I have this question and am having trouble proving transitivity for this:
For every $n \in \mathbb{N}$ let $\sim_n$ be the relation on $\mathcal{P}([n])$ specified by $A \sim_n B$ if and only if $A \subseteq B$ or $B \subseteq A$. Determine, with proof, all $n \in \mathbb{N}$ such that $\sim_n$ is an equivalence relation.
I am trying to separate it into four cases since for some $x,y,z \in \mathbb{Z}$ and assume $(x,y), (y,z) \in R$. Let $A = \mathcal{P}([x]), B = \mathcal{P}([y]), C = \mathcal{P}([z])$. This would imply that $A \subseteq B$ or $B \subseteq A$ and $B \subseteq C$ or $C \subseteq B$. I am having trouble proving the cases where 1. $A \subseteq B$ and $C \subseteq B$ and 2. $B \subseteq A$ and $B \subseteq C$. How do we prove that $A \subseteq C$ or $C \subseteq A$ in these cases?
There can be no $n>1$, for which $\sim_n$ is an equivalence relation, precisely because they would not satisfy transitivity. Because take $A=\{1,2,\cdots,k\},B=\{k+1,\cdots,n\}$, and $C=[n]$. Then $A\sim_n C$, and $B\sim_n C$, but $A\not\sim_n B$, because $A$ and $B$ are disjoint.