A question about partially ordered sets and their subsets I was reading about partially ordered sets and in the book, a theorem was proven. The theorem was that, given an poset, $(X, \le)$ there exists a set $Y$ of subsets of $X$ such that $(X, \le) \cong (Y, \subset)$. The proof went as follows:
"For each $a \in X$, let $Z_a = \{b \in X : b \le a\}$, and let $Y = \{Z_a : a \in X\}$. Define a map $\pi$ from $X$ to $Y$ by $\pi(a) = Z_a$. Clearly $\pi$ is a bijection. Moreover $a_1 \le a_2 \iff Z_{a_1} \subset Z_{a_2}$, so $\pi$ is an isomorphism between $(X, \le)$ and $(Y, \subset)$." 
I understand why these two sets are isomorphic, but I don't understand why $Y$ is a subset of $X$. If $(X, \le)$, then there is a relation on $X$ and a relation is defined to be a subset of the Cartesian product. If thats the case, then the relation set must be a set of ordered pairs. The set $Z_a$ is the set of all elements which have a partial order on $a$. $Y$ is the set of all $Z_a$'s. But, if every $Z_a$ is based on only that which has a relation on $a$, doesn't that break the ordered pairs (since they are in the form $(a,b)$, and with any $a$ considered, only the $b$ elements would be in the set) and imply that ordered pairs cant be in any $Z_a$ set? 
The only other interpretation I can think of is that $a$ itself is an ordered pair, because its a member of $X$, but then, I don't see how its possible for any $Z_a$ to have elements, given its definition.   
Am I misunderstanding something?  
 A: Indeed, the $Z_a$ do not consist of ordered pairs. But they needn't; the $Z_a$ are simply the elements on which $\subset$ becomes the relation.
As an example, consider $(\Bbb N, \le)$ with $1$ and $2$. Then $Z_1 = \{0,1\}$, and $Z_2 = \{0,1,2\}$, and both are in $y$.
Now $\subset$ becomes a relation on $y$, so that we have $Z_1 \subset Z_2$, or more formally, $(Z_1, Z_2) \in {\subset}$ (and $\subset$ itself is then to be considered, well, a subset of $y \times y$).
A: Actually, this $Y$ is a set of subsets of $X$. Note that the order relation in the second poset is set inclusion (perhaps it should be written as "$\subseteq$" as the partial order relation was written in a "less than or equal to" form).
This way of representing posets is directly related to the Dedekind-MacNeile completion of the poset $(X,\leq)$, and you may learn about this in the following Wikipedia link.
A: The proof doesn't make the claim that $y$ is a subset of $x.$ For a particular $a$ in $x,$ $Z_a$ is all things $\le a$ in the poset $(x,\le).$ If then $a$ is mapped by $\pi$ to this $Z_a$ it will be the case that, whenever $a \le b$ for elements $a,b$ of $x,$ it will be true also that $Z_a \subset Z_b.$ Check: any element  $u$ of $Z_a$ is $\le a$, and then from $a \le b$ using transitivity, also $u \le b$ making $u \in Z_b.$ [This check confirms $Z_a \subset Z_b.$]
A: This theorem means : if $X$ be any nonempty paritially ordered set and $Y$ is the power set of $X$ ( power set which is the set of all subset of $X$)
Then $Y$ is partially ordered set with the relation "subset or equal"
The proof doesn't depends on $X$ (if was it poset or not)
You can prove it as follows:
Every $A$ belong to $Y$ then $A$ is subset of $A$
Then the relation is reflexive...
