Two questions about ZFC Assume $ZFC$ is consistent and consider its model $M$. Let $a$ be an element of $M$ and $C=\{b:b\in a\}$ is a class of all elements from $a$:  


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*Is there such a logic formula $\phi(x)$ that $C=\{b:M\models\phi(b)\}$ (i.e. the class equality holds)?

*Let $D\subseteq C$ (i.e. $D$ is a subclass). Is there such an element $b$ from $M$ that $D=\{c:c\in b\}$?

 A: The answers are yes/no and no respectively. Throughout, I'll assume ZFC is consistent.
First, let's rephrase the question a bit more clearly:

Suppose $M$ is a model of ZFC (note that there are many such $M$ if ZFC is consistent) and $a\in M$. Is $$Set_M(a):=\{b: M\models b\in a\}$$ definable-with-parameters in $M$? Is $Set_M(a)$ definable-without-parameters in $M$? And if $X\subseteq Set_M(a)$, need $X$ be definable - with or without parameters - in $M$?



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*Note that by the Lowenheim-Skolem theorem there are such $M$ of arbitrary (infinite) cardinality.


The first question is trivial: just take the parameter $a$, and look at the formula $\varphi(x)\equiv x\in a$.
The second question has a negative answer in general, via a cardinality argument: there are only countably many parameter-freely-definable subsets of a model of ZFC, so any uncountable $M$ gives a counterexample (and appropriate countable elementary submodels of such yield countable counterexamples). However, there are models in which every element is parameter-freely-definable: these are the pointwise definable models. But they are rare.
The third question has a negative answer in general, again via cardinality: if $M\models $ZFC is countable, then $\mathcal{P}(\omega)^M$ is countable but the true powerset of $M$'s $\omega$ is uncountable, and no model whatsoever has the parameter-freely-definable version (since $\mathcal{P}(\omega^M)$ is always uncountable). However, for parameter-free-definability specifically it is consistent with ZFC (under very mild additional assumptions) that there are examples where the property does hold - for a massively overkill example, if $\kappa$ is inaccessible then $V_\kappa$ has this property.
