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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Given a non empty set \(X\), consider \(P(X)\) which is the set of all subsets of \(X\). Define the relation \(R\) in \(P(X)\) as follows: For subsets \(A,\; B\; in\; P(X)\), \(ARB\) if and only if \( A \subset B \) Is \(R\) an equivalence relation on \(P(X)\)?

$\begin{array}{1 1} \text{$R$ is an equivalence relation} \\ \text{$R$ is NOT an equivalence relation} \end{array}$

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  • A relation R in set A is called reflexive if $(a,a) \in R$ for every $a \in A$
  • A relation R in set A is called symmetric if $ (a_1,a_2) \in R =>(a_2a_1) \in R $ for $ a_1,a_2 \in A$
  • A relation K in set A is called transitive if $(a_1a_2) \in R \qquad (a_2a_3) \in R =>(a_1a_3) \in R\; for\; a_1,a_2,a_3 \in A$
$P(X) $ is the set of all subsets of set $X$ relation $R$ in $P(X)$ is defined by for subset $A,B$ in $P(X) ARB$ and only if A $\subset$ B
Since every set A is a subset of itself $A \subset A, ARA$ for all $A \in R(x) \rightarrow R$ is reflexive
Let $ARB \Rightarrow A \subset B$
But $B$ is not a subset of $C$ always, eg: $A=\{1,2\} \qquad B=\{1,2,3\}$
$\Rightarrow A \subset B \neq B \subset A$
Therefore Relation R is not symmetric
If $ARB$ and $BRC \rightarrow A \subset B$ and $B \subset C$.
$\Rightarrow A \subset C \rightarrow ARC \rightarrow R$ is transitive.
Hence $R$ is not an equivalance relation since it is not symmetric
answered Feb 28, 2013 by meena.p
edited Mar 20, 2013 by balaji.thirumalai

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