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Let $A$ be a set with $6$ elements, $R$ be a relation on $A$ and $n = |\{(x, y) \in A \times A : xRy\}|$.
(a) If $R$ is an equivalence relation on $A$, then what is the maximum value of $n$?
(b) If $R$ is a partial ordering on $A$, then what is the minimum value of $n$?
Explain your answers.

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2 Answers 2

Hint: (a) Prove that $xRy \;\forall x,y\in A$ is an equivalence relation. (b) Prove that $xRy \Leftrightarrow x=y$ is a partial ordering.

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Let $S=\left\{\left(x,y\right)\in A\times A \mid xRy\right\}$.

1) It is obvious that $S\subseteq A\times A$ and the relation for which $\forall x,y\in A, xRy$ is an equivalence relation so since $X\subseteq Y \implies \left|X\right|\le\left|Y\right|$, the maximum value of $n$ is $\left|A\times A\right|=\left|A\right|^2$.

2) Since you need reflexivity, you have $\forall x\in A, xRx$ so $\left\{(x,x)\mid x \in A\right\}\subseteq S$. And you can check easily that the axioms of a partial order are satisfied by the relation so that $\forall x,y \in A, xRy\iff x=y$. So you can take $S=\left\{(x,x)\mid x \in A\right\}$ and you therefore get $\left|\left\{(x,x)\mid x \in A\right\}\right|=\left|A\right|$ as minimum value for $n$.

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