Testwiki:Reference desk/Archives/Mathematics/2016 September 8

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September 8

Model theory

Let's say we are given two first-order formulas $\alpha ,\beta$ - each of which has two free variables. Let's assume that it follows from Peano system that for every $A,B,$ there exists $v$ satisfying both: $\alpha (A,v)$ - and $\beta (B,v)$.

Is it provable (maybe by Compactness theorem? ) that Peano system has a model in which, for every $A$ there exists $v$ satisfying both: $\alpha (A,v)$ - and $\beta (B,v)$ for every finite $B$?

HOTmag (talk) 17:48, 8 September 2016 (UTC)

Either I'm misunderstanding what you're asking, or it's trivial. Your premise is that PA proves $(\mathrm{\forall }w)(\mathrm{\forall }u)(\mathrm{\exists }v>u)\varphi (v,w)$? In any nonstandard model, fix $u$ nonstandard. Then for every $w$ there is a $v>u$ with $\varphi (v,w)$, by assumption. This $v$ is as desired.--2406:E006:384B:1:8C1E:B081:1ABF:3C81 (talk) 12:56, 9 September 2016 (UTC)

Thanks to your important comment, I've just changed slightly my original post. Please have a look at the current version of my question. HOTmag (talk) 13:31, 9 September 2016 (UTC)

I think this still isn't what you mean to ask. Let $\alpha (A,v)$ be a tautology, and let $\beta (B,v)$ be $B=v$. Then it's certainly true that for every pair $A,B$ there is a $v$ -- namely, $v=B$. But there is no $v$ that works for every finite $B$.--2406:E006:384B:1:8C1E:B081:1ABF:3C81 (talk) 14:34, 9 September 2016 (UTC)

That's what I meant. Thanks to your trivial example, I see I was wrong about my assumption. Thank you. HOTmag (talk) 15:21, 9 September 2016 (UTC)