Defining unordered pairs in set theory I am reading Naive Set Theory by Paul Halmos and am on section 3 (page 9) where he is talking about the axiom of pairing. In his explanation he states that a and b are two sets and A is the set containing a and b. He defines the unordered pair {a,b} as
$$ \{x \epsilon A: x=a \ \ or \ \ x=b\} $$
He then says that this set contains a and b.
I have three questions about this:


*

*How do we know that the set/unordered pair contains both of the sets a and b if the condition for the element x of the given set is that it is equal to a or equal to b (I'm interpreting the 'or' as the logical operator).  

*If A had no elements other than the sets a and b, does A={a,b}? 

*Is {a,b} read as 'the set containing a and b (and the empty set)'?
 A: *

*You can verify $a$ satisfies the condition $x=a\lor x=b$, so $a\in A$. Similarly, $b\in A$.

*Yes.

*It has $a$ and $b$ as elements and nothing else, and in particular $\emptyset\not\in\{a,\,b\}$ (unless $a=\emptyset\lor b=\emptyset$).

A: I think you mean that $A$ is a set containing $a$ and $b$ (but perhaps contains other sets, so there could be other sets containing both $a$ and $b$—like $\{a,b\}$ itself). There is not such thing as the set containing $a$ and $b$.
What there is, though, is the set containing exactly $a$ and $b$ and nothing else; what is defined as $\{a,b\}$. Don't be confused by the or in the definition: it is not a set containing $a$ or $b$, but the set whose elements $x$ are in $A$ and also satisfy the condition $x=a \vee x=b$. If $x=a$, then satisfies the condition, and so
$$x=a\in \{a,b\};$$
if $x=b$, then satisfies the condition, and so
$$x=b\in \{a,b\};$$
if none of $x=a$ or $x=b$ is true, then
$$x\notin \{a,b\},$$
so $\{a,b\}$ contains both $a$ and $b$, but nothing different from them both. Also, if both $x=a$ and $x=b$ are true, which implies $a=b$, then $x\in \{a,b\}$, too (the set contains only one element).
Your second affirmation is true, since two sets are equal if they have the same elements, or more precisely
$$A=B \iff (x\in A \iff x\in B).$$
Finally, $\{a,b\}$ does not necessarily contain $\emptyset$; just $a$ and $b$. It only turns out to be the case that
$$\emptyset \in \{a,b\}$$
if $a=\emptyset$ or $b=\emptyset$ (or both, of course).
You should not confuse the statements
$$x\in A$$
and
$$x\subset A.$$
While in common speech both could be read as '$x$ is contained in $A$', I'm only using this expression to mean the former, not the later. The former is also read as '$x$ is one of the elements of the set $A$'. The later, instead, means that every element of the set $x$ is also an element of the set $A$, but not necessarily that $x$ is itself an element of the set $A$ (this could be the case, though). 
Since it is never the case that some set is an element of $\emptyset$, it is true for any set $A$ that
$$\emptyset \subset A,$$
but not always
$$\emptyset \in A.$$
