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Given a class $C$ of arithmetical sentences, an arithmetical theory $T$ is said to be $C$-sound if all the theorems of $T$ which are in $C$ are true. For instance, $T$ is $\Sigma_1$-sound if all the $\Sigma_1$ theorems of $T$ are true.

Now, for some classes $C$, like the class of $\Sigma_1$ sentences, the statement "$T$ is $C$-sound" is expressible in the language of arithmetic. For other classes, like the class of all arithmetical sentences, $C$-soundness (which is just soundness) can't be definable in the language of arithmetic. But there are some classes, like the class of $\Pi_1$ sentences, which have a rarer property: not only is $C$-soundness definable in the language of arithmetic, it can be defined within $C$ itself.

However, I'm looking for an even rarer property:

does there exist a class $C$ for which $C$-soundness is definable within $C$, and which is also closed under negation?

Or, failing that:

does there exist a class $C$ for which "$T$ is not $C$-sound" is expressible within $C$?

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Could you clarify the contet of $T$? I had thought of it as fixed, but when you refer to expressing "$T$ is $C$-sound", it seems perhaps that you treat it more like a variable. After all, if $T$ is a true theory, then $T$ is $C$-sound for every $C$. For example, the empty theory is always $C$-sound for every $C$. So these would seem to be expressible by any tautology. But that wouldn't seem to be what you want; so I am unsure exactly what you want. –  Joel David Hamkins Jul 24 '13 at 3:34
    
@JoelDavidHamkins When I talk about C-soundness being definable in the language of arithmetic, I mean that "T is C-sound" is expressible in the language of arithmetic for all arithmetical theories T. –  Keshav Srinivasan Jul 24 '13 at 3:43
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If $T$ is arithmetical of even moderate complexity, then why should the assertion that $T$ is $\Pi_1$-sound be $\Pi_1$-expressible? Or are you only considering computably-axiomatizable theories rather than arithmetically axiomatizable theories? Also, I still find it unclear what you mean by saying "$T$ is $C$-sound" is expressible, since $T$ is a theory and not a number variable. Do you mean, given a number that is a TM program $p$ that accepts a theory, then the assertion "the theory accepted by program $p$ is $C$-sound" is expressible as a formula in variable $p$? –  Joel David Hamkins Jul 24 '13 at 3:50
    
I don't even know what "arithmetically axiomatizable" means. I'm just talking about recursively axiomatizable theories in the language of first-order arithmetic. As far as what I mean by C-soundness being expressible, yes we can say that "the theory encoded by p is C-sound" is expressible as an arithmetical formula in p. –  Keshav Srinivasan Jul 24 '13 at 7:16
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That's fine, and restricting to computably axiomatizable theories resolves all my issues. An arithmetically axiomatizable theory, in contrast---and this is what I had thought you meant by an "arithmetic theory"---is a theory with an arithmetically definable set of axioms; that is, it is $\Sigma^0_n$-axiomatizable for some $n$ rather than $\Delta^0_1$-axiomatizable as with computably axiomatizable theories. The theory of true arithmetic assertions, for example, is not arithmetically axiomatizable by Tarski's theorem, but of course any computably axiomatizable theory is. –  Joel David Hamkins Jul 24 '13 at 12:36

1 Answer 1

up vote 7 down vote accepted

The answer to both questions is negative.

Theorem. There is no class $C$ of formulas in the language of arithmetic, such that the assertion "$T$ is not $C$-sound" is uniformly expressible in $C$ for c.e. theories $T$ (regarded as an index of $T$ as a c.e. set).

Proof. Suppose $C$ is like that. By the Gödel fixed-point lemma, there is a sentence $\psi$ such that $$\text{PA}\vdash\psi\longleftrightarrow (\ \{\text{PA}+\psi\}\text{ is not }C\text{-sound }).$$ The assumption on $C$ ensures that the assertion "$T$ is not $C$-sound" is expressible in $C$, and so it follows that $\psi$ is PA-provably equivalent to a sentence in $C$, and we may assume without loss that $\psi$ is actually in $C$.

On the one hand, if $\psi$ is true, then $\text{PA}+\psi$ is true and therefore also sound, since every true theory is sound. But since $\psi$ asserts that $\text{PA}+\psi$ is not $C$-sound, and since this assertion is in $C$, then it must be true that $\text{PA}+\psi$ is not sound, a contradiction.

On the other hand, if $\psi$ is false, then because $\psi$ is in $C$, it is true that $\text{PA}+\psi$ is not $C$-sound. But in this case, $\psi$ would be true, contrary to our assumption. QED

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Thanks @JoelDavidHamkins! I suspected that Tarski's theorem would be involved. In fact, the whole reason I asked this question was trying to see what would happen if you restricted soundness to some arithmetically definable C-soundness property, but it looks like there's no evading the logic of Tarski. –  Keshav Srinivasan Jul 24 '13 at 14:28

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