Let $A=P(N)-\{\emptyset,N\}$ and $\subset$ is a partial order on A. Show that the set of minimal elements and maximal elements is countable.
My solution was to show that:
Set of minimal elements $=\{\{n\}:n\in N\}$
Sketch: if $X$ is minimal and $|X|\geq2$ then $\{f(0)\}\subset X$ where $f:2\rightarrow X$ is a one to one function. If $|X|=1 $ then the only proper subset would be $\emptyset$
Set of maximal elements $=\{N-\{n\}:n\in N\}$
Sketch: if $X$ is maximal and $|Q|\geq2$ where $Q\subseteq N$ then $N-Q\subset N- \{f(0)\}\subset X$ where $f:2\rightarrow Q$ is a one to one function. If $|Q|=1 $ then the only proper superset would be $N$
Hence both sets are countable with the obvious bijection with N
i.e. $n\rightarrow \{n\} and$ $n\rightarrow\{N-\{n\}\}$