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)'?

  • 2
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Dec 28 '18 at 7:56
  • $\begingroup$ Further to @AsafKaragila's point, in terms of subsets it "contains" $\{a\}$ etc. $\endgroup$ – 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 $a=\emptyset\lor b=\emptyset$).
  • $\begingroup$ @drhab Thanks; fixed. $\endgroup$ – J.G. Dec 28 '18 at 15:23

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