Re: Orlow cardinality question

imaginatorium@xxxxxxxxxxxxx said:
> Tony Orlow (aeo6) wrote:
> > imaginatorium@xxxxxxxxxxxxx said:
> > > Tony Orlow (aeo6) wrote:
> > > > Virgil said:
> > > > > In article <MPG.1d20c4e04d6ab043989e46@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > > >
> > > > > > > Such an assumption is your idea that there cannot be an infinite set
> > > > > > > of finite elements, for example. Or that the limit of a sequence must
> > > > > > > somehow become a part of that sequence.
> > > > > >
> > > > > > But I did not ASSUME there cannot be an infinite number of finite naturals. I
> > > > > > proved it mathematically
> > > > >
> > > > > Those proofs require assming the equivalent of of twhat they purport to
> > > > > prove.
> > > > >
> > > > That's simply not true. Respond to them specifically, without snipping and
> > > > otherwise amngling them.
> > >
> > > OK, I will amngle not.
> > I bet you don't even know what amngling is! ;)
> > I guess it's a sort of relf-seferential term.
> >
> > > But anyway, three is too many. If it were we who
> > > were the idiots, and you were trying to explain something, you might
> > > reasonably try three different explanations, but if you claim you have
> > > proofs in the normal mathematical sense, one is enough. If the proof is
> > > valid, the existence or otherwise of other proofs is irrelevant. So
> > > *you* choose one of your proofs, and I (prolly others too; why _do_
> > > cranks attract such attention? why is it all just so tantalisingly
> > > fascinating?) will point out the first error, without mangling or
> > > anything else. Then if you wish to you can revise it and try again, or
> > > just expand the claim that we show is false.
> >
> > Would that that were sufficient. But, since they all get rejected for reasons
> > that don't pan out, I have produced three. If three proofs from different areas
> > all agree, that does bolster the argument. So, you want to hack at one at a
> > time? Okay. Try the information theory one:
>
> OK. I will snip nothing, but I will interline comments enclosed in ((
> )).
>
> Preliminary remark: it is usual to state the precise result you wish to
> prove at the beginning, so I add this, to make sure I am not
> misunderstanding.
>
> Theorem: A set of (distinct) integers having infinite cardinality
> ["size" if you like] must include at least one infinite integer.
Why don't we say: "The size of the set of all whole numbers starting from 1 is
the value of its maximal element."
>
> [Is that OK? This is what you are trying to prove, so however fervently
> you believe it to be the case, you are not allowed to use it as an
> assumption within the proof.]
>
> > Given a set of symbols with size S, we can produce a set of all strings using
> > those symbols that have length L, and the size of this set will be S^L.
>
> ((Bad wording: it is the set that is of size S, not the symbols. But
> the meaning is clear. ))
>
> > Digital
> > number systems fall into this category, with S being the number base, which is
> > always finite (2 for binary, 10 for decimal). If we want to have an infinite
> > set of digital strings, therefore, S^L needs to be infinite, but S is finite,
> > so L, the length of the strings, needs to be infinite to have an infinite set
> > of such strings.
>
> (( There is a slight problem here: are you talking about a set of
> strings, every one of which is of the same length? If so, then it is
> indeed true that the string length must be infinite. However, this is
> not the simplest way to represent pofnats, which are normally written
> as strings of varying lengths - we write 1, not ...0001. There are two
> metacases here: you will accept that pofnats are normally written, er,
> the way they are normally written, which is the case I'll address here,
> or you insist that they must have an infinite sequence of preposed
> zeros. It really makes no difference, but if you insist I'm sure
> someone will do the slight amount of extra work involved in dealing
> with your nonstandard notation.
The leading and trailing zeroes are inherently implied by the use of digital
number systems, which employ infinite series of terms representing multiples of
powers of the number base. Such insignificant zeroes are omitted ingeneral
usage for lack of infinite space and general brevity, but are not non-existent
within the digital number systems.

When you deal with computers, the digits are all there, zeroes and all. That's
the model I am working from. If I add trailing zeroes, am I representing some
DIFFERENT natural number? This simplification of the argument is entirely
warranted, but your complication of it can be accomodated with a little extra
effort.
>
> Since pofnats are written with *finite* strings of digits (not starting
> with zero), there is no single value "L"; rather integers are written
> with various lengths in the sequence 1 (nine of these), 2 (90), and so
> on. The total number of pofnats is thus sum over the sequence of string
> lengths of the number of numbers of each length. Since (at least for
> pofnats) the number of numbers of each string length is finite, in
> order for there to be an infinite number of strings, the *number* of
> different string lengths must be infinite. [I hope you will agree with
> this!]
That didn't follow. The logic proceeds from N=S^L. You made a leap here that I
never did.
>
> Well, if all of the finite lengths (which of course are the pofnats)
> form an infinite set, then there are an infinite number of finite
> lengths, and thus an infinite number of strings, without requiring any
> single string to be infinite in length.
Did you just assume what you want to prove? That's a big IF, with a NO for an
answer.

> You of course believe this not
> to be the case; you believe that for a set of string lengths to form an
> infinite set of string lengths, at least one of the lengths must be
> infinite. But you cannote *assume* this here, because we are in the
> middle of what purports to be a proof of this very same claim. So you
> have not *proved* any need for infinite digit positions. Your "proof"
> is entirely circular. ))
Except that that wasn't my proof. Where is S^L? That was bogus, and only goes
to show that you didn't follow the proof at all, nor do you understand the
basics of digital number systems, and the role of 0's in them.
>
> >
> > By the definition of digital systems, where each digit as we move left
> > represents a multiple of the next higher power of the number base, any non-zero
> > digit an infinite number of positions to the left of the digital point
> > represents a multiple of the number base to an infinite power, or an infinite
> > value. Since a digital number system fully utilizes all combinations of digits
> > to produce its values, most of the infinite strings in the infinite set will
> > have non-zero digits in infinite positions, and represent infinite values.
> >
> > Therefore, since an infinite set of digital whole numbers requires the full set
> > of infinite strings of digits, and infinite strings of digits mostly represent
> > infinite values, most values in the infinite set of digital whole numbers have
> > infinite values.
>
> (( The rest would perhaps pass without much comment if the preceding
> were valid, but it is dreadfully woolly. I don't remember seeing *any*
> proof with "mostly" in it. Anyway, you would only need to prove the
> necessary existence of a single infinite integer to win a Fields medal,
> I think. ))

"Mostly" cetrainly means more than "sometimes", which is more than "never",
which is all I have to prove, that there is AT LEAST ONE infinite value in the
set, when what I have proved is that *most* values in the infinite set are
infinite.

I am too old for a Fields Medal, so I am going to have to settle for the
satisfaction of seeing people actually think properly for a change, and stop
departmentalizing their minds in this anti-reality movement that is so
prevalent. That's reward enough, although a nice book deal would help matters
so I can move on to creating HAL and correcting particle physics and cosmology.
:D
>
>
> > So, where in this argument do you see an error? Can S^L ever be infinite and
> > produce an infinite number of strings, without infinite S and/or L? Keep in
> > mind the relation to trees. S is the number of branchings at each node. L is
> > the number of levels in the tree, or branchings in each path.
>
> I was going to 'Mark the Spot' [ X ]: it's about where you say "so L,
> the length of the strings, needs to be infinite". But you are being
> very sloppy about distinguishing the length of a particular string from
> a set of lengths of strings. Unless you have to insist that numbers
> have leading zeros - if you do, then the posnats are the subset of
> these infinite strings in which an infinite sequence of zeros is
> followed by a finite sequence of digits with a non-zero first digit.
> This makes the argument very slightly messier, as far as I can see, but
> doesn't basically alter it. Anyway, as is actually obvious at a glance,
> the basic problem is that the argument is entirely circular.
No, YOUR argument here may be circular, but it wasn't mine. Why don't you try
working with the "slightly" messier version where the string lengths are
constant and numbers have leading and trailing zeroes, as defined by digital
number systems, and start over without changing the nature of the proof. Yes, I
insist on you following the proof as given, and not contorting it to suit your
purposes.
>
>
> Brian Chandler
> http://imaginatorium.org
>
>

--
Smiles,

Tony
.

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