# When is a subspace of a Scott space itself a Scott space?

Suppose $$P$$ and $$Q \subseteq P$$ are posets, and let $$\tau$$ and $$\rho$$ be their respective Scott topologies. Now $$Q$$ is also equipped with the subspace topology $$\tau\vert_Q$$ inherited from $$P$$. It is easy to see that:

$$\tau\vert_Q \subseteq \rho$$

I have not found an example when the reverse inclusion is not also true. So my question is:

When do the two topologies on $$Q$$ coincide?

I have found an example of posets $$P$$ and $$Q$$, as above where:

$$\tau\vert_Q \neq\rho$$

Let $$P=\mathcal{C}(\mathbb{N})$$, the power set of the natural numbers ordered by set inclusion. Let $$Q$$ be the subposet consisting of cofinite elements of $$P$$. That is:

$$Q=\{x\in P : \complement x \ \mbox{is finite} \}$$

where $$\complement x$$ denotes the complement of $$x$$ in $$\mathbb{N}$$.

Now $$Q$$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $$S$$ in $$Q$$ contains its supremum $$\bigcup S$$. To see this pick $$x_0\in S$$. If $$x_0 = \bigcup S$$, we are done. If not, pick $$y\in S\setminus\downarrow\{x_0\}$$. Since $$S$$ is directed there is $$x_1\in S$$ such that $$x_0\subsetneq x_1$$ and $$y\subseteq x_1$$. Repeating the process yields an ascending chain $$x_0, x_1\ldots$$ of elements that must be finite by ACC. So, for some $$n$$, $$\bigcup S = x_n \in S$$, as claimed.

Since every directed set in $$Q$$ contains its supremum, it follows that every up-set is open in its Scott topology $$\rho$$. So setting

$$B=\{\complement \{0\}, \mathbb{N}\}$$

we see that $$B\in\rho$$. I.e. it is open in the Scott topology on $$Q$$. Now $$P$$ is algebraic and its compact elements are precisely the finite subsets of $$\mathbb{N}$$. Thus:

$$\mathcal{B} = \{\uparrow\{x\} : x\in P \ \mbox{is finite}\}$$

is a basis for $$\tau$$, the Scott topology on $$P$$. Clearly each element of the basis contains an element of the form $$\complement \{n\}\in Q$$ where $$n\ge 1\in \mathbb{N}$$, and the same must be true for any $$C\in\tau$$. Since $$B$$ does not contain such an element it follows that for all $$C\in\tau$$, $$C\cap Q\neq B$$. We conclude that the two topologies $$\rho$$ and $$\tau\vert_Q$$ do not coincide.
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