Re: defining countable ordinals, how far can we go?

On Nov 11, 11:26 am, LauLuna <laureanol...@xxxxxxxx> wrote:
On Nov 11, 1:17 am, David Bernier <david...@xxxxxxxxxxxx> wrote:



It's easy to define or describe ordinals below omega^omega using
polynomials in Z[x] which
have no negative coefficients. For example, omega^2 + omega*3 + 7
corresponds to:
x^2 + 3*x + 7 . Obviously, we can have exponential towers
omega^(omega^omega) and
so on. Beyond that, there are the recursive ordinals, those alpha for
which there exists a computable
function f: NxN -> {0, 1} where f(m,n) = 1 iff m =n or m precedes n,
in some ordering
which happens to be a well-ordering isomorphic to the ordering by
"subset_or_equal"
on elements of alpha, whether we can prove it or not.

Using sentences in first order set theory notation, one can no doubt go
further.

Suppose all countable ordinals could be defined finitely in some way
(informal notion
of finite definability). Then we would have aleph_1 definitions, but
only aleph_0 finite
strings, which is a contradiction.

On the other hand, if we can define ordinals below a coutable ordinal
alpha finitely,
can we define alpha finitely? It may depend on what is meant by "define
finitely".

Probably the question of how far one can define finitely the countable
ordinals shouldn't be taken too seriously.

David Bernier

What you propose is essentially Berry's paradox.

'The least undefinable ordinal' defines no ordinal. Why? Because
informal definitions must be scattered along a hierarchy of logical
levels, to avoid paradox. Now, 'the least undefinable ordinal'
specifies no level of definability.

Consider 'the least undefinable ordinal at level n' and call 'D' that
definition. Then D is at level n+1.

This solves the paradox.

***************

1. potential

2. actual
.

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