Why does Kunen assume non-emptiness of the model for relative consistency proofs?

Question

Please bear with me, this question is difficult for me to phrase, and I am struggling with how best to describe my confusion.

I am currently re-reading Kunen's set theory text (the one that introduces independence proofs), and I am struggling to come to terms with Lemma 2.3 from Chapter IV $\S$2. It states:

Let $S,T$ be two sets of sentences in the language of set theory [we may as well consider them to be two sets of sentences in the same language; the actual language is immaterial for my question], and suppose that from $T$, we can prove that $\mathbf{M}$ (a class; alternatively, a predicate) is a model for $S$ and that $\mathbf{M}$ is non-empty. Then, $\text{Con}(T)\implies\text{Con}(S)$.

I am familiar with the method of forcing. In particular, the underlying mechanics of it essentially amount to:

1. Assume a countable model of $\mathsf{ZF}$. Call this new theory $\mathsf{ZF}^+$. i.e. $\mathsf{ZF}^+$ is the theory $\mathsf{ZF}$ plus the assumption that $\mathsf{ZF}$ has a model/is consistent. Call this model $\mathbf{M}\neq 0$.
2. Construct the model $\mathbf{M[G]}$ of $\mathsf{ZF}+\{\sigma\}$, where $\sigma$ is some sentence.

From this vantagepoint, it is straightforward to see $\text{Con}(\mathsf{ZF})\implies\text{Con}(\mathsf{ZFC}+\{\sigma\})$. And there is of course no need to restrict ourselves to a single sentence $\sigma$.

I suspect this is more or less what Kunen is trying to say in the aforementioned lemma. I only fail to see why $\mathbf{M}$ is non-empty. I think that it precludes the following from happening:

Let $S$ be the set of sentences $\{\forall x(x=x),\forall x(x\neq x)\}$. Then $\mathbf{M}=0$ models a contradiction and in particular $\text{Con}(T)\implies\text{Con}(S)$ is meaningless in the best case and false in the worst case since $S$ is inconsistent. Is this more or less what is being gotten at here? Or am I maybe missing something more elementary?

I guess I should also remark that I know that $S$ as I defined above isn't of the form $\sigma,\neg\sigma$, but without it, I am unable to rationalize the non-emptiness of $\mathbf{M}$.

Answer 1

I don't completely understand your rationale, but it seems vaguely on the right track.

The key thing is that Kunen is assuming the most common form of first-order logic where empty models are disallowed, and using a deductive system that reflects this. So, for instance, $\exists x\; x=x$ is provable in first-order logic, but false relative to the empty set.

So, if we didn't stipulate $\mathbf M\ne 0,$ we could let $T$ be literally any theory and $S = \{\lnot \exists x \; x=x\},$ and thus we could conclude via $\mathbf M=0$ that $\mathsf{Con}(T)\to \mathsf{Con}(S).$ But $S$ is inconsistent, so we conclude that every theory $T$ is consisent... obviously absurd.

Where this comes up formally is when we show that proofs from $S$ can be converted into proofs from $T$ relative to $\mathbf M,$ and thus if $S\vdash \psi,$ then $T\vdash \psi^{\mathbf M}$ (where $\psi$ is a contradiction in this case). The rationale is just like the soundness theorem for first order logic: in $T,$ we can prove any sentence of $S$ relative to $\mathbf M,$ and all the logical axioms and rules of inference of logic can be relativized to $\mathbf M$ as well. But, for $\mathbf M=0,$ the logical axioms can't be relativized to $\mathbf M.$