at the moment I try to figure out some details of Kunen's "Relative Interpretation" Definition (within the 2013 Edition of his "Set Theory", p. 99 to 100):

Definition If $\Lambda$ is some axioms in [the language] $\{\in\}$ for set theory, and $\mathcal{L}$ is a finite lexicon, a relative interpretation of $\mathcal{L}$ in $\Lambda$ consits of a non-empty class $A$ together with an assignment of a suitable semantic entity $s^\mathfrak{A}$ to each symbol of $\mathcal{L}$. Specifically,

If f is an $n$-ary function symbol [in $\mathcal{L}$] with $n > 0$, then $f^\mathfrak{A} : A^n \rightarrow A$ [...]

He mentions under this definition that in fact one does not assign to $f$ a function $f^\mathfrak{A}$ within $\{\in\}$ (since $A$ might be a proper class) but a formula $\phi(x_1, ... , x_n, y)$ such that \begin{align*} \Lambda \vdash \forall x_1, ..., x_n \in A . \exists ! y \in A . \phi(x_1, ... x_n, y) \end{align*}

... and so on for constants, etc. But how would one relativize \begin{align*} f(c) = c \end{align*}

as an example? Since $\phi_f(x,c) = \phi(c)$ does not make sense at all?

I took his example literally: let $\mathcal{L} = (0, +, \cdot)$ then I might define \begin{align*} \phi_0(y) & := \forall x\in A. x \notin y \\ \phi_{+}(x_1, x_2, y) & := y \in x_1 \vee y \in x_2 \wedge \neg (y \in x_1 \wedge y \in x_2)\\ \phi_{\cdot}(x_1, x_2, y) & := y \in x_1 \wedge y \in x_2 \end{align*}

but how to relativize $x_1 + x_2 = x_2 + x_1$ then?


PS: I'm aware of the more pedantic definition in Jech's Set Theory, but there it's only defined for $\mathcal{L} = \{\in\}$ and so it is in all the relevant questions on math.stack I have found so far.


In your example, $x_1+x_2=x_2+x_1$ gets relativized to be: $$\forall x_1\forall x_2\forall y\Bigl(\phi_+(x_1,x_2,y)\leftrightarrow\phi_+(x_2,x_1,y)\Bigr)$$

In general, replacing symbols by formulas only complicates things in terms of longer formulas, with more quantifiers and with longer proofs (well, depending on your inference rules, I guess).

It's a pain in the lower lower-back, but once you understand how it's done, it's not hard to see why this is true. And to understand how it's done, just look at the example above.

  • $\begingroup$ thanks! I thought of this, but what disturbs me: somehow, here, $=$ gets shifted to $\leftrightarrow$. This holds in a general way if one relativize via the given procedure? $\endgroup$ – aphorisme Jul 23 '14 at 18:41
  • $\begingroup$ Just think about this in general, $=$ is an equivalence relation. If $\varphi(x,y,z)$ is the formula $x+y=z$ then $x_1+x_2=x_2+x_1$ is translated to $\leftrightarrow$ quite immediately. Within the same language. $\endgroup$ – Asaf Karagila Jul 23 '14 at 18:50
  • $\begingroup$ ah... I see! Thanks again! $\endgroup$ – aphorisme Jul 23 '14 at 19:16
  • $\begingroup$ Again a question: The point here might be, that $=$ in $\mathcal{L}$ is somehow something different to the $=$ in $\{\in\}$, because it does compare different "things", or? $\endgroup$ – aphorisme Jul 23 '14 at 19:26
  • $\begingroup$ No, $=$ is part of the underlying logic. $\endgroup$ – Asaf Karagila Jul 23 '14 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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