so throughout my reading of model theory the idea of the "empty" theory has been put down as trivial, however I am curious as to why. Let us look at the following.

Suppose We have $L_=$, the language with equality but no other relation/function/constant symbol. Let $T$ be the empty theory (so it contains no $L_=$ sentences). Is it true that any two models of $T$ of the same cardinality $\kappa$ are isomorphic? That is, show that $T$ is $\kappa$-categorical for all cardinals $\kappa$.

My intuition wants me to say that the models of $T$ of cardinality $\kappa$ are just the underlying sets of a model, given the lack of relations/formulas. For countable models I can see how this would lead to an isomorphism $\phi : A \to B$ by setting $\phi(a_i)=b_i$ for all $i \in \mathbb{N}$, $a_i \in A$, $b_j \in B$. For uncountable sets I have having difficulty I am having slightly more issues though, due to the inability to index the elements, and due to my lack of set theory knowledge (attempting to look purely from a model theory perspective). Any help would be much appreciated.


Small distinction: it's really the language being empty (EDIT: by "empty" I mean "no nonlogical symbols," that is "no relation, function, or constant symbols") that matters, not the theory. If $T$ is a theory in the empty language, it's still the case that any two models of $T$ of the same cardinality are isomorphic.

You're quite right to be a little more careful when thinking about uncountable models - a lot of the time there's extra complications, or nice facts about countable models simply aren't true - but in this case the indexing's done for you!

Saying that "$\mathcal{A}$ has size $\kappa$" means "there is a bijection $f: \mathcal{A}\rightarrow\kappa$." So if $\mathcal{A}, \mathcal{B}$ are models of size $\kappa$, then we have maps $f, g: \mathcal{A}, \mathcal{B}\cong \kappa$, and this gives a bijection between $\mathcal{A}$ and $\mathcal{B}$: $i=g^{-1}f$. The choice of a bijection with $\kappa$ might seem suspicious, but it's perfectly fine - we're not claiming there is a unique isomorphism between $\mathcal{A}$ and $\mathcal{B}$ (indeed, this would be false of course), just that there is some isomorphism.

  • $\begingroup$ Your "small distinction" is a bit confusing: what you mean by an "empty language" is that the language has no non-logical symbols (where equality counts as a logical symbol). Apart from that, it's a good answer. $\endgroup$ – Rob Arthan Mar 16 '15 at 20:09
  • $\begingroup$ Generally, the "language" refers to the set of non-logical symbols (or at least that's how I've always seen it in model theory texts). $\endgroup$ – Noah Schweber Mar 16 '15 at 20:11
  • $\begingroup$ Another very common practice is to call the set of non-logical symbols (equipped with their arity) as the "signature" of a theory and use "language" to mean the set of all well-formed formulas over the signature. I always try to hedge my bets by writing "language with no non-logical symbols". $\endgroup$ – Rob Arthan Mar 16 '15 at 20:22
  • 1
    $\begingroup$ That's true, and "vocabulary" is also used. It's a good point - when in doubt, I should be clearer. I do have silly philosophical reasons for using "language" to denote only the nonlogical symbols, which accounts for my preference. :P $\endgroup$ – Noah Schweber Mar 16 '15 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.