Definable and quasi-definable over models. 
Definition: A set X is quasi-definable over C if is definable over all model M  containing C.
i) Show that a singleton $\{b\}$ is quasi-definable over C iff $b$ is algebraic over C
ii) Let $L = \{E\}$, T the theory of an equivalence relation with $n$ infinite classes, $n>1$. Show that equivalence classes are not definable over $\emptyset$ but if are quasi-definable over $\emptyset$.

For the point i) $\Rightarrow$ follows directly from the definition because as $\{b\}$ is quasi-definable then $b$ is algebraic since $\{b\}$ is finite. But the other direction is not clear for me because is true.
For the point ii) as  $L = \{E\} \ $   the only formulas over  $\emptyset$ 
are $ x=x $ or $xEx$ that defining the entire model and negations and conjunctions of these. So an equivalence class is not definable. 
but I could not see why are quasi-definable over  $\emptyset$.
I appreciate any hint. Thanks
 A: I assume you are working inside a monster model $\cal U$.
i)($\Leftarrow$) An element $b$ that is algebraic over $C$ belongs to every model containing $C$. In fact, suppose $\phi(b)\wedge\exists^{=n}x\,\phi(x)$ and let $M$ be any model containing $C$. Then $M\models \exists^{=n}x\,\phi(x)$. Hence $M$ contains all $n$ solutions of $\phi(x)$, so $b$ is among these.
ii) Note that there is a first-order sentence saying that there are $n$ non $E$-equivalent elements. This sentence has witnesses in any model containing $C$. This shows that every equivalence class is quasi-definable. Note also that there are $C$-automorphisms that swaps equivalence classes. Hence these cannot be definable.
Allow me a disgression.
Automorphisms are the key to understand this matter. The following proposition should answer your first question.
Proposition An element $b$ is algebraic over $C$ iff it has a finite orbit over $C$.
Proof ($\Rightarrow$) Suppose $\phi(b)\wedge\exists^{=n}x\,\phi(x)$ and then $\phi(fb)$ for every $C$-automorphism.
($\Leftarrow$) Let $M\supseteq C$ be any model (of small cardinality). Let $p(x)={\rm tp}(b/C)$. By homogeneity $p({\cal U})$ is the orbit of $b$ over $C$. If $p({\cal U})$ is infinite than it has the cardinality of $\cal U$. Then there is $c\notin M$ such that $c\models p(x)$. By homogeneity there is an $C$-automorphism $f$ such that $fb=c$. Then $b\notin f^{-1}[M]$. Note the latter is a model containing $C$. 
By the way note that a similar fact holds for definable sets.
Proposition A definable set $\varphi({\cal U},b)$ is quasi-definable over $C$ iff it has a finite orbit Aut$({\cal U}/C)$.
