Consider the following statement:
If $X$ is a set and $S$ is a collection of subsets of $X$ such that:
$\cap {A_i} \in S$ whenever $\{A_i\}_{i\in I}$ is a non-empty subcollection of $S$.
$ X\in S$
Then $(S,\subseteq)$ is a complete lattice.
I would like to take the dual of this statement. Is my attempt correct:
If $X$ is a set and $S$ is a collection of subsets of $X$ such that:
- $\cup {A_i} \in S$ whenever $\{A_i\}_{i\in I}$ is a non-empty subcollection of $S$.
- $ \emptyset\in S$
Then $(S,\subseteq)$ is a complete lattice.
I immagine that you know that the second statement is indeed true. If not, tell me, and I'll show you. I think you are asking when and how to apply the dual principle.
Note that the first statement also assures that $(S,\subseteq)^{\partial}=(S, \supseteq)$ is a complete lattice, where for a given ordered set $P$, $P^{\partial}$ is its dual, that is, the underlying set of $P$ equipped with the opposite order relation of $P$.
Let us call $\Phi$ the statement that says that conditions $1$ and $2$ of the first statement are sufficient for $(S,\subseteq)^{\partial}$ to be a complete lattice.
From the duality principal, the dual statement $\Phi^{\partial}$ holds true as well.
The statement $\Phi^{\partial}$ is exactly the second statement you wrote.
All that, it is to highlight that dual statements are only exceptionally equivalent, this case being one such case because an ordered set is a complete lattice iff its dual is a complete lattice, and the statement to be dualized is a statement about a complete lattice.
In particular the second statement you wrote is not the dual of the first but it is the dual of what I called $\Phi$. Nevertheless, it happens to be equivalent to the first.