Re: Cantorian pseudomathematics

MoeBlee said:
> Tony said:
> > Can we consider it an axiom that a^b is finite for finite a and b, but not for
> > infinite a or b and nonzero a and b?
>
> Forget about axioms here. We already have theorems. For x,y finite, x^y
> is finite. For either x or y infinite and x not=0 and y not=0, x^y is
> infinite.
>
> > If so, then perhaps we can agree that the
> > set of all strings of length L, with a finite alphabet of size S, is S^L, which
> > cannot be infinite unless L is infinite?
>
> Correct. We ALREADY agreed. It's a theorem of set theory.
>
> > If we consider digital number systems
> > in this context, and the fact that each digital number has an implicit infinite
> > leading string of 0's,
>
> What are we talking about? Real numbers with digital representation? If
> so, forget about "implicit" leading strings. Set theory doesn't have
> "implicit" things.

Do not dismiss this point. This is crucial, and handwaving dismissal is a
mistake here. Do you or do you not agree that, in digital number systems,
....0000000abc.xyz0000.... = abc.xyz, and that there is no difference between
their natural values? Can you see that every finite digital number is an
infinite string, with infinite strings of leading and trailing 0's?

>
> > can we not say that every set of strings of a given
> > length L includes as a subset the strings of all lesser lengths, since the
> > leading 0's don't change the value of the string?
>
> Forget about "leading zeros". We're talking about strings of length L.
> And no, the set of strings of length L does NOT have as a subset the
> strings of length less than L. The set of strings of length L has BY
> DEFINITION only strings of length L.

You are being dense again. The strings of length L represent natural values,
given the digital representation, right? Does or does not the set of strings of
length L include each and every natural value expressed by all strings of
length less than L? Yes or no?

>
> > If so, then for the set of
> > digital numbers to be actually infinite, we have to allow each string to have
> > an infinite number of bits, and not all leading 0's, or you haven't added any
> > new values to the set.
>
> I ALREADY showed you how, with just ONE symbol in the alphabet, we have
> an infinite number of strings. That's because strings don't have to be
> of just ONE length L, but of an UNLIMITED number of FINITE lengths, L1,
> L2, L3, ... Suppose the one symbol is '1'. Then we have these strings:
>
> 1
> 11
> 111
> .....
>
> Infinite number of strings on an alphabet of just one symbol.
>

Nice of you to bring this up. I have several comments.

First, please demonstrate Cantor's diagonal proof on the uncountability of the
reals, using unary.

Second, notice how the largest value in columns is always equal to the number
of values in rows. Always equal. Hmmmmm......

Third, unary numbers are a tad different than numbers in other bases. There are
no 0's, because they wouldn't add anything, and would cause all strings with
the same number of 1's to be equal, and not unique values. So, this has no
bearing on the leading and trailing 0's in binary and higher bases.

Fourth, and most important, you have no infinite strings there, only finite
strings, and a finite number of them, the same finite number as the finite
length of the longest string you have. You have an infinite set when you reach
an infinite string.
--
Smiles,

Tony
.

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