MHB How to Correctly Diagram Set Inclusions Among P, O, S, and E?

[Brian82784]

I've been trying to figure this practice problem out for a week now and I can't seem to come up with the correct diagram any help would be awesome.

 

[Evgeny.Makarov]

Hi and welcome to the forum!

Obviously, $\emptyset$ is a subset of all these sets and all sets are subsets of $U$. So, $\emptyset$ is the bottom element of the diagram and $U$ is the top one. Can you list other set inclusions among $P$, $O$, $S$ and $E$ (i.e., what is a subset of what)?

 

[Brian82784]

So this would be in a linear order?

 

[Brian82784]

I don't know why but this isn't clicking with me. I figured from the start this would be linear, but I wasn't sure.

 

[Evgeny.Makarov]

So this would be in a linear order?

Linear order happens when for every two sets, one is a subset of the other. Is this true for $P$ and $O$? You also have not listed inclusions...

 

[Brian82784]

You also have not listed inclusions... What do you mean by this?

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?

 

[Evgeny.Makarov]

You also have not listed inclusions... What do you mean by this?

I wrote in post #2:

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.

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.