In Tarskian semantics, can we assign a name to every object, in every model of every first order theory?

Question

Question: In Tarskain semantics, can we assign a name to every object, in every model, of every first order theory?

Definition of a name: In Tarskain semantics, a 'name' is what is referred to as a'constant symbol' in most logic texts. In particular, it is an individual non-logical constant (e.g. the '$a$' in $Pa$).

Motivation: The reason I ask this question is because in most logic texts, constant symbols are ascribed to each object without much explanation as to how, it is just assumed by the reader that it can be done, even if we have to enumerate an infinite list of formulae.

If it is not the case that every object can be named, in every model, of every first order theory, then this procedure can't always be done. This is interesting because one might naively assume in the case of models with countably infinite objects, that with countably infinite names all objects could be assigned one, because the cardinality of both collections are the same.

Answer 1

Obviously not. The pure FOL theory with no non-logical symbols and no axioms is satisfied by an uncountable model, but certainly there are only countably many formulae... Worse still, there is not even a single formula that uniquely picks out a single element in this model...

Answer 2

By way of contradiction, let us assume that we can name (assign a constant symbol to) every object, in every model of every first order theory.

Since we can name every object, consider a theory $T$ composed of an infinite set of formulae, comprising of a property for every name $(a,b,c,\dots)$ in the language $(1)$:

\begin{equation}\tag{1}\{Pa, Pb, Pc\dots\}\end{equation}

Then consider those same formula in a theory $T$, which also contains the sentence $(2)$.

\begin{equation}\tag{2}\exists x \lnot Px\end{equation}

Now each propositional atom cannot be satisfied unless that atom is false, so it cannot be the case that $\exists x \lnot Px$ is true.

However, by compactness, since each finite subset is satisfiable, the whole thing is.

This is a contradiction, so it must not be the case that for every first order theory, that we can name every object, in every model of the theory with Tarskian semantics.

Answer 3

Consider a language with a constant symbol $c$ and the theory: $\exists x(x=c) \land \forall y (y=c)$. This is the first-order theory of a model with one element so you can name every element in any model of this theory.