Question

If \(R = \{(1,1), (2,2), (1,2), (2,1), (3,3)\}\) and \(S = \{(1,1), (2,2), (2,3), (3,2), (3,3)\}\) are two relations on the set \(X = \{1, 2, 3\}\), the incorrect statement is:

• A. R and S are both equivalence relations
• B. \(R \cap S\) is an equivalence relation
• C. \(R^{-1} \cap S^{-1}\) is an equivalence relation
• D. \(R \cup S\) is an equivalence relation

Hint:

Check reflexive, symmetric and transitive for each combination separately; the union of two equivalence relations is not always transitive.

Answer 1

The Correct Option is D

Step 1: Think of equivalence relations as partitions.
Every equivalence relation on a set splits that set into disjoint groups, called equivalence classes, and every such grouping (partition) gives back an equivalence relation. This is a cleaner lens to check the four options.

Step 2: Identify the partitions for R and S.
R relates 1 and 2 to each other and leaves 3 alone, so R corresponds to the partition $\{1,2\}, \{3\}$. This is a valid partition of $X = \{1,2,3\}$, so R is an equivalence relation.
S relates 2 and 3 to each other and leaves 1 alone, so S corresponds to the partition $\{1\}, \{2,3\}$. This is also a valid partition, so S is an equivalence relation too.

Step 3: Intersections correspond to a common refinement.
$R \cap S$ keeps only the pairs both relations agree on, which is just $\{(1,1),(2,2),(3,3)\}$, everyone only related to themselves. That is the partition $\{1\},\{2\},\{3\}$, still a valid partition, so $R \cap S$ is an equivalence relation.
Since both R and S are symmetric, reversing all their pairs changes nothing, so $R^{-1} \cap S^{-1}$ gives the exact same set and is an equivalence relation for the same reason.

Step 4: Unions do not correspond to a partition in general.
$R \cup S$ tries to keep 1 and 2 together (from R) and 2 and 3 together (from S) at the same time. But grouping 1 with 2, and 2 with 3, forces 1 and 3 into the same group too if the result is to be a genuine partition, and the pair $(1,3)$ is missing from $R \cup S$.
So $R \cup S$ does not match any valid partition of $X$, which means it cannot be an equivalence relation. Concretely, it fails transitivity since $(1,2)$ and $(2,3)$ are present but $(1,3)$ is not.

Step 5: Match this back to the four statements.
Statement (A), about R and S, is correct. Statements (B) and (C), about the intersections, are correct. Statement (D), claiming $R \cup S$ is an equivalence relation, is the one that is wrong.
\[ \boxed{\text{The incorrect statement is (D)}} \]