Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can the statement

CH is not provable in ZFC

be formalized as en $\epsilon$-Formula $\phi$ s.t. $ZFC \vdash \phi $

If so why is it refered to as an "metatheorem".

share|improve this question
add comment

1 Answer 1

If ZFC is consistent, then no, it does not prove the assertion "CH is not provable in ZFC", since the non-provability of any assertion in a theory implies the consistency of that theory, and so this would mean that ZFC proves its own consistency, contrary to the incompleteness theorem.

However, you probably mean to refer to the relative consistency assertion

If ZFC is consistent, then CH is not provable in ZFC.

This indeed is formalizable as an assertion of number theory, and it can be proved in PA and indeed much weaker systems. In particular, yes, it can also be formalized in the language of set theory and proved in ZFC itself.

Such theorems are refered to as metatheorems because they are theorems about what is and is not provable. Generally the metatheorems are provable in extremely weak theories of arithmetic, simply because we do not generally make substantial mathematical assumptions in the metatheory, but only in the object theories.

But of course, inspecting such an assertion in the pure language of set theory would shed almost no illumination on the mathematics of the situation---it would be like inspecting the assembly language instruction set resulting from compiling a program written in a high-level language. But in principle, yes, such statements can be formalized this way.

share|improve this answer
    
As someone who spent quite a lot of time with hex editors and disassemblers, I can actually testify that you can learn a lot about the code itself if you put your mind into it. It's just not very easy to do... –  Asaf Karagila Apr 11 '13 at 7:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.