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Step-by-Step Solution
Step 1: Understand the Problem and Key Definitions
We are given a set $S = \{1, 2, 3, \ldots, 10\}$ and its power set
$P(S)$. Two relations $R_{1}$ and $R_{2}$ are defined on $P(S)$ as
follows:
$A\,R_{1}\,B$ if
$$(A \cap B^{c}) \cup (B \cap A^{c}) = \emptyset.$$
$A\,R_{2}\,B$ if
$$(A \cup B^{c}) = (B \cup A^{c}).$$
We must determine whether each of these relations is an
equivalence relation. Recall that a relation $R$ is an equivalence
relation if it satisfies:
Reflexivity: $A\,R\,A$ for all $A$.
Symmetry: $A\,R\,B \implies B\,R\,A$.
Transitivity: $A\,R\,B$ and $B\,R\,C \implies A\,R\,C$.
Step 2: Check if $R_{1}$ is an Equivalence Relation
2.1 Reflexivity
We check if $A\,R_{1}\,A$ for every $A \in P(S)$. By definition:
$A\,R_{1}\,A$ means
$$(A \cap A^{c}) \cup (A \cap A^{c}) = \emptyset.$$
Since $A \cap A^{c} = \emptyset$ for any set $A$, the union is
clearly $\emptyset$. Hence $R_{1}$ is reflexive.
2.2 Symmetry
Next, we check if $A\,R_{1}\,B$ implies $B\,R_{1}\,A$.
Given $A\,R_{1}\,B$:
$$(A \cap B^{c}) \cup (B \cap A^{c}) = \emptyset.$$
This expression is symmetric in $A$ and $B$, because swapping
$A$ and $B$ yields:
$$(B \cap A^{c}) \cup (A \cap B^{c}) = \emptyset,$$
which is the same statement. Thus $B\,R_{1}\,A$ holds. So
$R_{1}$ is symmetric.
2.3 Transitivity
We check if $A\,R_{1}\,B$ and $B\,R_{1}\,C$ imply $A\,R_{1}\,C$.
Suppose:
$$(A \cap B^{c}) \cup (B \cap A^{c}) = \emptyset,$$
$$(B \cap C^{c}) \cup (C \cap B^{c}) = \emptyset.$$
From $(A \cap B^{c}) \cup (B \cap A^{c}) = \emptyset$, no element
is in $A$ but not in $B$ or vice versa. Therefore $A = B$.
Similarly, from the second condition, $B = C$. Consequently,
$A = B = C$, which directly leads to
$$(A \cap C^{c}) \cup (C \cap A^{c}) = \emptyset,$$
proving $A\,R_{1}\,C$. Hence $R_{1}$ is transitive.
Conclusion for $R_{1}$
Since $R_{1}$ is reflexive, symmetric, and transitive, it
qualifies as an equivalence relation.
Step 3: Check if $R_{2}$ is an Equivalence Relation
3.1 Reflexivity
We check if $A\,R_{2}\,A$ for every $A \in P(S)$. By definition:
$A\,R_{2}\,A$ means
$$(A \cup A^{c}) = (A \cup A^{c}),$$
which is obviously always true. Hence $R_{2}$ is reflexive.
3.2 Symmetry
Next, we check if $A\,R_{2}\,B$ implies $B\,R_{2}\,A$.
Suppose:
$$(A \cup B^{c}) = (B \cup A^{c}).$$
Interchanging $A$ and $B$ yields
$$(B \cup A^{c}) = (A \cup B^{c}),$$
which is the same equality. Hence $B\,R_{2}\,A$ holds, and
$R_{2}$ is symmetric.
3.3 Transitivity
Finally, we check if $A\,R_{2}\,B$ and $B\,R_{2}\,C$ imply
$A\,R_{2}\,C$. Assume:
$$(A \cup B^{c}) = (B \cup A^{c}),$$
$$(B \cup C^{c}) = (C \cup B^{c}).$$
From the first equation, no element can be exclusively in $A$
but not in $B$, and vice versa, forcing $A = B$. From the second
equation, similarly, $B = C$. Consequently $A = C$, which implies
$$(A \cup C^{c}) = (C \cup A^{c}).$$
Therefore, $A\,R_{2}\,C$, proving transitivity of $R_{2}$.
Conclusion for $R_{2}$
Since $R_{2}$ is reflexive, symmetric, and transitive, it also
qualifies as an equivalence relation.
Step 4: Final Conclusion
Both $R_{1}$ and $R_{2}$ satisfy all three properties of
equivalence relations. Therefore, the correct statement is that
both $R_{1}$ and $R_{2}$ are equivalence relations.
