A: Goodstein’s argument over the natural numbers

For any given natural number $m$, we can express the Goodstein sequence $G(m)$ for $m$ in a form where each term of the sequence is expressed in it’s hereditary representation:

[1]     $G(m) \equiv \left\{g_{1}(m)_{[2]},\ g_{2}(m)_{[3]},\ g_{3}(m)_{[4]}, \ldots \right\}$

where the first term $g_{1}(m)_{[2]}$ denotes the unique hereditary representation ofthe natural number $m$ in the natural number base $[2]$:

e. g. $g_{1}(9)_{[2]} \equiv 1.2^{(1.2^{1.2^{0}}+1.2^{0})} + 0.2^{(1.2^{1.2^{0}}+0.2^{0})} + 0.2^{1.2^{0}} + 1.2^{0}$

and if $n > 1$ then $g_{(n)}(m)_{[n+1]}$ is defined recursively from $g_{(n-1)}(m)_{[n]}$ as below.

B: The recursive definition of Goodstein’s Sequence over the natural numbers

For $n > 1$ let the $(n-1)^{th}$ term $g_{(n-1)}(m)$ of the Goodstein natural number sequence $G(m)$ be expressed syntactically by its hereditary representation as:

[2]     $g_{(n-1)}(m)_{[n]} \equiv \sum_{i=0}^{l} a_{i}.n^{i_{[n]}}$

where:

(a) $0 \leq a_{i} < n$ over $0 \leq i \leq l$

(b) $a_{l} \neq 0$

and for each $0 \leq i \leq l$ the exponent $i$ too is expressed syntactically by its hereditary representation $i_{[n]}$ in the base $[n]$; as also are all of its exponents and, in turn, all of their exponents, etc.

We then define the $n^{th}$ term of $G(m)$ as:

[3]     $g_{n}(m) = \sum_{i=0}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} )-1$

C: The hereditary representation of $g_{n}(m)$

Now we note that:

(a) if $a_{0} \neq 0$ then the hereditary representation of $g_{n}(m)$ is:

[4]          $g_{n}(m)_{[n+1]} \equiv \sum_{i=1}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} ) + (a_{0}-1)$

(b) whilst if $a_{i} = 0$ for all $0 \leq i < k$, then the hereditary representation of $g_{n}(m)$ is:

[5]          $g_{n}(m)_{[n+1]} \equiv \sum_{i=k+1}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} ) + {c_{k}}_{[n+1]}$

where:

$c_{k} = a_{k}.(n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}} - 1$

$= (a_{k} - 1).(n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}} + \left\{(n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}} - 1\right\}$

$= (a_{k}-1).(n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}} + n\left\{(n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}-1} + (n+1)^{k_{[n\ \hookrightarrow\ (n+1)]}-2} \ldots +1\right\}$

and so its hereditary representation in the base $(n+1)$ is given by:

${c_{k}}_{[n+1]} \equiv (a_{k}-1).(n+1)^{k_{1[n+1]}} + n\left\{(n+1)^{k_{2[n+1]}} + (n+1)^{k_{3[n+1]}} \ldots +1\right\}$

where $k_{1[n+1]} \equiv k_{[n\ \hookrightarrow\ (n+1)]}$ and $k_{1} > k_{2} > k_{3} > \ldots \geq 1$.

D: Goodstein’s argument in arithmetic

For $n > 1$ we then consider the difference:

$d_{(n-1)} = \left\{g_{(n-1)}(m)_{[n]}- g_{n}(m)_{[n+1]}\right\}$

Now:

(a) if $a_{0} \neq 0$ we have:

[6]          $d_{(n-1)} = \sum_{i=0}^{l} (a_{i}.n^{i_{[n]}}) - \sum_{i=1}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} ) - (a_{0}-1)$

(b) whilst if $a_{i}=0$ for all $0 \leq i < k$ we have:

[7]          $d_{(n-1)} = \sum_{i=k}^{l} (a_{i}.n^{i_{[n]}}) - \sum_{i=(k+1)}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} ) \\- (a_{k} - 1).(n+1)^{k_{1[n+1]}} - n\left\{(n+1)^{k_{2[n+1]}} + (n+1)^{k_{3[n+1]}} + \ldots +1 \right\}$

Further:

(c) if in equation (6) we replace the base $[n]$ by the base $[z]$ in the term:

[8]          $\sum_{i=0}^{l} a_{i}.n^{i_{[n]}}$

and the base $[n+1]$ also by the base $[z]$ in the term:

[9]          $\sum_{i=k+1}^{l} (a_{i}.(n+1)^{i_{[n\ \hookrightarrow\ (n+1)]}} ) + (a_{0}-1)$

then we have:

[10]          $d^{\prime}_{(n-1)} = \sum_{i=0}^{l} (a_{i}.z^{i_{[n\ \hookrightarrow\ z]}}) - \sum_{i=1}^{l} (a_{i}.z^{i_{[n\ \hookrightarrow\ z]}}) - (a_{0}-1)$

$= 1$

since $(i_{[n\ \hookrightarrow\ (n+1)]})_{[(n+1)\ \hookrightarrow\ z]} \equiv i_{[n\ \hookrightarrow\ z]}$;

(d) whilst if in equation (7}) we replace the bases similarly, then we have:

[11]          $d^{\prime}_{(n-1)} = \sum_{i=k}^{l} (a_{i}.z^{i_{[n\ \hookrightarrow\ z]}}) - \sum_{i=(k+1)}^{l} (a_{i}.z^{i_{[n\ \hookrightarrow\ z]}}) \\- (a_{k}-1).z^{k_{1[(n+1)\ \hookrightarrow\ z]}} - n\left\{z^{k_{2[(n+1)\ \hookrightarrow\ z]}} + z^{k_{3[(n+1)\ \hookrightarrow\ z]}} + \ldots +1\right\}$

$= a_{k}.z^{k_{[n\ \hookrightarrow\ z]}} - (a_{k} - 1).z^{k_{1[(n+1)\ \hookrightarrow\ z]}}) - n(z^{k_{2[(n+1)\ \hookrightarrow\ z]}} + z^{k_{3[(n+1)\ \hookrightarrow\ z]}} + \ldots +1)$

$= z^{k_{1[(n+1)\ \hookrightarrow\ z]}}-n(z^{k_{2[(n+1)\ \hookrightarrow\ z]}} + z^{k_{3[(n+1)\ \hookrightarrow\ z]}} + \ldots +1)$

where $k_{1[(n+1)\ \hookrightarrow\ z]} \equiv k_{[n\ \hookrightarrow\ z]}$,

and $k_{1[(n+1)\ \hookrightarrow\ z]} > k_{2[(n+1)\ \hookrightarrow\ z]} > k_{3[(n+1)\ \hookrightarrow\ z]} > \ldots \geq 1$.

We consider now the sequence:

$G(m)_{[z]} \equiv (g_{1}(m)_{[2\ \hookrightarrow\ z]},\ g_{2}(m)_{[3\ \hookrightarrow\ z]},\ g_{3}(m)_{[4\ \hookrightarrow\ z]}, \ldots )$

obtained from Goodstein’s sequence by replacing the base $[n+1]$ in each of the terms $g_{n}(m)_{[n+1]}$ by the base $[z]$ for all $n \geq 1$.

Clearly if $z > n$ for all non-zero terms of the Goodstein sequence, then $d^{\prime}_{(n-1)} > 0$ in each of the cases—equation (10) and equation (11).

(Since we then have:

$d^{\prime}_{(n-1)} \geq (z^{k} - (z-1)(z^{(k-1)}+z^{(k-2)}+z^{(k-3)}+\ldots+1))=1$

in equation (11).)

The sequence $G(m)_{[z]}$ is thus a descending sequence of natural numbers, and must terminate finitely in $\mathbb{N}$ if $z > n$.

Since $g_{n}(m)_{[(n+1)\ \hookrightarrow\ z]} \geq g_{n}(m)_{[n+1]}$ if $z > n$, Goodstein’s sequence $G(m)$ too must terminate finitely in $\mathbb{N}$ if $z > n$.

Obviously, since we can always find a $z > n$ for all non-zero terms of the Goodstein sequence if it terminates finitely in $\mathbb{N}$, the condition that we can always find some $z > n$ for all non-zero terms of any Goodstein sequence is equivalent to the assumption that any Goodstein sequence terminates finitely in $\mathbb{N}$.

E: Goodstein’s argument over the ordinal numbers

We shall see next how Goodstein mirrors the above argument over the ordinals and curiously concludes the Theorem that, since we now have $\omega > n_{o}$ for any finite ordinal $n_{o}$, therefore any Goodstein sequence over the natural numbers must terminate finitely since a corresponding Goodstein sequence of transfinite ordinals cannot be an infinitely descending sequence of ordinals!