# Step-by-Step Solution Let \(\sim\) be a **preference relation** on a set \(X\). Define \(I(x) = \{y \in X : y \sim x\}\). We are to show that the collection \(\{I(x) \mid x \in X\}\) is a **partition** of \(X\), i.e., 1. For all \(x, y \in X\), either \(I(x) = I(y)\) or \(I(x) \cap I(y) = \emptyset\). 2. For every \(x \in X\), there is \(y \in X\) such that \(x \in I(y)\). Let's proceed step by step. --- ## Step 1: **Properties of the Relation \(\sim\)** A **preference relation** is typically assumed to be an **equivalence relation**: - **Reflexive**: \(x \sim x\) for all \(x\) - **Symmetric**: \(x \sim y \implies y \sim x\) - **Transitive**: \(x \sim y, y \sim z \implies x \sim z\) If the relation is not explicitly said to be an equivalence relation, the problem would not work as written. So we assume this. --- ## Step 2: **Show Each \(I(x)\) is non-empty** By reflexivity, \(x \sim x\) for all \(x\). Thus \(x \in I(x)\), so every \(I(x)\) is **non-empty**. --- ## Step 3: **Show Every Element Is in Some \(I(y)\)** For any \(x \in X\), take \(y = x\). By reflexivity, \(x \in I(x)\). So, for every \(x\), there is a \(y\) (namely, \(x\)) such that \(x \in I(y)\). --- ## Step 4: **Show the Sets are Pairwise Disjoint or Equal** Suppose \(I(x) \cap I(y) \neq \emptyset\). That is, there exists \(z \in I(x) \cap I(y)\). So \(z \sim x\) and \(z \sim y\). We want to show \(I(x) = I(y)\). Let \(w \in I(x)\). So \(w \sim x\). But \(z \sim x\) and \(z \sim y\), and the relation is symmetric and transitive. - \(w \sim x\), \(x \sim z\) ⇒ \(w \sim z\) (transitive). - \(w \sim z\), \(z \sim y\) ⇒ \(w \sim y\) (transitive). Thus, \(w \in I(y)\), so \(I(x) \subseteq I(y)\). Similarly, any element in \(I(y)\) is also in \(I(x)\), so \(I(x) = I(y)\). Thus, if \(I(x) \cap I(y) \neq \emptyset\), then \(I(x) = I(y)\). Otherwise, \(I(x) \cap I(y) = \emptyset\). --- ## Step 5: **Conclusion — Partition** - **Cover:** Every \(x \in X\) is in some \(I(y)\). - **Disjoint or Equal:** \(I(x) = I(y)\) or \(I(x) \cap I(y) = \emptyset\). Hence, \(\{I(x) \mid x \in X\}\) is a **partition** of \(X\). --- ## **Final Answer** \[ \boxed{ \left\{ I(x) \mid x \in X \right\} \text{ is a partition of } X. } \] **Proof:** - Every \(x\) is in some \(I(y)\) (namely \(I(x)\)), so the sets cover \(X\). - If \(I(x) \cap I(y) \neq \emptyset\), then \(I(x) = I(y)\); otherwise, \(I(x) \cap I(y) = \emptyset\). Thus, the collection forms a partition of \(X\).