Conference rusure::math

1302.0. "Non-standard analysis" by HERON::BUCHANAN (combinatorial bomb squad) Thu Sep 27 1990 07:21

	An interesting note from a regular contributor to the net.   The
discussion concerns 'proofs' that 4.9999... = 5.   The 10x - x argument that
is referred to is:

	Suppose x = 4.9999...
	Then  10x = 49.999...
	10x - x = 45
	So x = 45/9 = 5

	What follows is Wlod's rebuttal of KRamden's criticism of Wlod.   The
beginning and the end are slightly emotional.   The value of the note is that
it is a lucid example of a relatively advanced topic.

Regards,
Andrew.

-------------------------------------------------------------------------------

In article <12041:Sep2604:27:[email protected]> [email protected] (Dan Bernstein) writes:
>In article <[email protected]> [email protected] (Wlodzimierz Holsztynski) writes:
>  [ on the widespread misconception that 4.9999... < 5 ]
>> Why misconceptions? Perhaps a shortcoming of the formal education.
>> It only shows that the non-standard analysis is more intuitive than
>> the classical analysis.
>
>Say what? This has nothing to do with nonstandard analysis.

Say what?? You don't see it. But it has ev'rything to do with the
non-standard analysis in the context. It's so funny that you've spent so
much energy on a negative activity and that the need to prove someone wrong
has blinded you.

> The sum of 9/10^j, j > 1, is 1 in the nonstandard reals.

The issue was not well defined and one can argue either way.
I mean that you can argue, as you did, in a way which explains
nothing, or one can shed the light on the psychological issue
we were dealing with.  Thus I am going to justify the intuition
that .999... < 1 in the context of non-standard analysis, in
a precise mathematical way, once we agree on the interpretation
(which is always necessary before you do math, and it should be done
in a sensible way - sometimes there is more than one, but yours
was not one of them).

I will also show that the  "10x - x" argument, which shows that
4.999... = 5 in classical model fails in the non-standard one.

Let  N = {1,2,...} be the set of positive integers.  Let  F  be
a non-principal ultrafilter in  N, i.e.  F  is a family of subsets
of  N  such that (i)the empty set nor any one-element set do not
belong to F  (ii) intersection of any two members of F belongs to F,
and (iii) for any subset  A  of  N  either  A  or  N\A belongs to F.

It follows that if  A  is a subset of N, and  A  contains a subset
which belongs to  F,  then  A  belongs to  F  (it's a nice exercise,
a mood improver).

Two real functions (sequences)  x,y : N --> R  are called almost
equal if  x|A = y|A  for a set  A  from  F, i.e. if the set of all
n such that x(n)=y(n)  belongs to F.  This is an equivalence relation.
We can identify the (standard) real numbers with the classes of the
constant sequences  x : N --> R.  We introduce the arithmetic
operations and ordering in the most natural way, coordinate by coordinate.
For instance we say that  [x] < [y]  if  the set of  n  for which
x(n) < y(n)  belongs to  F (we could say, if  x < y almost everywhere).
It is pleasant to observe that you get a linear ordering (rather than
merely partial).  You can also see that  if  [x] != 0  then  there
exist  x' almost equal to  x  such that  x'(n)  is never 0  -  just
modify  x only where  x(n)=0;  then define  x'(n)=1,  and otherwise
let  x'(n) = x(n).  Now you can define  1/[x] = [1/x'].  Now it is
easy to believe, or rutine and just tedious to prove that the set of
all classes, R', forms an ordered field, which extends the standard R
(after you identify reals with the classes of constant sequences).

Now let  x(1) = .9  and  x(2) = .99  etc.  Obviously  [x] < [1].
Or  let  y(1) = 4  and  y(2) = 4.9  and y(3) = 4.99  etc.
Again  [y] < [5].  Hence students could claim the non-standard
intuition.  They could even make fun of the  "10x -x" proof.  They
were right feeling uneasy and "cheated",  because in their model
this proof is false.

Indeed,  [10]*[y] = [10*y] = [z],  where  z(1) = 40,  z(2) = 49,
z(3) = 49.9,  z(4) = 49.99  etc.  Now  [z] - [y] = [d], where
d(1) = 36,  d(2) = 44.1,  d(3) = 44.91,  d(4) = 44.991,  d(5) = 44.9991
etc.  It follows  that  [d]/[9] = [y'],  where

y'(1) = 36/9 = 4
y'(2) = 44.1/9 = 4.9
y'(3) = 44.91/9 = 4.99
y'(4) = 44.991/9 = 4.999  etc.

We see that  y' and y  are one and the same sequence, hence
( [10]*[y] - [y] ) / [9]  =  [d]/[9]  =  [y]  and of course not [5],
as one new from the begining.

I went thru this proof anyway just to show that even in the classical
framework it was a bad proof.  One can't proof that  4.999... = 5
without going into the nity-gritty of the limit notion. And that's what
I really have proved here (meta-mathematically speaking).
	
>
> ... but I found the above description as an easier way to
                      ^^^^^^^^^^^^^^^^^
He means thru the power series'

>introduce people to what's going on. For some reason it seems to annoy
                                                      ^^^^^^^^^^^^^^^^^
>the hell out of set theorists. Don't ask me why.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Where, the hell :-), did you get that impression?  I am sure that you're
just making it up for an extra effect.  (A majority of mathematicians
enjoy having different models, constructions, proofs, approaches -
that's to a great extend what mathematics is about.)

>
>---Dan

>Let  N = {1,2,...} be the set of positive integers.  Let  F  be
>a non-principal ultrafilter in  N, i.e.  F  is a family of subsets
>of  N  such that (i)the empty set nor any one-element set do not
>belong to F  (ii) intersection of any two members of F belongs to F,
>and (iii) for any subset  A  of  N  either  A  or  N\A belongs to F.

	One thing not tackled in the base note...

	Can you prove the existence of some such F, constructively or
non-constructively?

Regards,
Andrew.

    4.99999... = 5
    That depends on the Machine floating point Epsilon value :-)
    so now both sides are happy .
    /naser

	re .1,

>>	One thing not tackled in the base note...
>>
>>	Can you prove the existence of some such F, constructively or
>>non-constructively?

	You can prove the existence of some such F nonconstructively
	assuming Zorn's lemma, or even constructively if someone hands
	you a choice function on the collection of all nonempty subsets
	of the positive integers.

	This is Tarski's ultrafilter theorem, or the prime ideal theorem,
	that the axiom of choice implies that every filter over a
	Boolean algebra can be extended to an ultrafilter.  The converse
	does not hold, though; the prime ideal theorem does not imply
	the axiom of choice.  It does imply, though, that every set can
	be given a total linear ordering.

	Dan

.0>	Let  N = {1,2,...} be the set of positive integers.  Let  F  be
.0>	a non-principal ultrafilter in  N, i.e.  F  is a family of subsets
.0>	of  N  such that (i)the empty set nor any one-element set do not
.0>	belong to F  (ii) intersection of any two members of F belongs to F,
.0>	and (iii) for any subset  A  of  N  either  A  or  N\A belongs to F.

    .1>>	One thing not tackled in the base note...
    .1>>
    .1>>    Can you prove the existence of some such F, constructively or
    .1>>    non-constructively?

    This is the heart of the matter.

    Thanks Dan (.3), I recalled from course in logic and model theory a 
    decade ago, the non-principal ultrafilter "construction" was based 
    on AOC (axiom of choice) and thought it was in fact equivalent to it
    which I now know is wrong. 

    Its precisely these unintuitive applications of AOC that 
    make me real nervous. Its clear to me that if I hand a choice to 
    explain either the concepts of convergence of sequences and series 
    or the non-standard reals with its reliance on non-principal 
    ultrafilter constructions, the former wins hands down. Non-standard
    analysis proponants will argue that because of their theory
    the "delta x" or infinitessimals calculus arguments of Liebnitz 
    can be made rigorous. Conceding this, I STILL believe its at too 
    heavy a cost...I just can't see/draw this ultrafilter.

    -Franklin

        If you limit yourself to the constructible reals and
        sequences of them (or to the ordinal definable ones) then
        you don't need the axiom of choice to construct the
        nonstandard model.
        
        Dan

>.3	
>       You can prove the existence of some such F nonconstructively
>	assuming Zorn's lemma, or even constructively if someone hands
>	you a choice function on the collection of all nonempty subsets
>	of the positive integers.
        
        Let me phrase that better.  Suppose N is the set of positive
        integers, X is the set of all subsets of N, and Y is the set
        of all subsets of X.  I meant a function c with domain Y such
        that c(A) is an element of A for every nonempty A in Y.
        
        It is easy to define a function c' with domain X such that
        c'(B) is an element of B for every nonempty B in X.  Just let
        c'(B) be the least positive integer in B.  The ultrafilter
        ("some such F") can be constructed from c, not from c'.
        
        Dan