Re: Cantor Confusion
- From: Andy Smith <andy@xxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 05 Jan 2007 15:04:01 EST
Are you familiar with the Peano axioms? Do you know what mathematical
induction is?
Yes, and yes, even though my training is in physics.
It's as easy proof by induction that every natural number can be represented by
a finite number of digits.
Yes, but I am suspicious of rules of inference and axioms
when applied to infinite sets. Many things don't work or
give indeterminate answers when applied to infinite
series.
On which step does it appear?
The last one! (NB, iteration formula was wrong, should have had Nm = N(m-1)+ N(m-1) + N(m-1)+ N(m-1)+ 2)
That's why we have axioms.
Yes, but I am not wholly convinced about their validity to infinite sets
or to the application of otherwise irrefutable rules of
inference in these circumstances.
To answer my own question, the difference between enumerating the
rationals and the reals is one of process. The standard
scheme for mapping the rationals means that one can pick
any rational whatever and easily compute its corresponding
mapping integer. In contrast, with my suggested scheme for
mapping the reals, it is easy to pick any irrational
number and demand its indes - and the answer is only
available when all the counting is complete. So, fair
enough.
But I would turn around the question. Suppose we have
the set of all integers {N} as a completed thing. It is easy
enough to visualise that - e.g. label all the terms
in the series 1 + 1/2 + 1/4 + 1/8 ..., and consider the
interval [0,2] which then contains, in a bag as it were, all
of them.
View them as binary numbers, and reflect {N} about the
binary point 010 becomes .010... etc.
What is the largest fraction present in the reflected
set? Is it .111111... ?
Presumabaly you would say not. You would say, it can be as close to
111111... as you like, but is always epilson away from it.
But, this is a complete set - there is no process. So, if
111111... is not the largest member, what is it?
You didn't comment on the application of Cantor's diagonalisation argument
to {N} - again, with the inverted system above, that
would seem to suggest, if the argument is valid (and
in this case, the infinite matrix is definitely diagonal)
that there are integers not present in {N}.
But I'm a physicist.
.
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