I saw a proof of the disjointedness of two equivalence classes. The proof was as follows:
Suppose we have two equivalence classes $[a]\not=[b]$. We'll show they are disjoint. Suppose $x\in[a]\cap[b]$. Then, $x\sim a$ and $x\sim b$. By symmetry, $a\sim x$, and hence, by transitivity, $a\sim b$. Therefore, $[a]=[b]$, which is a contradiction.
My problem with this proof is that it is possible that the two equivalence classes are unequal but one is a subset of the other. For example, under the following relation: $a \sim b$ iff $a \equiv b$ (mod 3), $A = [2] \cup [3]$ is an equivalence class, $[2] \neq A$ but $[2] \cap A \neq \emptyset $ and therefore $A$ and $[2]$ are not disjoint. We have two non-disjoint equivalence classes that are unequal. Is something wrong with the proof? If yes, how do I fix the proof above? If not, why does my example not apply? And if nothing needs fixing, what am I missing?
For context, the definition of equivalence classes I have in the instructional material is $\forall x \in A$, if $ x \sim y$, $ y \in A$. The proof I found was not in the instructional material; it was said to be trivial to prove. However, for the reason I highlighted above, I could not find a good proof. I considered refining the definition by using the term distinct equivalence classes, but there would be nothing to prove. Even then, I feel like there is a 'localisation behaviour' with equivalence classes I wish to express but cannot. How can I refine the intuition of disjointedness with equivalence classes?
@XanderHenderson's comment about your counter-example ($[2] \cup [3]$ is not an equivalence class) is key, but I will say that the proof, as you presented it, seems to finish a little early.
"Two equivalent elements have the same equivalence class" is true, but a proof should be provided (even though the proof is extremely short). But most texts make a few remarks about equivalence classes when they introduce them, and they will often demonstrate that fact as part of the remarks.
Also, if you're making your proof more rigorous, you shouldn't be implying, like you do in your title, that "equivalence classes are disjoint". You should say "equivalence classes either coincide or are disjoint". Then your proof can take two non-disjoint equivalence classes and prove that they are identical. It's basically what you've already written, but I think it's clearer.