Intrinsic characterization of sets of subsets of A representing some ordering in A? I'm currently working my way through Naive Set Theory by Paul Halmos and am confused by what sort of answer might satisfy this exercise on page 23:

Find an intrinsic characterization of those sets of subsets of $A$ that correspond to some order in $A$.

This exercise is in Section 6: Ordered Pairs, and comes after a discussion of how to define the ordering of a quadruplet $\{a, b, c, d\}$ by generating a set where every element is a set that contains


*

*the element in question along with

*every element that comes before the element in the supplied ordering.


So the set representing the ordering $c, b, d, a$ would be:
$C = \{\{c\}, \{c, b\}, \{c, b, d\}, \{c, b, d, a\}\}$
I can make various statements about this sort of set of subsets, like


*

*every element of $C$ has an ordering that's a subset of $C$
But I'm sure this doesn't count as an intrinsic characterization.
What does an intrinsic characterization of sets of subsets of $A$ look like? How can I think about the answer to this question?
Thank you!
 A: Let (S,<=) be a partial order.
For all a in S, let D(a) = { x in S : x <= a }.
Then (S,<=) is order isomorphic to { D(a) : a in S } with the subset order.  
I suppose that order isomorphism is the intrinsic characterization of (S,<=) by subsets of S to which he is alluding. 
A: Halmos mentions that for any $a$ in an ordered set $(A, \le)$ we can construct $\mathcal{O}(a) = \{\,x \in A \mid x \le a\,\}$, and thus the set $\mathcal{O}(A) = \{\,\mathcal{O}(a)\mid a \in A\,\}$. He then describes a process by which we can recover the original order on $A$ by taking the smallest set (by inclusion) of $\mathcal{O}(A)$, extracting the element of $A$ it represents, then taking the next smallest set, which has added one element of $A$, etc. This description of the reverse process implies to me that the order $\le$ he's referring to is actually a well-ordering, where $A$ has a least element and we can remove elements of $A$ and still have a least element.
By a characterization, I think he means a set of properties that is necessary and sufficient for a set $\mathcal{O}$ to have if it is $\mathcal{O}(A)$ for some $A$. By "intrinsic", I think he means this characterization shouldn't reference the order operation $\le$.
I thought about this yesterday, and have got it down to 3 properties:

*

*$\bigcup \mathcal{O} = A$.

*Any nonempty $\mathcal{S} \subseteq \mathcal{O}$ has a minimal element $M$ such that $M \subseteq X$ for all $X \in \mathcal{S}$. (In other words, $\mathcal{O}$ is well-ordered by inclusion).

*For any two elements $x$ and $y$ of $\bigcup \mathcal{O}$ (where $x \neq y$) there is a set $X \in \mathcal{O}$ that contains either $x$ or $y$, but not both.

The first property assures that $\mathcal{O}$ isn't missing any elements of $A$. It's not really necessary if you consider $\bigcup \mathcal{O}$ as an implicit definition of $A$, but you need it if you're matching a given $A$. The second guarantees we can follow the process of successively picking and removing the smallest element of $\mathcal{O}$. And the third guarantees distinct elements of $A$ can be separated by the derived order.
To show that an $\mathcal{O}$ with these properties defines a well-ordering of $A$, I found it easier to work with a strict order $\lt$ on $\bigcup \mathcal{O}$ defined by $x \lt y$ iff there is some set $X \in \mathcal{O}$ that contains $x$ but not $y$. Irreflexivity follows directly from the definition of $\lt$. Asymmetry can be derived by contradiction from property 2 by considering the minimal set between a set in $\mathcal{O}$ that contains $x$ but not $y$ and one that contains $y$ but not $x$. Transitivity is derived by using property 2 to show that a set in $\mathcal{O}$ containing $x$ but not $y$ must be a subset of one containing $y$ but not $z$. Trichotomy ($x \lt y$ or $x = y$ or $y \lt x$) can be derived directly from property 3. And well-ordering requires both 2 and 3. For well-ordering, we get the minimal element of some nonempty $A' \subseteq A$ by finding the minimal $X \in \mathcal{O}$ that contains at least one element of $A'$. If it contains more than one element of $A'$, then property 3 shows it's not minimal, which contradicts property 2, so it must have exactly one element of $A'$, which is then shown to be the minimal element of $A'$.
Whether or not this is the most concise or understandable characterization I don't know (it's why I searched for this question). I'm answering it long after it was asked, since it does come up in searches, and I would have found a different answer useful.
