Is the product topology of scott topologies a scott topology? Defnition:
Let $(A, \leq)$ be a poset(partially ordered set) and $O \subseteq A$. Then

*

*$O$ is a directed set if and only if for any $x, y \in O$, there exists $z \in O$ such that $x \leq z \mathrm{\ and\ } y \leq z$.

*$O$ is an upper set if and only if for any $x \in O$, $y \in A$, $x \leq y$ implies $y \in O$.

*$O$ is inaccessible by directed joins if and only if for any directed set $S \ (\subseteq A)$ with a least upper bound $\sup S$, $\sup S \in O$ implies $S \cap O \neq \emptyset$.

*$O$ is scott open if and only if $O$ is an upper set and inaccessible by directed joins.

*The scott topology of A is the topology whose open sets are scott open sets.

Question: Let $(A_i, \leq_i)$ be posets for $i = 1,2,\dots,n$ and $A := A_1 \times \cdots \times A_n$ a poset with the partial order defined as follows:
$$ x \leq y \iff x_i \leq_i y_i\ (\mathrm{for\ all}\ i = 1,2,\dots,n) \\
\mathrm{for}\  x = (x_1, \dots, x_n),\ y = (y_1, \dots, y_n) \in A.$$
Then, does the product topology $\mathcal{O}$ of the Scott topologies of $A_i$ coincide with the Scott topology $\mathcal{O}'$ of $A$?
(I checked an open set in $\mathcal{O}$ is an open set in $\mathcal{O}'$, but I can't show the converse.)
Thank you in advance.
 A: The answer is no. Let $A_1=\mathbb{N}\times(\mathbb{N}\cup\{\infty\})$, where $(a,b)\leqslant (a',b')$ iff $a=a'$ and $b\leqslant b'$, or $b'=\infty$ and $b\leqslant a'$; this is sometimes called the "Johnstone space". Let $A_2=\mathcal{O}_{A_1}$ be the Scott topology on $A_1$, where $\leqslant$ is given by $\subseteq$. Now define the subset $X=\{(a,U):a\in U\}\subseteq A_1\times A_2$. Note that $X$ is open in the Scott topology on $A_1\times A_2$; indeed, if $(a,U)\in X$ and $(a,U)\leqslant (a',U')$, then $a\in U$ and $a\leqslant a'$, whence $a'\in U$, and $U\subseteq U'$, whence $a'\in U'$, whence $(a',U')\in X$, so that $X$ is an upper set. To see that $X$ is inaccessible by directed joins, suppose $\{(a_i,U_i)\}_{i\in I}$ is a directed set with least upper bound and that $(a,U):=\sup_{i\in I}(a_i,U_i)\in X$. Then $\{a_i\}_{i\in I}$ is a directed subset of $A_1$ with supremum $a$, and $a\in U$, whence $U\cap\{a_i\}_{i\in I}\neq\varnothing$; say $a_{i_0}\in U$. Now $U=\sup_{i\in I}U_i=\bigcup_{i\in I}U_i$, so since $a_{i_0}\in U$, there exists some $i_1$ with $a_{i_0}\in U_{i_1}$. Since $\{a_i,U_i\}_{i\in I}$ is a directed set, there exists some $i_2$ such that $(a_{i_2},U_{i_2})\geqslant (a_{i_k},U_{i_k})$ for each $k\in\{0,1\}$, and now $(a_{i_2},U_{i_2})\in X$, whence $X\cap\{(a_i,U_i)\}_{i\in I}$ is non-empty, as desired. Thus $X$ is open in the Scott topology on $A_1\times A_2$. However, $X$ is not open in the product topology on $A_1\times A_2$; see for example Theorem II-4.10 and Exercise II-4.26 in the book Continuous Lattices and Domains.
