Partial orders on downward closed sets [closed]

Question

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' (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.

Answer 1

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 $\bigcap\mathcal{B}=\inf\mathcal{B}$.