Equivalence relation R is $|R|\geq n+2$

Question

I have been asked to proove this following statement:

If a given equivalence relation $R$ on $A=\{1,2,3,\dots,n\}$ has less than $n$ equivalence classes, then $|R|\geq n+2$

Answer 1

To prove the claim it suffices to show that there exists at least one equivalence class with two or more elements. That's true because if there are two elements $m,k \in A$ such that $m \neq k$ and $mRk$ then by reflexivity $kRm$ holds which combined by the fact that $pRp$ for every $p \in A$ gives us that $|R| \geq n+2$. Now let's prove that this is indeed the case. Suppose that for every $m,k \in A$ with $m\neq k$ we have $(m,k) \not\in R$. Then we would have exactly one equivalence class for each element of $A$ so we would have exactly $n$ equivalence classes, which is a contradiction. Therefore, there exists at least one equivalence class with two or more elements.