**Puzzle.** How many partitions does the empty set have, and what are they?

There is one partition of the empty set, and it is the empty set itself.

Checking from definition:

**Definition.** A **partition** of a set \\(X\\) is a set \\(P \subseteq P(X)\\) such that:

1. Each set \\(S \in P \\) is nonempty.

2. Distinct sets \\(S, T \in P\\) are disjoint: that is, if \\(S \ne T\\) then \\(S \cap T = \emptyset \\).

3. The union of all the sets \\(S \in P\\) is \\(X\\): that is,

$$ X = \bigcup_{S \in P} S $$

Clearly \\(\varnothing \subseteq P(\varnothing)\\), as the empty set is a subset of any other set.

1. is true as all elements of the empty set are nonempty.

2. is true as there are no distinct sets belonging to the empty set.

3. \\(\varnothing = \bigcup_{S \in \varnothing} S\\) is true provided we defined the union of an empty family of sets to be the empty set (which is probably the only reasonable definition)

Now what left to be done is to prove that the empty set is the only partition of the empty set. Let's make use of one-to-one correspondence between partitions of a set and its equivalence relations. Since \\(\varnothing \times \varnothing = \varnothing\\) and \\(\varnothing\\) is the only subset of \\(\varnothing\\), \\(\varnothing\\) is the only binary relation defined on \\(\varnothing\\). All properties of equivalence relation are trivially true in case of the empty relation since they start with _for all_ and there are no elements in the empty set. Since there is only one equivalence relation of the empty set there has to be exactly one partition.