This question was inspired by Joel David Hamkins's excellent [question](https://mathoverflow.net/questions/205985/is-there-a-leibnizian-model-with-no-definable-elements-in-a-finite-language) on Leibnizian structures with no definable elements. Let $n$ be a positive integer. Is there an infinite structure in a finite language which is Leibnizian, and has exactly $n$ definable elements? The structure has to be infinite, because otherwise we can just take the linear order with $n$ elements.



An infinite Leibnizian structure in a finite language with precisely $n$ definable elements
*Turning my comment into an answer to move this off the unanswered queue:*


Take a Leibnizian structure with no definable elements and just add $n$ constants to it (and make the structure "otherwise boring" on those elements - e.g. no relation involving one of those elements should hold, and any function involving one of those elements as an input should do something boring like output the first coordinate always). The result will be a Leibnizian structure with exactly $n$ definable elements.


One natural follow-up question is whether there is a relational Leibnizian structure $\mathcal{A}$ with exactly one definable element $a$ such that $\mathcal{A}\setminus\{a\}$ is **not** Leibnizian. The answer to this is **yes**: following [Joel's construction](https://mathoverflow.net/a/206058/8133), let $A\subseteq\mathbb{Z}$ be such that $(\mathbb{Z};<,A)$ is Leibnizian and has no definable elements. We now replace the unary predicate $A$ with a binary predicate $R$ and a fresh element $\*$: let $\mathcal{A}$ be the $\{R,<\}$-structure with domain $\mathbb{Z}\sqcup\{\*\}$, where $<$ is interpreted as usual and $$R^\mathcal{A}=\{(x,y): x=\*\wedge y\in A\}.$$ Clearly $\mathcal{A}$ is Leibnizian with unique definable element $\*$ and removing $\*$ from $\mathcal{A}$ results in *(something equivalent to)* the very non-Leibnizian structure $(\mathbb{Z};<)$. A similar trick can produce for any $k<n$ a Leibnizian structure with exactly $n$ definable elements which becomes non-Leibnizian if we remove at least $k$ of those definable elements.


Of course in a sense all of these are cheats. My next guess at the "right" question is whether there is a Leibnizian structure which is not mutually interpretable with any Leibnizian structure without definable elements. I suspect the answer to that question is negative, but I'm not sure.


