Given a Non-Empty Set X, Consider P(X), the Power Set of X. Define A R B if and Only if A ⊆ B. Is R an Equivalence Relation on P(X)? Justify Your Answer

NCERT Class 12 Mathematics — Relations and Functions, Miscellaneous Exercise.

Problem

Given a non-empty set $X$, consider $P(X)$, the power set of $X$ (the set of all subsets of $X$). Define the relation $R$ on $P(X)$ as follows: for subsets $A, B \in P(X)$,

$$A\,R\,B \iff A \subseteq B$$

Is $R$ an equivalence relation on $P(X)$? Justify your answer.

Key insight. A relation is an equivalence relation only if it satisfies all three of reflexivity, symmetry and transitivity — proving two of the three isn’t enough, and a single explicit counterexample is all it takes to rule out symmetry.

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Approach

Check the three defining properties of an equivalence relation — reflexivity, symmetry, transitivity — one at a time, directly from the definition $A\,R\,B \iff A\subseteq B$. Reflexivity and transitivity follow immediately from basic subset properties. Symmetry is where it breaks down, and the cleanest way to show that is a concrete counterexample rather than a general argument.

Solution

Step 1 — Check reflexivity

For any $A \in P(X)$, is $A\,R\,A$, i.e. is $A \subseteq A$? Every set is a subset of itself, so yes — this holds for every $A$. Hence $R$ is reflexive.

Step 2 — Check symmetry

For $R$ to be symmetric, $A\,R\,B$ (i.e. $A\subseteq B$) would have to imply $B\,R\,A$ (i.e. $B \subseteq A$) for every pair $A, B$. But $A\subseteq B$ does not in general force $B\subseteq A$ — take $X=\{1,2,3\}$, $A=\{1\}$ and $B=\{1,2\}$. Then $A\subseteq B$, so $A\,R\,B$ holds. But $B\not\subseteq A$ (since $2\in B$ but $2\notin A$), so $B\,R\,A$ fails. This single counterexample is enough: $R$ is not symmetric.

Step 3 — Check transitivity

Suppose $A\,R\,B$ and $B\,R\,C$, i.e. $A\subseteq B$ and $B\subseteq C$. Subset containment is itself transitive: if every element of $A$ is in $B$, and every element of $B$ is in $C$, then every element of $A$ is in $C$, so $A\subseteq C$, i.e. $A\,R\,C$. This holds for every such triple, so $R$ is transitive.

Step 4 — Combine the three results

$R$ is reflexive and transitive, but it fails symmetry. Since an equivalence relation requires all three properties to hold, $R$ is not an equivalence relation on $P(X)$.

Answer

Common mistakes

• Stopping after checking reflexivity and transitivity. Both hold here, which can tempt a student into concluding $R$ is an equivalence relation without ever testing symmetry — but all three properties are required, and this relation fails exactly one of them.
• Trying to “prove” symmetry fails in general without a concrete example. A clean, specific counterexample (like $A=\{1\}\subseteq B=\{1,2\}$ but not the reverse) is both simpler and more convincing than an abstract argument, and is exactly what’s expected in a “justify your answer” question.

Practise next

• On the same power set $P(X)$, define $A\,R\,B$ iff $A\cap B = \varnothing$. Check whether this relation is reflexive, symmetric, or transitive.