I am trying to understand section 3 of Naive Set Theory by Halmos.

First assume that there is a set, then by axiom of specification, empty set exists. And then he motivates the necessity of axiom of pairing by the following reasoning:

For all we know there is only one set and that one is empty. Are there enough sets to ensure that every set is an element of some set? Is it true that for any two sets there is a third one that they both belong to? What about three sets, or four, or any number? We need a new principle of set construction to resolve such questions. The following principle is a good beginning.

Axiom of pairing: For any two sets there exists a set that they both belong to.

My question is, how does Axiom of pairing help if we only have one set to begin with?

  • $\begingroup$ The two sets don't have to be distinct. $\endgroup$ Commented Sep 23, 2017 at 5:22
  • $\begingroup$ Pairing doesn't imply distinct. If you have the empty set, you can have the empty set twice. So there is a set containing the empty set. And you can look twice to see that it contains the empty set. Linguistically dubious, I know. But mathematically sound. $\endgroup$
    – fleablood
    Commented Sep 23, 2017 at 5:23

1 Answer 1


Applying the axiom of pairing to $\varnothing$ and $\varnothing$ gives $\{\varnothing\}$, which is different from $\varnothing$. Now we have two sets to play with, and we can apply the axiom to $\{\varnothing\}$ and $\varnothing$ to get a third set, $\{\varnothing,\{\varnothing\}\}$. And so on and so forth.

The axiom of pairing and the later axiom of union together allow for the construction of arbitrarily complex finite sets from just the empty set. These complex sets may have certain properties, which may be identified with certain mathematical constructs – natural numbers, maps, etc. In short, the axiom allows for the representation of all the familiar objects in mathematics as sets (albeit very intricate ones).

  • 2
    $\begingroup$ Well not arbitrary. Pairing can't create an infinite set from finite ones. In fact it can't even create sets with cardinality larger than 2. $\endgroup$ Commented Sep 23, 2017 at 5:27

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