A/S is a refinement of A/R

Question

Suppose $A$ is a finite set. $R$ and $S$ are equivalence relations on $A$.

Suppose $S\subseteq R$.

I have to show that $A/S$ is a refinement of $A/R$.

I have started with : for every $K_1\in A/S$ there is $K_2\in A/R$ such that $K_1\subseteq K_2$.

but I'm not sure how to choose my $K_2$.

Answer 1

Your $K_1$ is an equivalence class in $A/S$. We want the corresponding equivalence class in $A/R$. Here's a sensible way to identify that class: take any representative $a \in A$ of $K_1$, i.e., take some arbitrary $a \in K_1$. Then let $K_2$ be the equivalence class of $a$ in $A/R$.

The goal now is to prove that if we take any $x \in K_1$, the equivalence class of $x$ in $A/R$ is $K_2$. The way we'll do this is by relating $x$ to $a$ and using the properties of equivalence classes. Since $x \in K_1$, $xSa$, and thus $xRa$ (since $S \subseteq R$), so $x$ and $a$ are in the same equivalence class in $A/R$, which is $K_2$. Thus if $x \in K_1$, $x \in K_2$, which proves $K_1 \subseteq K_2$.