If $S$ and $R$ are equivalence relations on $A$, and $A/R=A/S$, then $R=S$

Question

This is for proof verification. We have that $A/R=A/S$ where $A/B$ denotes the set of all equivalence classes of elements of $A$, or $A$ modulo $B$, where $A$ is a set and $B$ is a relation. Let $X$ and $Y$ be subsets of $A$ so that $x\in X$ and $y\in Y$. $[x]_R$ is the equivalence class of an element in $A$ w.r.t a relation $R$ on $A$.

$A/R=A/S\Longleftrightarrow \{[x]_R :x\in A\}=\{[x]_S : x\in A \} $

and

$[x]_R=\{z\in A : zRx \} $ and $[x]_S=\{z\in A : zSx\}$.

We know that for a set to be equal, all of the elements in the set must be equal, so we must have that $[x]_R=[x]_S $ for all $x\in A$. By the same argument, $zRx=zSx$ for all $z\in A$, so $(z,x)\in R$ and $(z,x)\in S$. This implies that the relation sets are equal because each coordinate in the relation set $R$ is in the relation set $S$ and vice versa, that is, they are subsets of each other so they must be equal. I refer to "the relation set" because I am using the definition $R\subset A\times A$ from the book 'How to Prove It'. Sorry if this is incorrect terminology.

Answer 1

The hypothesis that $A/R=A/S$ tells you that $[x]_R\in A/S$ for each $x\in A$, so it tells you that $[x]_R=[y]_S$ for some $y\in A$, but it doesn’t immediately tell you that $[x]_R=[x]_S$; for that you have to work just a little harder. You know that $x\in[x]_R$, so if $[x]_R=[y]_S$, then $x\in[y]_S$, and therefore $xSy$. That in turn implies that $[x]_S=[y]_S$, and now you can conclude that $[x]_R=[x]_S$.

The rest of your argument is correct but could use a few more details and a little more clarity, perhaps something like this.

Suppose that $\langle x,y\rangle\in R$. Then $x\in[y]_R=[y]_S$, so $x\in[y]_S$, and therefore $\langle x,y\rangle\in S$. Thus, $R\subseteq S$, and a similar argument shows that $S\subseteq R$, so $R=S$.