Working in first order logic, let $\mathcal{M} = (M,I)$ be an $\mathcal{L}_A$-structure, where $A$ consists only of the unary function symbol $F_1$ (here, $M$ is the universe and $I: \{x_i\} \rightarrow M$ the interpretation of variables in $M$). I was asked to come up with
- an example of an infinite $\mathcal{M}$ in which every singleton $\{m\}$ in $M$ is definable, and
- an example of an infinite $\mathcal{M}$ in which no nonempty finite subset of $M$ is definable.
I was hoping to clear up some of my confusion related to the meaning of "$A$ consists only of $F_1$" which seems to be holding me up:
If I take this to mean that $A$ also includes the symbol $\hat{=}$, then can I not take care of (1) right off the bat and make $\{m\}$ trivially definable from parameters in the set $\{m\}$ using $\varphi(x_1, x_2) := (x_1 \hat{=} x_2)$? Then only those $\mathcal{M}$-assignments $\nu$ with $\nu(x_2) = m = \nu(x_1)$ witness membership in $\{m\}$. And I don't see how any automorphism of $\mathcal{M}$ can expand the set definable via $\varphi[m,m]$ beyond $\{m\}$, since $m$, itself a parameter here, must be a fixed point of any such automorphism.
But then under this reading, how do I block against definability of singletons in the second case whatever the $\mathcal{M}$? The $\varphi(x_1, x_2)$ above seems to work whatever the $\mathcal{M}$ unless I am grossly misunderstanding the satisfaction notion for formulas of the form $(a \hat{=} b)$.
[And even so, I don't see how I could read "$A$ consists only of $F_1$" to mean that $A$ excludes the symbol "$\hat{=}$". If I excluded $\hat{=}$ from $A$, how would anything be definable? I wouldn't even have $\mathcal{L}_A$-formulas to define things with, since giving up $\hat{=}$ and predicate symbols I would not even be able to form the atomic formulas, right?]