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{aligned} { \downarrow }( \alpha ) &= { \left \{ a \mid a \sqsubseteq ^ \circ \alpha \right \} } \\ &= { \left \{ a \mid \forall f:A \to \Sigma ~ f(a) \to f( \alpha ) \right \} } \\ &= \bigcap _{f:A \to \Sigma } { \left \{ a \mid f(a) \to f( \alpha ) \right \} } \end{aligned} Because complete types are internally complete, the result would follow if we can show that

S = { \left \{ a \mid f(a) \to f( \alpha ) \right \} } is complete. Using similar reasoning to The invariant point is an upper bound for the initial lift algebra, we may show thatS can be computed as a pullback of powers of\Sigma , whence it follows thatS is complete as well.