2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Nov 19, 2021 at 22:11

2 Answers 2

5
$\begingroup$

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: $$\varnothing/{\sim} = \{[x]\mid x\in \varnothing\} = \varnothing.$$

$\endgroup$
2
$\begingroup$

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 $$\emptyset /{\sim}=\emptyset.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .