In Dana Scott's paper called "Continuous Lattices" from 1971, I am lost in the following passage. This passage occurs in some exposition after his Proposition 2.4.
Before going any deeper, however, we should clear up another point about topologies. Suppose that $D$ is any $T_0$ space which becomes a complete lattice under its induced partial ordering. Then it is evident from our definitions that every set open in the given topology is also open in the topology induced from the lattice structure. Question: when do the two topologies agree? Answer: A sufficient condition is:
$$y = \bigsqcup \{ \sqcap U | y \in U\} $$
for all $y$, where $U$ ranges across the given open sets. Because if $V$ is open in the lattice sense and $y \in V$, then $\sqcap U$ ... $V$ is a union of open sets and is thus open in the given topology.
I am completely lost with regards to this passage. Where he says "it is evident from our definitions that every set open in the given topology is also open in the induced topology", I rather have no clue why it is true, and I find it obvious that any set open in the induced topology is open in the original topology!
To explain: I interpret the "induced partial order" as the order imposed on a topology by letting $x \leq y$ iff for all open $U$, $ x\in U \implies y \in U$. He introduces this definition as the first sentence of the second section, "Continuous Lattices." The "induced topology" on a partial order is composed of the sets that are upper sets and are inaccessible from taking the joins of directed sets outside the set.
Scott points out earlier that "every limit of a directed set is a lub, but not every lub is a limit" which is the reason for the induced topology in the first place. It should be nontrivial to show that the original sets of the given topology are all Scott-open even with the assumption that the order forms a complete lattice. What am I missing? All I want to see is a short explanation of how, if $U$ is an open set of $D$ in the original topology, then a directed set has a join in $U$ iff it eventually enters $U$.
The answer is simple: Scott made a mistake. In a correction released after the original paper (which can be found appended to the paper as it's published in Toposes, Algebraic Geometry and Logic), he points out that there are a wide variety of counterexamples to his original statement and fixes the proof.