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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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