Re: On the diagonal argument (3)

[Graham Cooper]

On May 21, 10:41 pm, Gus Gassmann <horand.gassm...@googlemail.com>
wrote:
> On May 20, 11:58 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
>
>
> > We know that a bijection can be given between the rationals and the
> > naturals.  We can encode rationals as binary strings, consider the complete
> > list of rationals (now complete in the sense of countability), and the
> > diagonal argument shows that not all binary strings represent rational
> > numbers.  Now suppose that a bijection between reals and naturals is given
> > (*).  As above for the rationals, the diagonal argument would show that not
> > all binary strings represent real numbers.  (IOW, that a real number is not
> > *any* binary string.)
>
> > One could read this as such:  given any type of number, we can encode it as
> > a binary string, and the diagonal argument shows that not all binary strings
> > represent such number.  So, the diagonal argument is rather, and more
> > fundamentally, showing that there is no possible complete list  of all
> > binary strings, conversely, that given any encoding (an encoding is always
> > "countable") of numbers to binary strings, there are binary strings that do
> > not encode any of those numbers: they possibly represent "meta-numbers", in
> > an escalation that can never end.  In a word, there are more strings than
> > numbers (any numbers).
>
> *IF* the reals were countable, then you could infer this. However, as
> long as you construct the string according to the rules (consisting
> only of digits, perhaps a sign at the left end, and a decimal point
> somewhere), the string _looks_ like a real number, and if it looks
> like a real number, it *is* a real number, because the initial
> substrings (those start start at the leftmost position) form a Cauchy
> sequence, and thus have a limit, which is the entire string and
> therefore is a real number. So you have constructed a real number on
> the list, i.e., the list could not possibly have contained all the
> real numbers to begin with.
>
> > (*) After all, a bijection between reals and naturals seems possible when
> > one provides an exact definition of real number, and, more to the point, an
> > explicit *encoding* of such real numbers to binary strings.  The hypothesis
> > that real numbers correspond to the encompassing set of all binary strings
> > rather leads, by the diagonal argument (as read above), to an ultimate set
> > of numbers and non-numbers, the non-numbers being the never empty set of
> > numbers we have not yet defined (not-a-number's, properly).  I'd rather call
> > these
> > ultimate numbers the meta-numbers, obviously a super-set of the reals, but
> > are they really so, i.e. numbers?  (A number is all we can do with it.)
>
> Not only are they numbers, they are *real* numbers.

No they're a topologocial model, a superset of anything accessible.

The only reason uncountable theory has taken favour is because you
lumped uncountable Omega in there!

You wouldn't call AD(N2NPERMUTATION(DIGITPOS)) a missing row would
you?

xxOxx..
xxxOx..
Oxxxx..
xOxxx..
xxxxO..
..

So why do you call AD(X=Y(DIGITPOS)) a missing row?


Oxxxx..
xOxxx..
xxOxx..
xxxOx..
xxxxO..
..


Herc
--
http://tinyurl.com/AntiDiagonals
http://tinyrul.com/MissingSet

[David Bernier]

[...]

Speaking of topology, it's useful in proving the intermediate
value theorem or IVT:
IVT: If f: [0, 1] -> Reals is continuous, a = min(f(0), f(1)),
b = max(f(0), f(1)), then if s is any real
satisfying a <= s <= b, there is a t in [0,1]
where f(t) = s [exactly ].

Topology is necessary because all we know of the function 'f'
is that it is continuous function on the unit interval.
There are all kinds of bizarre continuous real-valued functions
on [0, 1] ; so in rigourous analysis, one uses properties of
the reals to prove the IVT.

Anciently, there was the question whether a non-constant
polynomial function over the domain of complex numbers always
had at least one complex number root. A now standard theorem,
the fundamental theorem of algebra, states that "Yes, that's
the case". In rigourous analysis, properties of the real numbers
and the complex numbers are used to show this.

A special case is a polynomial:
X^k + C_{k-1} X^(k-1) + ... + C_{1} X + C_0 ,
where the coefficents C_{k-1} ... C_{1}, C_0 are real numbers,
_and_ k is an odd number positive integer ( k is one of 1, 3, 5, etc.).

Suppose I tell you that, loosely speaking,
lim_{X -> +oo} f(X) = +oo and lim_{X -> -oo} f(X) = -oo ;

is that enough for you to believe that at some point 'a' on
the real number line, f(a) = 0 (that the graph of f
crosses the x-axis in the Cartesian plane?)

David Bernier

[LudovicoVan]

"David Bernier" wrote in message news:jpf82p$r7r$1...@dont-email.me...

> in rigourous analysis, one uses properties of
> the reals to prove the IVT.

If I get the sense of your objection, there is no such problem: there are
many flavours of non-standard analysis, and they are of course at least as
rigorous as the standard one (and not necessarily more difficult).
Conversely, you don't (necessarily) need the standard reals to be able to do
analysis.

-LV

[Michael Press]

In article <jpf82p$r7r$1...@dont-email.me>,

In particular that an interval is connected.
Then let A = {x: f(x) < s} b = {x: f(x) > s}.
A and B are both open by definition of continuous function.
A union B =/= [0,1] because [0,1] is connected.
Hence there is a z in [0,1] such that f(z) = s.

Proving real intervals are connected is a bit of work.

--
Michael Press