Let $P = (V, \sqsubseteq)$ be a partial order and $\mathfrak{D}(P)$ denote the class of downward-closed subsets of the partial order $P$ (i.e, the class of $A \subseteq V$ such that $y\in A \;\&\; x \sqsubseteq y$ implies $x \in A$).


Let a partial order be called *downward complete*, if every non-empty subset has an infimum, *upward complete* if every non-empty subset has an supremum, and *complete* if both downward and upward complete.


In [La Matematica della Verita'](https://books.google.co.uk/books/about/La_matematica_della_verit%C3%A0_Strumenti_ma.html?id=2t2rAAAACAAJ&redir_esc=y) (2006): page 177 , theorem 6.7.7 states that the following three conditions are equivalent:


1. $(V, \sqsubseteq)$ is complete.
2. $(V, \sqsubseteq)$ is downward complete and $\exists z \forall x(x \sqsubseteq z)$
3. $(V, \sqsubseteq)$ is upward complete and $\exists z \forall x(z \sqsubseteq x)$


It is then stated that, conditions 4. and 5. (below), together with the equivalence of 1.-3. just stated, imply that $(\mathfrak{D}(P), \subseteq)$ is a complete partial order:


4. $V \in \mathfrak{D}(P)$ and
5. $\bigcap \mathcal{B} \in \mathfrak{D}(P)$, for any non-empty subfamily $\mathcal{B}$ of $\mathfrak{D}(P)$


I'm wondering if there is a mistake. If not, it is not clear to me how 3. and 4. plus the equivalence of 1.-3. ensure $(\mathfrak{D}(P), \subseteq)$ is a complete partial order.



Partial orders on downward closed sets
Conditions 4 and 5 show that $(\mathfrak{D}(P),{\subseteq})$ satisfies condition 2 in the list: because $V\in\mathfrak{D}(P)$ we have the second part of 2; and 5 says that $(\mathfrak{D}(P),{\subseteq})$ is downward complete, because $\bigcap\mathcal{B}=\inf\mathcal{B}$.


