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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?

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  • $\begingroup$ As you write $Y$ is a set of subsets of $X$, so it's not a subset of $X$ but a subset of The power set $\mathcal{P}(X)$. $\endgroup$ – Emilio Novati Feb 8 '15 at 14:05
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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.

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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 $y \times y$).

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  • $\begingroup$ Why is 0 an element of $Z_{a_1}$? $\endgroup$ – Julian Jefko Feb 8 '15 at 15:25
  • $\begingroup$ @JulianJefko Because $0 \in \Bbb N$ (note that this convention is not universal) and $0 \le 1$. $\endgroup$ – Lord_Farin Feb 8 '15 at 15:26
  • $\begingroup$ But, how do you know that 0 is a b that has a partial order on 1, if that relation was never specifically defined? How does 0 satisfy the definition of $Z_{a_1}$? $\endgroup$ – Julian Jefko Feb 8 '15 at 15:30
  • $\begingroup$ @JulianJefko There you go. I added that I consider the standard ordering on $\Bbb N$. $\endgroup$ – Lord_Farin Feb 8 '15 at 15:32
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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 $Z_a \subset Z_b.$]

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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...

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