[Math Noise] (fwd)

Jim Choate ravage at EINSTEIN.ssz.com
Mon Jan 20 07:22:46 PST 1997



> Infinity does not have a predecessor, so it makes no sense to
> count back from it a finite number of steps.

If infinity does not have predecessors (ie is immune to normal arithmetic
operations) then it is not possible for a sequence to approach it by adding
a finite amount to succesive terms in order to approach it. This means that
a sequence can not meaningfuly be asymptotic with infinity (meaning I have to
be able to draw a asymptote, at least in theory, in order to demonstrate the
limit).

> If one constructs the Ordinals, which are isomorphism classes of
> well-ordered sets, and the Cardinals, which are equivalence
> classes of equipotent sets, one will automatically end up with
> all sorts of transfinite numbers.

If infinity is not a number, how is it possible to have a definite number
(ie transfinite) which is larger than it?

My contention is that number theory as you present it is playing fast and
loose with the concept of infinity not being a number or visa versa.


                                                  Jim Choate
                                                  CyberTects
                                                  ravage at ssz.com







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