I have two equivalence relations $A$ and $B$. If $xAy \implies xBy$, how can I show that $A$ has no fewer equivalence classes than $B$?
I'm imagining partitioning a plane with boundaries, and how $A$ has to respect all the same boundaries as $B$, but can add its own, but I don't know how that idea translates into a proof.
Because if $x$ and $y$ belong to the same equivalence class with respect to $A$, then they also belong to the same equivalence class with respect to $B$. So, the equivalence class of $x$ with respect to $A$ is a subset of the equivalence class of $x$ with respect to $B$. In other words, each equivalence class with respect to $B$ is an union of equivalence classes with respect to $A$.
