Set of upper sets for preorders

Question

I understand the definition of an upper set for a preorder:

an upper set in $P$ is a subset $U$ of $P$ satisfying the condition that if $p \in U$ and $p \leq q$, then $q \in U$.

I am confused on finding the set of upper sets of preorder P: $U(p)$

We can give the set U an order by letting U $\leq$ V if U is contained in V

I am given the following preorder P:

The upper set preorder looks like this (ignore the dotted lines):

I am confused about this diagram. Why doesn't it include the following relations?

• {b} pointing to {b, a}
• {c} pointing to {c, a}

{b} is contained in {b,a} and {c} is contained in {c, a}

Answer 1

You are right that $\{b\} \subseteq \{a, b\}$, but the latter is not an upper set (as mentioned in the comments by Andreas Blass), so that is why it is not in the picture. This is because $a \leq c$ while $c \not \in \{a, b\}$. Similarly $\{a, c\}$ is not an upper set, because it does not contain $b$. So the picture would contain the relations you mention, if those sets would have been in the picture at all.

To check that the second picture contains all upper sets you just have to check all possible subsets by hand. There are eight sets to check:

• the four that are already in the picture;
• the sets $\{a, b\}$ and $\{a, c\}$, which are not upper sets as we just saw;
• the set $\{a\}$ is also not an upper set because it does not contain $b$ while $a \leq b$ (or similar argument using $c$);
• the empty set $\emptyset$, as Idéophage correctly points out in the comments this would technically satisfy your definition of an upper set and should thus be in the picture, however it is not uncommon to require upper sets to be nonempty in which case $\emptyset$ is also not an upper set.