I have started to grow a forest!

Every complete

For every map

- First we show that
f_ \infty is an upper bound forf . Fixingi: \omega , we need to show thatf~i \sqsubseteq ^ \circ _A f_ \infty . Because\overline{f} extendsf , it suffices to show\overline{f} ~i \sqsubseteq ^ \circ _A f_ \infty . Using the fact that every map is monotone with respect to the specialization order, the result holds becausei \sqsubseteq ^ \circ _{ \overline{\omega} } \infty . - Let
\alpha be an upper bound forf . We need to show thatf_ \infty \sqsubseteq ^ \circ \alpha . If the principal lower set{ \downarrow }( \alpha ) is complete, we have the following lifting situation:In the above

\tilde{f} is the unique extension off considered as a map\omega \to { \downarrow }( \alpha ) . By uniqueness of\overline{f} as the extension off : \omega \to A ,\tilde{f} is equal to\overline{f} considered as maps\overline{\omega} \to A . Consequently we have thatf_ \infty = \overline{f} ( \infty ) = \tilde{f} ( \infty ) , so the result follows by observing that\tilde{f} ( \infty ) \in { \downarrow }( \alpha ) .It remains to show that

{ \downarrow }( \alpha ) is complete. We can express the principal lower set as follows:\begin{align*} { \downarrow }( \alpha ) &= \Set{a | a \sqsubseteq ^ \circ \alpha} \\ &= \Set{a | \forall f:A \to \Sigma ~ f(a) \to f( \alpha )} \\ &= \bigcap \limits _{f:A \to \Sigma } \Set{a | f(a) \to f( \alpha )} \end{align*} Because complete types are closed under small intersections, the result follows by the closure property of admissible subsets.