Ytukyg

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:

1. 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).




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




3. Is {a,b} read as 'the set containing a and b (and the empty set)'?

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
$$x=ain {a,b};$$
if $x=b$, then satisfies the condition, and so
$$x=bin {a,b};$$
if none of $x=a$ or $x=b$ is true, then
$$xnotin {a,b},$$
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 $xin {a,b}$, 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
$$A=B iff (xin A iff xin B).$$

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
$$xin A$$
and
$$xsubset A.$$
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
$$emptyset in A.$$

1. You can verify $a$ satisfies the condition $x=alor x=b$, so $ain A$. Similarly, $bin A$.


2. Yes.


3. It has $a$ and $b$ as elements and nothing else, and in particular $emptysetnotin{a,,b}$ (unless $a=emptysetlor b=emptyset$).