# Correctness of relation definition constructed from partition set

I came across this in a university textbook.

Let $A$ be a nonempty set and $S$ is its partition. We define a relation $R$ as $$R = \{(a,b) \in A \times A ~|~ \exists M \in S \land a \in M \land b \in M\}$$

Then $R$ is equivalence relation on $A$ and $S$ is its partition.

Then in the textbook follows a proof of the statement, of which the first sentence is

Because all sets in $S$ are pairwise disjoint, the definition of relation $R$ is correct.

My question is: why wouldn't the definition of $R$ be correct if the sets of $S$ weren't disjoint? Of course that then $S$ wouldn't be a proper partition of $A$, but does it somehow affect the correctness of definition of $R$?

• Maybe he means only that $R$ is a set of disjoint sets, and thus it can be an equiv rel. – Mauro ALLEGRANZA Jul 10 '17 at 13:58
• I do not understand your comment. $R$ is not a set of disjoint sets, it is a relation. – Gogis Jul 10 '17 at 14:05
• Yes, but the relation $R$ is an equivalence rel iff it partitions the set $A$ in disjoint sets $[x]_R$ called equiv classes. If the original partition $S$ is not "disjoint" we have that there are $M, N \in S$ such that for some $o$: $o \in M,N$. If so, $o \in [a]_R$ for all $a \in M$ and also $o \in [b]_R$ for all $b \in N$. Conclusion: the equiv classes "induced" by $M$ and $N$ are not disjoined. – Mauro ALLEGRANZA Jul 10 '17 at 14:11
• I see. Maybe it was meant this way. But it's confusing, since it says that the definition is correct. I don't think that $R$ being an equivalence relation is a part of the definition of $R$. – Gogis Jul 10 '17 at 14:17
• Correct... he want to prove that $R$ is an equiv rel that partitions the set $A$ in the same way as the original partition $S$. – Mauro ALLEGRANZA Jul 10 '17 at 14:20