Herc
>> >>
>> >> then why the hell r u pretending to be a great mathematician and
>> >> logician, u poor lost soul, and insisting on being an argumentative
>> >
>> > and logic to 3rd year of course.
>>
>> oh, since you are such an intelligent and well-educated person in
>> the area of diagonalization-proofs, then maybe you could do me and
>> most probably other people a great favour by translating cantor's
>> diagonalization-proof, or at least his set-theoretical proof for
>> the uncountability of real numbers into the language of logic you
>> claim to have learned to 3rd year? you know, per definition a proof
>> in logic does only consist of "A and B -> C" respectively "A or
>> B -> C" or similar terms such that these things are either proven
>> to be true within the same proof, or are an axiom of ZFC. i.e. you
>> create logical expressions and assign them to variables, then those
>> variables are the input to some axiom or another expression proven
>> to be true which does expect such parameters as input, and you keep
>> this going until you get an expression showing that no bijection
>> does exist between natural numbers and their powerset (i.e. the
>> expression "for all F: F is a function mapping natural numbers
>> to powerset -> either there exists a set of subsets of N which
>> is never output by F (for all natural numbers), or there exist
>> 2 natural numbers not equal to eachother which output the same
>> set of subsets of N) on the right side of the last "->".
>>
>> of course this proof does have a finite length, and judging by
>> the shortness of Cantor's proofs one could easily imagine that
>> such a task wouldn't take more than half a day or a day. I am
>> too stupid to do such a thing, and I did not study 3 years of
>> logic (neither 1 full year), thereby I wouldn't be capable to
>> produce such a proof even if I invested several years into it,
>> but judging by your postings to this thread I got the impression
>> that you are better educated than me and actually accurately
>> trained in such things so that doing me that favour would
>> take a few hours only, most presumably the hours needed for
>> digging up within your notes a pre-existing proof you did
>> already perform several years ago. by doing me this favour
>> you would also increase the range of your tools for proving
>> that said proof is wrong, as you easily could point out a
>> part of the proof which does not use any facts proven before
>> nor any axiom, and nobody could ever argue against the power
>> of your arguments since putting this exact proof into a
>> machine verifying that kind of proofs would actually yield
>> that the proof does contain an error. further you would
>> also do a favour to the science of mathematics which does
>> build up on such theories and the idea that all of them
>> can get expressed (together with their proof) as some
>> finite sequence of logic-symbols with a very small set
>> of (classes of) axioms as the basis for the proof. maybe
>> you would then even show those mathematicans that some of
>> their theories do rely on axioms which are not part of ZFC?
>>
>> since you did already observe the similarities between the
>> diagonalization-proof and the proof for the halting-problem,
>> such a formalization of your theories (and proofs) might
>> even help people to understand your arguments in that area?
>> --
>> Better send the eMails to netscape.net, as to
>> evade useless burthening of my provider's /dev/null...
>>
>> P
>
> Not only have I formalised Cantor's proof, I knocked off about 50 steps off others' formalisms!
>
> L(x, y) = the yth digit of the xth real.
this is not really the language of logic, but since you so believe in
such a function's definability...
>
> 1. L(a,a) = L(a,a) (obvious)
> 2. exists b, L(a,b) = L(b,b) (provable from 1, with b=a)
> 3. forall a, exists b, L(a,b) = L(b,b) (generalization of 2)
> 4a. not(exists a, forall b, L(a,b) != L(b,b) (negation of 3)
> 4b. not(exists a, forall b, L(a,b) = !L(b,b) (! is some suitable digit change function)
> 5. exists r, not(exits a, forall b, L(a,b) = r(b)) (r(b)= !L(b,b))
neither you did define the set in which the function r is contained.
So, suppose R:=set of enumerations of each subset of N, such that an
element r of that set R is a function with r(a) in N and forall b:
(b!=a -> r(b)!=r(a)), then L(0):=R,L(a):=L(a-1)-C(L(a-1)),L(a,b):=C(L(a))(b)
where C is the choice-function we use for the set R and its subsets.
then your !(r(a)):=r(C(N-a,r(a))) where C is a choice-function on the
set N which takes a natural number as a hint. I think these definitions
can easily be understood as equivalent to a definition of reals between
0 and 1 and their enumeration if you restrict R to contain only functions
with r(a):=10^a times some number between 0 and 9.
>
> "There exists a real that no member of list L matches at every digit."
> Of course the free b in (r(b)= !L(b,b)) puts doubt on such an algebraic derivation.
why? it is just a definition of an r in R which is not in the list.
if you indeed can create a list L, and if you really can create a
function "!", then there does exist an r in R with r(b):=!L(b,b).
what I fail to understand is the step from (1-3) to 4, I just fail
to see how to argue here in the language of logic. from 1 we know
that b:=a in 2, and we can see that this does work for all a, but
even though the negation 4a is true, it is only so because one could
choose b:=a and thereby the function "!" is not guaranteed to exist.
don't you know any formulazation of this proof where the function
"!" is defined explicitely such that it does yield a value some r
would yield, or where it at least is proven to exist? in whole this
proof doesn't seem formal enough for mee! it does attempt to prove
that no surjection from N to R does exist in a constructive way
(i.e. that some r in R is never equal to some q(b):=L(a,b) forall
a in N), but as Herc said, in the 1st and 2nd step the variable
"a" is a free variable, and in the last step b is a free variable,
and in logic free variables are not even allowed for defining some
function (or predicate), or am I wrong? how does one define functions
within the language of logic? for example 2nd order logic with ZFC?
how does one define a choice-function for finite sets? Herc, since
you had 3 years of logic, what did you learn on this topic?
-- Better send the eMails to netscape.net, as to evade useless burthening of my provider's /dev/null... P
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