Consider the first order language ‎$‎‎‎\mathcal{L}=\{\in,\subseteq\}‎$ and ‎$‎‎\{\in\}$-theory ‎$\text{ZFC}$.‎ ‎Is ‎the‎re a formula ‎$‎‎\psi ‎(x,y)‎ \in \{\subseteq\}-Form‎$ ‎with ‎the ‎following ‎property?‎‎

‎$‎‎ZFC\cup\{‎\forall ‎x‎\forall ‎y~(x\subseteq y‎\leftrightarrow ‎‎\forall z (z\in x\rightarrow z\in y))‎‎\}\vdash$$ ~\forall ‎x‎\forall ‎y~(x\in y‎\leftrightarrow \psi ‎(x,y)‎‎)‎‎$‎

  • $\begingroup$ How about $\varphi(x,y)\equiv y\in x$, $\psi(x,y)\equiv y\in' x$? $\endgroup$ – Christian Remling Apr 23 '14 at 23:44
  • $\begingroup$ Is this a homework problem? The formulas $x\notin y$ and $x\notin'y$ seem to work. $\endgroup$ – Joel David Hamkins Apr 23 '14 at 23:44
  • $\begingroup$ The previous comments were for the original version of the question (see edit history). $\endgroup$ – Joel David Hamkins Apr 24 '14 at 1:14

The edited question is much more interesting than the original.

The answer is no, you cannot define $\in$ from $\subseteq$. To see this, note from this previous MO answer that there are nontrivial automorphisms of $\langle V,\subseteq\rangle$. Namely, let $f:V\to V$ be any permutation, and let $\pi(x)=f[x]$, the pointwise image of $x$ under $f$. It follows that $x\subseteq y\iff \pi(x)\subseteq\pi(y)$. But we may easily choose $f$ so that $\pi$ does not respect $\in$, that is, so that for some $x,y$ we have $x\in y$ but $\pi(x)\notin \pi(y)$. For example, let $f$ swap $\emptyset$ with some other set nonempty set $a$, and let $x=\emptyset$ and $y=\{\emptyset\}$, so that $x\in y$, but $\pi(x)=f[\emptyset]=\emptyset$ and $\pi(y)=f[\{\emptyset\}]=\{f(\emptyset)\}=\{a\}$ so that $\pi(x)\notin\pi(y)$.

It follows that we cannot define $\in$ from $\subseteq$, since $\psi(x,y)\iff \psi(\pi(x),\pi(y))$ for any formula using only $\subseteq$, since it is an $\subseteq$-automorphism.

  • $\begingroup$ It is a nice approach. It uses existence of $\subseteq$-automorphisms of the universe and non-existence of such automorphisms for $\in$ but what about the case that one replaces $\subseteq$ by another relation $E$ such that $\langle V,E\rangle$ has no automorphism? Should we begin producing models of ZFC for any possible $\{E\}$ formula or there is a shortcut solution? $\endgroup$ – user45939 Apr 24 '14 at 2:24

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