Given a subset $S$ of a partially ordered set $P$ associated with a partial order denoted $\le $, then $S$ is called an upper set if : $$\forall x \in S,\forall y \in P :x \le y \implies y \in S$$
The lower set can be defined analogously.
Prove the following theorems:
I) Every partially ordered set is an upper set (lower set) of itself.
Setting $S=P$ in the previous definition yields:
$$\forall x \in P,\color{blue}{\forall y \in P} :x \le y \implies \color{blue}{y \in P}$$ Implies $P$ is an upper set of itself.
$$\color{red}{\forall x \in P},\forall y \in P :x \le y \implies \color{red}{x \in P}$$ Implies $P$ is a lower set of itself.
This is indeed a relation which is trivially true.
II) The intersection and union of upper sets is an upper set.
Given two subsets $S,A$ of two (not necessarily equal) partially ordered sets $(P,\le),(V,\le)$ respectively, the intersection of these two sets is either an empty set or it is not, if the first case happens then their intersection is trivially an upper set, for the second case their intersection is a non-empty set , since this set contained in both $P,V$ implies it should be an upper set. The union can be proven analogously.
III) The complement of any upper set ( resp. lower set) is a lower set ( resp. upper set).
Given a subset $S$ of a partially ordered set $(P,\le)$, by definition:
$$\forall x \in S,\forall y \in P :x \le y \implies y \in S$$
The complement of $S$ is a set $S^c$ containing elements not contained in $S$, e.g.:
$$S^c=\left\{s:s∉S \right\}$$
This is where I have a problem and cannot continue.
IV) The minimal elements of any upper set form an antichain. Conversely any antichain A determines an upper set.
I know that the set of maximal/ minimal elements of a partially ordered set is an antichain,but I cannot claim since an upper set $S$ is a subset of a partially ordered set $(P,\le)$, hence $S$ is a partially ordered set itself, because if $S$ is an empty subset then the reflexivity property of partial order relation $\le$ would not follow, so how should I prove that?(and the converse)
Can someone check the validity of my proofs and also prove the other ones?
I)
Seems fine
II)
I'd say this is too hasty to be a proof. First of all, I would only consider upper sets of the same partially ordered set, so let $A,B\subset P$ be upper sets. If $x\in A\cap B$ and $y\geq x$, then $y\in A$ because $A$ is an upper set and $y\in B$ because $B$ is an upper set, so $y\in A\cap B$. Thus $A\cap B$ is an upper set. On the other hand, if $x\in A\cup B$ and $y\geq x$, then if $x\in A$ we get $y\in A$ and if $x\notin A$, then $x\in B$, so $y\in B$, which shows that $y\in A\cup B$.
III)
Let $S$ be an upper set, and let $x\in P\setminus S=S^c$. If $y\leq x$, we have to prove that $y\in S^c$, so assume the contrary: if $y\notin S^c$, then $y\in S$, so by $S$ being an upper set and $y\leq x$ we get $x\in S$, contradiction.
IV)
If $x,y$ are two minimal elements of any upper set, and $x$ and $y$ are comparable, then $x\leq y$ because $x$ is minimal, as well as $y\leq x$ because $y$ is minimal. Thus any two comparable minimal elements of an upper set must be equal to each other.
The difficult part is showing that any antichain determines an upper set: given two different antichains, closing them upwards will yield two different upper sets. Let $A,B$ be two antichains, let \begin{align*} S=\{x\in P\mid \exists y\in A(y\leq x)\}\\ T=\{x\in P\mid \exists y\in B(y\leq x)\} \end{align*} be the upper sets determined by $A$ and $B$ respectively, and without loss of generality let $x\in A$ and $x\notin B$. If there is some $y\in B$ such that $x\leq y$, then we see that $x\in S$ and $y\in S$, but $x\notin T$, since $T$ contains no elements below $y$ (otherwise $B$ would contain some element distinct from $y$ that is comparable to $y$). On the other hand, if there is some $y\in B$ such that $y\leq x$, then $y\in T$ and $y\notin S$, for similar reasons. Finally, if $x$ is incomparable to all elements of $B$, then $x\in S$ and $x\notin T$, since $x$ is not comparable to any element of $B$.
Note that an empty partially ordered set still satisfies the reflexivity property: for all elements $x$ in the empty set we have $x\leq x$ vacuously.