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