Suppose we have a first order set theory $T$ that is finitely axiomatizable. Is there a general way to formalize a set theory $T^+$ that extends $T$ and that is slightly stronger than $T$ and that is not finitely axiomatizable? I need $T^+$ to only have the set ontology, i.e. doesn't prove existence of proper classes.
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asked Jun 28 at 18:25
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$\begingroup$ What does "slightly stronger than $T$" mean? Would $$T+Con(T)+Con(T+Con(T))+...$$ work? (Note that for any "reasonable" theory $T$, the infinite consistency extension $T+Con(T)+...$ is never finitely axiomatizable by the second incompleteness theorem.) Of course this trick assumes that $T$ isn't too unsound; if you want one which always works, with more effort we can find a family of Rosser-type sentences $(\sigma_i^T)_{i\in\omega}$ which are uniformly independent of each other over $T$ (= no finite set implies one not in the set), and consider $T\cup\{\sigma_i^T: i\in\omega\}$. $\endgroup$– Noah SchweberCommented Jun 28 at 19:13
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$\begingroup$ No! That's too much stronger. Something like $T+Con(T)$ or something near this vicinity. $\endgroup$– Zuhair Al-JoharCommented Jun 28 at 19:16
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$\begingroup$ $$ \begin{align} & T+Con(T)+Con(T+Con(T))+... \\ {} \\ & T+\operatorname{Con}(T)+\operatorname{Con}(T+\operatorname{Con}(T))+\ldots \\ {} \\ & T+\operatorname{Con}(T)+\operatorname{Con}(T+\operatorname{Con}(T))+\cdots \end{align} $$ @NoahSchweber You will have noticed that the formatting of the three dots in the first line above looks conspicuously different from what you see in the second line above. Do you know that what is in the first line will strike some people the way a spelling error does? And $Con$ in the first like differs from $\operatorname{Con}$ in the second. And${}\,\ldots$ $\endgroup$– Michael HardyCommented Jun 28 at 20:41
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$\begingroup$ $\ldots\,$and the third line also differs from the first two. $\endgroup$– Michael HardyCommented Jun 28 at 20:41
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2$\begingroup$ @MichaelHardy OK? For what its worth, of your three examples the first two are equi-good to me and the third is the only one which strikes me as wrong. I think these sorts of typographical preferences are going to vary between people, and aren't really worth much energy. $\endgroup$– Noah SchweberCommented Jun 28 at 20:47
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1 Answer
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For any "reasonable" theory $T$, we can find a computable sequence of sentences $(\sigma_i^T)_{i\in\omega}$ such that
$T\cup\{\sigma_i^T: i\not=n\}\not\vdash\sigma_n^T$ for each $n$ (so the extension $T\cup\{\sigma_i^T: i\in\omega\}$ is not finitely axiomatizable), and
the theories $T$ and $T\cup\{\sigma_i^T: i\in\omega\}$ have the same consistency strength (precisely: $\mathsf{PA}$, or even much less, proves $Con(T)\rightarrow Con(T\cup\{\sigma_i^T: i\in\omega\})$).
I believe this is folklore, but I'll try to track down a reference; for now, see the discussion here (although that only treats the case of generating a single sentence). Note that these new $\sigma$-sentences are in the original language of $T$ itself, so we never wind up e.g. introducing classes to our set theory.
answered Jun 28 at 19:19
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$\begingroup$ if we assume $T$ can define the naturals. Then if I add a new constant $K$ to the language of $T$, and add the axiom schema that for any formula $\phi$ in the language of $T$ we have $(\exists !n : \phi(n))\to \forall n \, (\phi(n) \to n < K)$. where $n$ range over the naturals. Clearly $K$ is a non-standard natural that is indefinable in the language of $T$. Would that qualify as a theory that extends $T$ (albeit in a minimally different language) and that is not finitely axiomatizable? Or it can be finitely axiomatized? $\endgroup$ Commented Jul 1 at 18:43