Let $D$ be a cpo with the Scott topology, then Kleene's fixed point theorem states that every continuous function $f:D\rightarrow D$ has a fixed point:
$$ \operatorname{Fix}(f) = \bigsqcup_{n\in\mathbb{N}^{>0}} \{ f^n(\bot)\}$$
I am trying to prove that the function:
$$\operatorname{Fix}:D^D\rightarrow D$$
is continuous but only seem to be confusing myself. Does anyone have any suggestions?
Suppose $U\subseteq D$ is open in the Scott topology. Then $$\text{Fix}^{-1}(U) = \{f\in D^D\mid f^n(\bot)\in U\text{ for some $n\in \mathbb{N}^{>0}$}\} = \bigcup_{n\in \mathbb{N}^{>0}} U_n,$$ where $U_n = \{f\in D^D\mid f^n(\bot)\in U\}$.
Clearly $U_n$ is upwards closed, since if $f\leq g$, then $f^n(\bot) \leq g^n(\bot)$, so $f\in U_n$ implies $g\in U_n$. And suppose $(g_i)_{i\in I}$ is a directed family in $D^D$ such that $g = \sup_I g_i \in U_n$. Then $g^n(\bot) = \sup_I g_i^n(\bot)\in U$, since the $g_i$ are continuous, so there is some $j\in I$ such that $g_j^n(\bot)\in U$, and hence $g_j\in U_n$. This proves that $U_n$ is inaccessible by directed joins.