an example of the Scott topology

Question

Let $X=A\cup B$ where $A$ is set of all positive integers w.r.t the Scott topology and $B$ is singleton set $\{b\}$. A set $U$ is open in $X$ iff $U$ is open in $A$ or $A\subseteq U$ and $U\cap B \neq\emptyset$.

The space $X$ is a continuous poset w.r.t the Scott topology. Isn't it? Can someone explain to me?

Answer 1

I’m going to assume that we’re talking about the Scott topology on $A$ induced by the usual order. $U\subseteq A$ is Scott open iff $U$ is an upper set, and if $D$ is a directed subset of $A$ that is disjoint from $U$, then $\sup D\notin U$. What are the upper sets in $A$? For each $n\in A$ let $A_n=\{k\in A:k\ge n\}$; each $A_n$ is an upper set, and it’s not hard to show that these are the only upper sets in $A$. Suppose that $A_n$ is one of these upper sets, $D\subseteq A$, and $D\cap A_n=\varnothing$. Then $D\subseteq\{k\in A:k<n\}$, so $\sup D=\max D\le n-1$, and therefore $\sup D\notin A_n$. Thus, each $A_n$ is open in the Scott topology on $A$, and that topology is $\tau_A=\{\varnothing\}\cup\{A_n:n\in A\}$.

The only set $U\subseteq X$ such that $A\subseteq U$ and $U\cap B\ne\varnothing$ is $X$ itself, so the given topology on $A\cup B$ is $\tau=\tau_A\cup\{X\}$. Suppose that we extend the usual order on $A$ to $X$ by setting $b<n$ for each $n\in A$. The upper sets in $X$ with this order are then exactly the members of $\tau$, and each $U\in\tau$ has the property that if $D\subseteq X$ is disjoint from $U$, then $\sup D\notin U$, so $\tau$ is the Scott topology on $X$ generated by this order. (Basically we’ve made $b$ into $0$.)

Now let $x,y\in X$ with $y\le x$ and suppose that $D\subseteq X$ is such that $x\le\sup D$. Then $D$ is finite (since $\sup D$ exists), so $\sup D=\max D$, and $y\le x\le\max D\in D$. Thus, $y$ is way below $x$. For $x\in X$ let $P_x=\{y\in X:y\le x\}$. We’ve just shown that $P_x$ is the set of elements way below $x$, and clearly $\sup P_x=x$, so $X$ is a continuous poset.

(Note that both $A$ and $X$ are linearly ordered, so every subset is directed.)

Answer 2

I think the topology that you define is not the Scott topology. The open sets are $X, \emptyset, \uparrow\{n: n\in A\}$.

Since for any elements in $A$ is non-comparable with $\{b\}$ so $\uparrow\{b\}=\{b\}$. Moreover, for every directed subset $D$ such that $\sup D$ exists and $\sup D \in \{b\}$, then $\sup D= b$. Hence $D\cap \{b\}\neq\emptyset$. So, The Scott topology on $X$ is $\{X, \emptyset, \uparrow\{n: n\in A\},\{b\}\}$.

So your topology is coarser than the Scott topology.