Subset (Comprehension, Separation) Axiom and Definability I am reading Moshe Machover's book, Set Theory, Logic, and Their Limitations, and on p. 19 he states that if $A\cup B$ is a set, then $A$ and $B$ are too by the Subset Axiom.  But this confuses me.  Clearly, $A$ and $B$ are both subsets of $A\cup B$.  But how do we know that they are definable?  (Similarly, if $A$ is a class of sets, and $\cup A$ is a set, then I take it that $A$ is supposed to be a set by Powerset and Subset.  But my question re-arises here.)  I'm obviously missing something basic.  Thanks, in advance.
 A: Classes are defined as "extensions" of properties [page 16].
The "universe of discourse" is made of sets and individuals.

A class that is not a set is called a proper class; a proper class is not 
  an object, and therefore cannot be a member of any class [page 16]. 

Thus, sets are those that some axiom of the theory claims that they exists, like e.g. 

3.2. Axiom of Pairing (A2) : For all objects $a$ and $b$ the class $\{ a, b \}$ is a set [page 17], 

or some theorem of the theory proves that they exists, like e.g. 

3.9. Theorem : $\emptyset$ is a set [page 18]. 

We have that: 3.14. Definition [page 19] defines the union (or join) $A \cup B$ of two classes $A$ and $B$, which - of course - is a class. 
We do not know that, in general, $A \cup B$ is a set, but we can prove it:

3.15. Theorem: $A \cup B$ is a set iff both $A$ and $B$ are sets. 

The first part: if $A,B$ are set, then also $A \cup B$ is, is proved by Axiom of Pairing (A2) and Axiom of Union set (AU).
The second part: if $A \cup B$ is a set, then also $A,B$ are, is proved through:

3.6. Axiom of Subsets (AS) : If $B \subseteq A$ and $A$ is a set then so is $B$.

We know that $A \cup B = \{x : x \in A \text { or } x \in B \}$ is a set; but, e.g. $B \subseteq A \cup B$, and thus by AS also $B$ is a set (the same for $A$).  
For $B \subseteq A \cup B$, we have to apply: 

3.4. Definition: Let $A$ and $B$ be classes. If every member of $B$ is also a member of $A$, we say that $B$ is a subclass of $A$, briefly: $B \subseteq A$. 

Clearly, if $a \in B$, then $a \in B$ or $a \in A$, and thus $a \in A \cup B$. 
A: Assuming your set theory is sufficiently strong, this is not true:
Let $M$ be a countable transitive model of some large fragment of your set theory. Then there is a subset $X \subseteq \omega$ such that $X \not \in M$ (since $M$ is countable). If $\omega \setminus X$ were in $M$, then so would $X$ be (because we can correctly calculate complements in $M$). Hence $X, \omega \setminus X \not \in M$, but $\omega = X \cup (\omega \setminus X) \in M$. 
