It turns out that the empty set is a subset of

*any*set no matter what that set is. In this post I will provide a proof of this. But first let me tell you what I mean by

*subset*.

Let

*A*and

*B*be two nonempty sets.

*A*is a subset of

*B*if all elements of

*A*are also elements of

*B*. For example let

*A*= {1,3,5,7} and

*B*= {0,1,2,3,4,5,6,7}.

*A*is a subset of

*B*, because every element of

*A*is also an element of

*B*. Now consider the set

*C*= {1,3,5,7,9}.

*C*is

*not*a subset of

*B*, because there is an element of

*C*that is

*not*an element of

*B*.

*C*is an example of a set that is not a subset of

*B*. This will be an important part of the proof to show that the empty set is a subset of any set.

For those of you who are confused by the proof of statement (ii) let me put it in words.

Let

*A*be any nonempty subset of*U*. By way of contradiction suppose the empty set is*not*a subset of*A*. It follows that there exists an element*x*in the empty set such that*x*is not an element of*A*. But the empty set has nothing in it; therefore*x*cannot be an element of the empty set. [By the definition of the empty set] It follows that for all*x*in*U*,*x*cannot be in the empty set. Therefore there are no elements in the empty set that are not in the set*A*(this is our equivalent definition of what it means to be a subset). Thus the empty set must be a subset of*A*.
I've probably made this proof more complicated than it should be, but once you understand it, it is truly an elegant proof!