First, let me say that this is a really great question.
It seems to me that any increasing base-bumping function would give the
same Goodstein result that you eventually hit $0$. That is, I claim that for any increasing
sequence of bases $b_1$, $b_2$ and so on, if we define the
Goodstein sequence by starting with any number $a_1$, and
then if $a_n$ is defined, we write it in complete base
$b_n$, replace all instances of $b_n$ with $b_{n+1}$,
subtract $1$, and call the answer $a_{n+1}$. The theorem
would be that at some point $n$ in the construction, we
have $a_n=0$.
The proof of the original theorem proceeded by associating
any number $a$ in complete base $b$ with the countable
ordinal obtained by replacing all instances of $b$ with the
ordinal $omega$ and interpreting the resulting expression
in ordinal arithmetic. They key fact is that the ordinal
associated with $a$ in base $b$ is strictly larger than the
ordinal associated in base $b+1$ with the number obtained
by replacing all $b$'s with $b+1$'s and subtracting $1$.
If we replace $b$ with some larger $b'$ and
do the same thing, then it appears that this key fact still goes
through, since it was proved by observing what happens when the subtract-$1$ part causes a complex term to be broken up with coefficients below the new base. Thus, the newly associated ordinals would still
be descending, so they must hit $0$, but this happens only
if the numbers themselves hit $0$.
No comments:
Post a Comment