Today we’re going to follow Hinman’s proof of Tarski’s theorem. The proof is by contradiction, and employs diagonalization. First, assuming clauses (i)-(iii) on definability, a universal -section for the class of
-definable sets is established and it’s diagonal is shown to not be definable over
.
A -section is a set defined for any set
and any relation
such that for each
,
.
is universal for a class
if, and only if, every member of the class
is a
-section. The diagonal set of
is
.
Second, assuming the definability of , there is a formula
such that for all
,
. But since the function
is effectively computable, it is definable and so,
, for some
. But this means that,
and
,
which means that is definable. This is a contradiction. So,
is not definable and not effectively countable. Nor is
effectively countable.
In the next update (probably on a plane!) I’ll discuss this theorem and get into Hellman’s example.