# Does the empty set permit zero or one equivalence classes?

Let $$\sim$$ be an equivalence relation on $$\varnothing$$ (it must in fact be the empty relation). Does $$\varnothing\big/\sim$$ have no elements, or one, namely $$\varnothing$$? Put differently, does $$\varnothing$$ get partioned into one equivalence class, namely $$\varnothing,$$ or no equivalence classes, because equivalence classes are defined relative to elements of the original set, of which there are none?

I think the answer is the latter, i.e., there are no equivalence classes, because equivalence classes cannot be empty by definition. But I must admit, even after checking the definitions, I'm a bit confused.

• Remark: in mathematics, to preserve the correspondence between equivalence relations and partitions, we define partitions to be sets of non-empty sets. In computer science, it is common to allow partitions to include the empty set. So for (some) computer scientists, $\emptyset$ has two partitions, while for (most) mathematicians it has only one. The two communities agree that there is only one equivalence relation on the empty set. Nov 19, 2021 at 22:11

If $$\sim$$ is an equivalence relation on $$X$$, then $$X/{\sim} = \{[x]\mid x\in X\},$$ where $$[x] = \{y\in X\mid x\sim y\}$$ is the equivalence class of $$x$$. When $$X$$ is empty, there is no $$x\in X$$, so there are no equivalence classes: $$\varnothing/{\sim} = \{[x]\mid x\in \varnothing\} = \varnothing.$$
Equivalence classes of an equivalence relation on a set $$A$$ are non-empty subsets of $$A$$. If $$A=\emptyset$$, there are no non-empty subsets, hence no equivalence classes. In other words $$\emptyset /{\sim}=\emptyset.$$