© All Rights reserved @ LearnWithDash
Step-by-Step Explanation
1. Understanding the Relation
We have a set $A$ partitioned into disjoint subsets $A_1, A_2, \ldots, A_k$ such that
$A = A_1 \cup A_2 \cup \cdots \cup A_k \quad \text{and} \quad A_i \cap A_j = \varnothing \quad \text{for } i \neq j.$
The relation $R$ on $A$ is defined by:
$R = \{(x,y) : \text{there is an } i \text{ with } 1 \leq i \leq k \text{ such that } \bigl(y \in A_i \iff x \in A_i\bigr)\}.
Equivalently, $(x,y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_i$ in the partition.
2. Reflexivity
To check reflexivity, we need to see if every element $a \in A$ is related to itself under $R$.
Condition: $(a,a) \in R$ for all $a \in A$.
Since any $a \in A$ must lie in exactly one of the $A_i$ (because the $A_i$'s are disjoint and cover $A$), we have:
$a \in A_i \iff a \in A_i, \text{ which is always true.}$
Hence, $(a,a) \in R$ for every $a \in A$. Therefore, $R$ is reflexive.
3. Symmetry
To check symmetry, we need to verify that whenever $(x,y) \in R$, we also have $(y,x) \in R$.
Condition: If $(x,y) \in R$, then $(y,x) \in R$.
If $(x,y) \in R$, by definition, $x$ and $y$ lie in the same $A_i$. That is, $x \in A_i$ exactly when $y \in A_i$. This symmetric condition also shows that $y$ and $x$ lie in the same $A_i$ if and only if $x$ and $y$ do. Hence, $(y,x) \in R$.
Thus, $R$ is symmetric.
4. Transitivity
To check transitivity, we need to ensure that whenever $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$.
Condition: If $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$.
If $(x,y) \in R$, then $x$ and $y$ lie in the same $A_i$. Similarly, if $(y,z) \in R$, then $y$ and $z$ lie in the same $A_i$. This implies $x, y, z$ all belong to the same $A_i$. Therefore, $x \in A_i \iff z \in A_i$, leading to $(x,z) \in R$.
Hence, $R$ is transitive.
5. Conclusion: $R$ is an Equivalence Relation
Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation on $A$. Indeed, each $A_i$ forms one equivalence class, and every element of $A$ belongs exactly to one such class.
