Q.Given a non empty set $X$, consider $P(X)$ which is the set of all subsets of $X$. Define the relation $\mathbf{R}$ in $P(X)$ as follows: For subsets $A, B$ in $P(X)$, $A \mathbf{R} B$ if and only if $A \subseteq B$. Is $\mathbf{R}$ an equivalence relation on $P(X)$? Justify your answer.

The key idea is that set inclusion ($\subseteq$) defines a partial order, not an equivalence relation, because it fails symmetry.

Step 1: Reflexivity

For any $A \in P(X)$, $A \subseteq A$ always holds. So $\mathbf{R}$ is reflexive.

Step 2: Symmetry (the critical test)

If $A \subseteq B$, does it force $B \subseteq A$? No — take $A = \emptyset$, $B = X$ (with $X$ non-empty). Then $A \subseteq B$ is true, but $B \subseteq A$ is false. So $\mathbf{R}$ is not symmetric.

Step 3: Transitivity

If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$ always holds. So $\mathbf{R}$ is transitive.

Since symmetry fails, $\mathbf{R}$ cannot be an equivalence relation.

The core idea here is to check the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity. The relation is defined by set inclusion — one of the most natural partial orders on subsets. An equivalence relation, by contrast, captures the idea of "sameness" in some sense, while inclusion captures "containment" or "being a subset of". These are fundamentally different kinds of relations.

Let’s test each property step by step.

1. Reflexivity: For any subset $A \in P(X)$, is it true that $A \mathbf{R} A$?

   The condition is $A \subseteq A$, which is always true — every set is a subset of itself.

   So $\mathbf{R}$ is reflexive.

2. Symmetry: For any $A, B \in P(X)$, if $A \mathbf{R} B$, must we have $B \mathbf{R} A$?

   $A \mathbf{R} B$ means $A \subseteq B$. For symmetry, we would need $B \subseteq A$ to hold automatically. But this is false in general.

   For example, take $X = \{1, 2\}$, $A = \{1\}$, $B = \{1, 2\}$. Then $A \subseteq B$ is true, but $B \subseteq A$ is false.

   So $\mathbf{R}$ is not symmetric.

3. Transitivity: For any $A, B, C \in P(X)$, if $A \mathbf{R} B$ and $B \mathbf{R} C$, does it follow that $A \mathbf{R} C$? $A \mathbf{R} B$ means $A \subseteq B$, and $B \mathbf{R} C$ means $B \subseteq C$. From set theory, if $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$. So $\mathbf{R}$ is transitive.

Since symmetry fails, $\mathbf{R}$ cannot be an equivalence relation. It is, in fact, a partial order on $P(X)$ — reflexive, antisymmetric, and transitive — but that’s a different story.

Method: Deciding whether a relation is an equivalence relation

To classify a relation, test reflexive, symmetric, transitive — and remember that finding ONE property that fails is enough to say "not an equivalence relation".

Steps

Step 1: Reflexive — check $a\,R\,a$ for all $a$. For $\subseteq$, every set satisfies $A\subseteq A$.

Step 2: Symmetric (the decisive test here) — check whether $a\,R\,b$ forces $b\,R\,a$. For $\subseteq$, produce a counterexample: $A=\varnothing,\ B=X$ gives $A\subseteq B$ but not $B\subseteq A$, so symmetry fails.

Step 3: Transitive — check $a\,R\,b$ and $b\,R\,c \Rightarrow a\,R\,c$. For $\subseteq$ this holds.

Step 4: Conclude — since symmetry fails, the relation is not an equivalence relation (it is in fact a partial order: reflexive, antisymmetric, transitive).

Common Mistakes

Mistake 1: Thinking the subset relation $\subseteq$ is symmetric.

Why it's wrong: $A\subseteq B$ does not give $B\subseteq A$ unless $A=B$; e.g. $\varnothing\subseteq X$ but $X\not\subseteq\varnothing$. Correct approach: give one such counterexample to show symmetry fails, so $R$ is not an equivalence relation.

Mistake 2: Assuming reflexive + transitive alone makes an equivalence relation.

Why it's wrong: all three of reflexive, symmetric, transitive are required; $\subseteq$ has the first and third but not the second (it is a partial order). Correct approach: verify symmetry explicitly before concluding.