Kristen said:
You also have not listed inclusions... What do you mean by this?
I wrote in post #2:
Evgeny.Makarov said:
Can you list other set inclusions among $P$, $O$, $S$ and $E$ (i.e., what is a subset of what)?
It's a good idea, if a response to your post is not clear, to ask questions about it right away. Otherwise we may give you a lot of recommendations and be under impression that you got them while this may not be so.
Kristen said:
So are you saying that it wouldn't be linear because P and O are not matching. P has a 2 whereas O has a 1?
You need to master the concept of set inclusion. A set $A$ is a called subset of a set $B$, and this is denoted by $A\subseteq B$, if every element of $A$ is also an element of $B$. For example, $\{1,3\}\subseteq\{1,2,3,4\}$, but $\{1,3\}\nsubseteq\{2,3,4,5\}$ because $1\in\{1,3\}$, but $1\notin\{2,3,4,5\}$. In this topic, we don't say that sets are matching; it's not a technical term.
What can be said about $P$ and $O$? Yes, $P$ has a 2 whereas $O$ has a 1, but this does not mean by itself that $P\nsubseteq O$ and $O\nsubseteq P$. Maybe $P$ also has 1 and $O$ has 2. The fact is that this is not the case: $1\in O$, but $1\notin P$, which means $O\nsubseteq P$. Similarly, $2\in P$, but $2\notin O$, so $P\nsubseteq O$. The sets $P$ and $O$ are what is called incomparable under $\subseteq$. In a linear order, meanwhile, all sets are comparable.
To finish the diagram, I suggest you write all pairs of sets among $P$, $O$, $E$ and $S$ such that the first one is a subset of the second. For example, $S\subseteq P$ and so on. Then arrange the sets so that each subset is below its superset. As has been said, $\emptyset$ is a subset of everything, so it is the bottom element in the diagram.
