Can a theory be not satisfiable? Idk if I expressed myself right, can it be that a theory has no model?

Question

According to lecture on FO logic: A Σ-theory T is defined by a set of Σ-sentences.

According to another lecture on mathematical logic (predicate logic): A theory is a satisfiable set T ⊆ FO (τ) of sentences, which is closed under |=, meaning it holds for all τ-sentences ψ with T |= ψ, that ψ ∈ T.

I am asking because of this true/false statement (*): There exists a signature Σ and a Σ-theory T such that no Σ-formulas are T-satisfiable.

A formula is T-satisfiable, if there exists a structure to satisfy both T and this formula.

The statement could be true if the theory is unsatisfiable, meaning it has no model. If the theory has a model that satisfies it, you could pick the formula to be one of the sentences of the theory.

Answer 1

Can a theory be not satisfiable?

If you restrict the definition of theory to "satisfiable set" of sentences, the answer is NO.

But, in general, we have that a formula (set of formulas) is unsatisfiable iff it is inconsistent.

Thus, a single-axiom "theory", like e.g.$\{ \ \forall x (x \ne x) \ \}$ has no model.

Every inconsistent set of sentences is unsatisfiable, i.e. has no model.

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The "standard" use of theory in logic is:

a set of sentences in a formal language.

See also David Marker, Model Theory: An Introduction (Springer, 2002), page 14:

Let $\mathcal L$ a language. An $\mathcal L$-theory $T$ is simply a set of $\mathcal L$-sentences.