1
$\begingroup$

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)'?

$\endgroup$
  • 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
3
$\begingroup$

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 $$\emptyset \in A.$$

$\endgroup$
3
$\begingroup$
  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 $a=\emptyset\lor b=\emptyset$).
$\endgroup$
  • $\begingroup$ @drhab Thanks; fixed. $\endgroup$ – J.G. Dec 28 '18 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.