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

$$\{x \epsilon A: x=a \ \ or \ \ x=b\}$$

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)'?

• The word "contains" is ambiguous and sometimes means "as an element" and sometimes "as a subset". If $x$ is an element of $X$, then it is not necessarily a subset. So $X$ contains $x$ as an element, but not necessarily as a subset. – Asaf Karagila Dec 28 '18 at 7:56
• Further to @AsafKaragila's point, in terms of subsets it "contains" $\{a\}$ etc. – J.G. Dec 28 '18 at 8:14

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 $$\emptyset \in A.$$

1. You can verify $$a$$ satisfies the condition $$x=a\lor x=b$$, so $$a\in A$$. Similarly, $$b\in A$$.
2. Yes.
3. It has $$a$$ and $$b$$ as elements and nothing else, and in particular $$\emptyset\not\in\{a,\,b\}$$ (unless $$a=\emptyset\lor b=\emptyset$$).
• @drhab Thanks; fixed. – J.G. Dec 28 '18 at 15:23