Re: November 25 is Infinite Clause day!!

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 11/22/04 

Date: Mon, 22 Nov 2004 02:01:05 GMT

In sci.logic, herc777@hotmail.com
<herc777@hotmail.com>
 wrote
on 21 Nov 2004 14:17:40 -0800
<1101075460.302763.81790@z14g2000cwz.googlegroups.com>:
> -Not to mention HERC777's understanding. The question is not whether
> -the diagonal number's digits are within the list (most likely,
> -they are
>
> this is your 1st error, you overlook the fact all digit sequences to
> infinite length are computable the conclusion of Cantor's proof relies
> on the existence of a new sequence of digits which is clearly non
> existent.

Not sure all infinite digit sequences are computable. One problem:
the set of IDS is a power set (10^N); card(10^N) > card(N).

Since all UTMs depend on integer codes (usually broken down into
instructions), this yields digit sequences that cannot be generated
by a TM.

>
>
> -We also need to construct Diag. This is fairly simple,
> -and one can have many variants, so we need just pick one:
> -
> -if List(N,N) = 4 then Diag(N) = 5 else Diag(N) = 4
>
> this is your second error, your construction can never be realised.

Correct, it's non-computable. This doesn't mean it can't exist;
there are a large number of non-computable functions out there.

> its like saying construct the set of all sets that don't contain
> themselves, if its a member of itself its not a member, if its not a
> member of itself its a member. you can theorise the construction but
> you cannot refute your basic premises of set theory because of one
> theorem, regardless if a contradiction is formed.
>
> this is all you are doing.
> let R = 1, 2, 3, 4...
> let p <> R1, p <> R2, p <> R3...
> therefore R is incomplete.
>
> your claim that p is a valid construction but its not rigorous.

Neither p nor Diag is a valid construction as such (in the sense that
it can be realized by a TM). Diag merely needs to exist in some form.
Cf. Cantor's first proof.

>
> Its like your Godel proof, you are trying to prove ! (proof(X) <-> X)
> proof(X) <=> Exists a proof of X.
>
> so you allow this theorem, G <-> ! proof(G) but you don't allow this
> meta-consistency theorem X <-> proof(X). Either of these theorems will
> work but they are mutually exclusive, no one here has shown a PROOF
> that ! (proof(X) <-> X).

That's because there is no such proof. All Godel proved is that
there are true unprovable statements.

> In AI the meta-constistency theorem is called
> Truth Maintenance, only proven facts are allowed and never need to be
> revoked. The alternate system is called belief revision. Godel's
> proof only works in one automated theorem system.
>
> Same with the set of reals, assume all possible combinations of digits
> are present in a countable list, with UTM(n, 0) they actually are. Now
> examine the construction of diag, it is a flawed construction! QED.

That it is. Diag cannot be constructed using a UTM. However, that
doesn't mean it's not a real.

>
>
> -Does an N exist such that for every positive M in J,
> - Diag(M) = List(N,M)?
> -
> -The answer clearly is no (if M=N Diag(N) != List(N,N) by
> construction).
>
> This is your 3rd error. Why is it when I write 'obviously' it gets
> picked up yet everyone uses 'clearly' around here. What you all
> actually mean by 'clearly' is "clearly it's in the text book!" Your
> construction breaks the premise of a complete list, it doesn't
> contradict it.

So you're stating that another member in the list is actually equal
to Diag?

Which one, then? Or are you hypothesizing that List is not
a denumerable mapping?

>
> An infinite number of people toss a coin RANDOMLY an infinite number of
> times. After 10 people have tossed a coin 5 times each, the 1st 3
> places in the sequence are (tend to be) all covered.
>
> hhh..
> hht..
> hth..
> htt..
> thh..
> tht..
> tth..
> ttt..
>
> there are 2 duplicates due to random spread.
>
> After several million people have tossed a coin 30 times each, by the
> binomial distribution, all combinations have been covered up to 15
> tosses.
>
> 000000000000001..010101010
> 000000000000010..010101010
> 000000000000011..010101010
> 000000000000100..010101010
> 000000000000101..010101010
> 000000000000110..010101010
> 000000000000111..010101010
> 000000000001000..010101010
> ...
> 111111111111110..010101010
> 111111111111111..010101010
>
> <- - - - - - - ->|<- - - - ->
> all combinations still random
> covered
>
> The number of coins that get covered increases without bound,
> logarithmically. The length of 'covered combinations' is approx.
> log(people doing coin tosses). As the countable list approaches
> infinite length..
>
> As #people -> oo, number of digits covered -> log(oo).
>
> Therefore an infinite list contains all sequences of {H, T} of infinite
> length.
>
> 1 HTHTHTHTHT
> 2 HHHHHHHHH
> 3 TTTTTTTTTTT
> 4 THTHTHTHTHT
> ..
>
> Take the diagonal! HHTH...
> Invert it! TTHT..
>
> ALL SEQUENCES ARE PRESENT, TTHT.. is not a new sequence.
>
> INFINTIE people toss coins infinite times each, can you with 100%
> certainty form a NEW {H, T} sequence? CLUE : NO!! Inverting the
> diagonal is not a new sequnce of tosses, and modifying the real
> diagonal does not a new real make.

OK, dumb question. Are you hypothesizing that card(2^N) = card(N)?
2^N can be mapped to the set of all fractional expansions, where
the n'th digit is 0 if n is not in the set S, and 1 if n is.
(Recall that the set S is in 2^N, therefore a *subset* of N.)

>
> Herc
> 

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.

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