Let P (S) denote the power set of S = \ 1, 2, 3, , 10\ . Define the...

Given,

$S = \left\{1, 2, 3, \ldots \ldots 10\right\}$

$P\left(S\right) =$ power set of $S$

$A{R}_{1}B ⇒ \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right) = ϕ$

Now for reflexive property, replacing~$B$~with~$A$~we get,

\left(A \cap A^{c}\right) \cup \left(A^{c} \cap A\right)$~which~$ϕ$~always,</p><p>Now checking symmetric we will interchange~$A \text{ and } B,</p><p>So,~ \left(B \cap A^{c}\right) \cup \left(B^{c} \cap A\right)~which is same as~ \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right), hence the relation is symmetric,

So,~${R}_{1}$~is reflexive, symmetric

Now checking for transitive
\left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right) = ϕ;</p><p>Now from diagram the elements in~ \left(A \cap B^{c}\right) \cup \left(A^{c} \cap B\right)$~will be,</p><p>\left\{a\right\} \cup \left\{b\right\}$~which is given as empty set~$ϕ$

Hence, we can say that,~\left\{a\right\} = ϕ = \left\{b\right\} ⇒ A = B$</p><p>Now taking,~$( B \cap C^{c} ) \cup ( B^{c} \cap C ) = ϕ ∴ B = C$</p><p>$∴ A = C$equivalence.</p><p><img src="/api/proxy-image?url=https${R}_{2} ≡ A \cup B^{c} = A^{c} \cup B$</p><p>Now for reflexive replacing~$B \to A$~we get,</p><p>$A \cup A^{c} = A^{c} \cup A$~which is true,</p><p>And for symmetric interchanging~$A ⇔ B$~we get,</p><p>$B \cup A^{c} = B^{c} \cup A$~which is again true,</p><p>Hence,~${R}_{2} \to$Reflexive, symmetric</p><p><img src="/api/proxy-image?url=https$A \cup B^{c} = A^{c} \cup B ⇒ \left\{a, c, d\right\} = \left\{b, c, d\right\}$</p><p>On comparing both side, we get~\left\{a\right\} = \left\{b\right\} ∴ A = B$

And,~$B \cup C^{c} = B^{c} \cup C ⇒ B = C$

$∴ A = C ∴ A \cup C^{c} = A^{c} \cup C$

$∴$ Equivalence

Hence, both given relation are equivalence.