Models of the empty theory 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. 
 A: 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.
