$X \neq \emptyset$ is a set and on $P(X)$ (the power set of $X$), we know that "$\subseteq$" subsuming relation is a partially ordered set.
For $\forall a\in X$, how can I show the maximal element of subset $A=P(X) \setminus \{X, \emptyset\}$ of $X\setminus \{a\}\in P(X)$ It is a made up problem and I am unable to find anything resembles to this question. And I am asked to find the $A$'s minimal elements, too.
An element is said to maximal if it is not contained in any other element. Here, for any a in X, X\ {a} is only contained in X. Since X is not present in A, X\ {a} will not be completely contained in any other set belonging to A. Hence it is maximal for each a in X.
An element is minimal if it isn't bigger than any element in that set. Here for any a in X, {a} doesn't contain any smaller set in it ( though it contains the empty set, the empty set is not present in A. So you can't consider it to be contained in {a} ). Hence every singleton is a minimal set.
