I was doing this exercise:
Let $\sim$ be an indifference relation on X. For all $x \in X$ define $I(x) = \{y \in X: x\sim y\}$ Show that the set (of sets) $\{I(x): x \in X\}$ is a partition of $X$, i.e.
i) for all $x,y \in X$ either $I(x) = I(y)$ or $I(x) \cap I(y) = \emptyset$
ii) for every $x \in X$, there is $y \in X$ such that $x \in I(y)$
So, correct me if I'm wrong,
$I(x) = \{y \in X: x\sim y\}$ is the set of all the images of $x$, i.e. it contains all the $y$ such that they have an indifference relation with $x$ and $I(x) \subseteq X$
$\{I(x): x \in X\}$ is the collection of all different I-equivalence classes
then my answers to the two points are:
i) Since equivalence classes are disjoint we can either have:
(1) they belong to the same equivalence class, so it doesn't matter which element we choose as representative for that class, i.e. $I(x) = I(y)$ for all $x,y \in X$
(2) they belong to different equivalence classes so they belong to disjoint sets and $I(x) \cap I(y) = \emptyset$
ii) By completeness either $x \sim y$ or $y \sim x$
Are my answers wrong? How can I write them "formally"?
i)
If $z\in I(x)\cap I(y)$ then $x\sim z\wedge y\sim z$.
The relation $\sim$ is symmetric so that $x\sim z\wedge z\sim y$.
The relation $\sim$ is transitive so that $x\sim y$.
Now if $u\in I(y)$ or equivalently $y\sim u$ it follows again by transitivity that also $x\sim u$ or equivalently $u\in I(x)$.
Proved is now that $I(y)\subseteq I(x)$ and similarly it can be proved that $I(x)\subseteq I(y)$.
ii)
$y=x$ serves well here since $x\in I(x)$. This because the relation $\sim$ is reflexive.