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}\$$.