Let $A\neq\emptyset$ a set and $R\subseteq A\times A$ equivalence relation s.t the complementary relation $\bar{R}=(A\times A)\setminus R$ is transitive. Prove that $|A/R|=1$ (cardinality of quotient set is 1).
There has to be a contradiction if the cardinality is bigger then 1 but I haven't found it yet. I'll be glad for any hint.
If I am not mistaken, the quotient set is the set of equivalence classes under $R$. Thus the assumption $\vert A / R \vert =1$ means there is exactly one such class, i.e. any two elements $a, b \in A$ satisfy $(a, b) \in R$. So that's what we need to prove. Now, I will write $aRb$ for $(a, b) \in R$ from here on out. Suppose then that there are more than one class. Then there exists $c \in A$ such that $aRc$ is false. Thus $(a, c) \in \bar R$, or $a \bar{R} c$. I claim that we then have $c \bar{R} a$ as well. For if not, then $cRa$, and by the symmetry of $R$, $aRc$, a contradiction. So now the transitivity of $\bar R$ implies both $a \bar{R}a$ and $c \bar{R} c$, contradicting the reflexitivity of $R$, I.e. that $xRx$ for all $x \in A$. Thus there is but one equivalence class under $R$. QED.