Recall the notion of upper set from Example 1.45, and let \\((P,\leq)\\) be a preorder.

1. Show that the set \\(\uparrow p:=\\{p'\in P\mid p\leq p'\\}\\) is an upper set, for any \\(p\in P\\).

2. Show that this construction defines a monotone map \\(\uparrow:P^\mathrm{op}\to\mathcal{U}P\\).

3. Draw a picture of the map \\(\uparrow\\) in the case where \\(P\\) is the preorder \\((b\geq a\leq c)\\) from Exercise [1.48](https://forum.azimuthproject.org/discussion/1954).

Example 1.45:

Given a preorder \\((P,\le)\\), an _upper set_ in \\(P\\) is a subset \\(U\\) of \\(P\\) satisfying the condition that
if \\(p \in U\\) and \\(p \le q\\), then \\(q \in U\\). ``If \\(p\\) is an element then so is anything bigger.''
Write \\(\mathcal U(P)\\) for the set of
upper sets in \\(P\\). We can give the set \\(\mathcal{U}\\) an order by letting \\(U \le V\\) if \\(U\\) is
contained in \\(V\\).

For example, if \\((\mathbb{B},\leq)\\) is the booleans (Example 1.33), then its preorder of uppersets \\(\mathcal U\mathbb{B}\\) is
\\[\boxed{\begin{matrix}\lbrace\mathrm{false},\mathrm{true}\rbrace\\\\\uparrow\\\\\lbrace\mathrm{true}\rbrace\\\\\uparrow\\\\\varnothing\end{matrix}}\\]

Just \\(\\{\mathrm{false}\\}\\) by itself is not an upper set, because \\(\mathrm{false}\leq\mathrm{true}\\).