Thanks to Benya for the result and to Rafael for helping write this. This result is included in the latest draft of the new Definability of Truth paper.
Working in the framework presented in Definability of Truth, we prove that every reflective, coherent is supported only on non-standard models. We construct a finitely additive measure over the language which is supported on standard models and satisfies many of our desiderata. This shows that 's non-standard support is a direct result of requiring countable additivity.
##Background Note that some of the notation has changed from Definability of Truth: is written as . This is to distinguish this as a function from the language to the reals, whereas , for some , is a function from the powerset of the set of all theories to the reals. Benya has shown that , so we abuse notation and use the same symbol for both functions.
We now write for the inner language symbol instead of , as they are importantly different entities. Recall that is a function from the language to , and is a function in the metalanguage. We have modified the definition of sufficiently that we discuss it in the next section.
as a three place relation
Since we are now working in the realm of Peano arithmetic, we can weaken our definition of our inner probability distribution. Instead of defining as a function in the language, it is sufficient for to be a three place relation which arbitrarily approximates some metalanguage function.
Definition 1. Let be the extension of the language (which contains at least the language of arithmetic) by a relation symbol . A standard theory is a complete theory over which extends the theory of the standard natural numbers (over ) and satisfies the following two conditions: and
Intuitively, is a three-valued relation which approximates some function arbitrarily well. We imagine to mean that , where is the function that we require to exist above.
##Disbelief in Standard Theories
We would like to construct coherent, reflective distributions which are supported on standard theories of Peano Arithmetic. We prove that not only do such distributions not exist, but every coherent, reflective distribution must assign zero probability to any set of standard theories.
Theorem 1. Let be a consistent theory which extends Peano Arithmetic. Let be a coherent, reflective probability distribution over . Then, must assign probability zero to any set of standard theories of .
Proof. Consider the sentence defined by . Then, . Applying the reflection principle, we get that Now, so we apply countable additivity and De Morgan's laws to get that
This means that assigns probability zero to any set of theories which do not prove that . In particular, no standard theory can prove this statement, as otherwise it would contain a number smaller than one which was also greater than any standard rational smaller than one.
We can trace the cause of the result in the last section back to the countable additivity condition. To show this, we construct a finitely additive measure which fulfills many of our desiderata and is supported on standard theories.
Definition 2. Define the set to be the set of all standard theories. A is defined as . We define the base theory as .
Definition 3. Define a function such that .
Theorem 2. Existence of Finitely Additive Measure Supported on Standard Theories. is well-defined, finitely additive, and satisfies a version of the reflection principle:
Proof. We first show that is well-defined. By consistency of the reflection principle, we have that there exists a coherent, reflective over our base theory .
Say that . Then, by definition, and are in exactly the same complete theories, so must be logically valid. Therefore, by completeness, proves , so by Gaifman coherence, . Hence, is well defined.
We now check that is a finitely additive measure. Clearly, and . We need only check that it is finitely additive.
If , then is logically valid, so, by completeness, proves . Thus, by Gaifman coherence,
By the way we defined it, clearly satisfies the modified reflection principle we gave.