I read that, with the Scott topology, suprema of sequences are topological limits (See page 1 of [this article](https://akjournals.com/view/journals/10473/88/1-2/article-p35.xml)).


Let $(X, \le)$ be a DCPO, and $D$ be a directed subset of $X$.
I can easily see that the identity net on $D$ converges to $\bigvee D$.


How about the other direction? If the identity net on $D$ converges to $x$, does it imply that $x = \bigvee D$?



Scott topology: Suprema of sequences are topological limits
$\newcommand\LL{\mathcal L}$**Brief answer:** The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL\_D$ of limits of the identity net on $D$ we have
\begin{equation\*}
 \LL\_D=L\_s:=\{y\in X\colon x\le s\}, 
\end{equation\*}
where
\begin{equation\*}
 s:=s\_D:=\sup D=\bigvee D. 
\end{equation\*}


One can similarly see that the statement "Suprema of sequences are also topological limits, i.e., $\sup x\_n=x \iff \lim x\_n=x$" in the paper linked in your post is incorrect in general (and almost always). E.g., if $X=(-\infty,0]$ with the usual order and $x\_n=0$ for all $n$, then $\sup x\_n=0$, whereas the set of limits of $(x\_n)$ in the Scott topology is the entire set $X=(-\infty,0]$.




---


**Details:**


First, a review of definitions.


Let $X$ be a set with a partial order $\le$.


* A point $b\in X$ is an *upper bound* of a subset $A$ of $X$ if $x\le b$ for all $x\in A$.
* A nonempty subset $D$ of $X$ is *directed* if any finite subset of $D$ has an upper bound in $D$.
* A point $s\in X$ is the *supremum* of a subset $A$ of $X$ and denoted by $\sup A$ or $\bigvee A$ if $s$ is the least upper bound of $A$, that is, if $s$ is an upper bound of $A$ and for any upper bound $b$ of $A$ we have $s\le b$.
* A subset $A$ of $X$ is an *upper set* if for any $x\in A$ we have $U\_x\subseteq A$, where $U\_x:=\{y\in X\colon x\le y\}$.
* A subset $A$ of $X$ is a *lower set* if for any $x\in A$ we have $L\_x\subseteq A$, where $L\_x:=\{y\in X\colon y\le x\}$.
* The pair $(X,\le)$ of a set $X$ with a partial order $\le$ is a *directed-complete partial order (dcpo)* if each directed subset $D$ of $X$ has a supremum. In what follows, $(X,\le)$ will be a dcpo.
* A subset $G$ of $X$ is *open in the Scott topology*, or *Scott-open*, if $G$ is an upper set and for any directed subset $D$ of $X$ such that $\sup D\in G$ we have $D\cap G\ne\emptyset$. Equivalently, a subset $F$ of $X$ is *closed in the Scott topology*, or *Scott-closed*, if $F$ is a lower set and for any directed subset $D$ of $F$ we have $\sup D\in F$.
* The identity net $I\_D$ on a directed subset $D$ of $X$ is the map $D\ni x\mapsto I\_D(x):=x\in X$.
* The set $\LL\_D$ of limits of the identity net $I\_D$ is defined by the following condition on $x\in X$:


\begin{equation\*}
\begin{aligned} 
x\in\LL\_D\iff &\text{for any Scott-open $G$ such that $x\in G$ } \\ 
&\text{there is some $z\_G\in D$ such that $U\_{z\_G}\subseteq G$.}
\end{aligned} 
\end{equation\*}




---


Let now $D$ be any directed subset of $X$.


Suppose that $x\in\LL\_D$.
We want to show that then $x\in L\_s$. Suppose the contrary. Then $x\in V:=X\setminus L\_s$, and the set $V$ is Scott-open (since the set $L\_x$ is Scott-closed for any $x\in X$). On the other hand, $z\_V\in D$ and $s=\sup D$, so that $z\_V\le s$, that is, $s\in U\_{z\_V}$. Also, by the definition of $\LL\_D$, we have $U\_{z\_V}\subseteq V$. So, $s\in U\_{z\_V} \subseteq V=X\setminus L\_s$, which contradicts the obvious fact that $s\in L\_s$.


We see that $x\in L\_s$ for any $x\in\LL\_D$, that is,
\begin{equation\*}
 \LL\_D\subseteq L\_s. \tag{1}\label{1}
\end{equation\*}


Let us now show that $s\in\LL\_D$. Take any any Scott-open $G$ such that $s\in G$. Since $G$ is Scott-open and $s=\sup D\in G$, we have $D\cap G\ne\emptyset$. Take now any $z\_G\in D\cap G$. Since $z\_G\in G$ and the Scott-open set $G$ is an upper set, we get $U\_{z\_G}\subseteq G$. Since $z\_G\in D$, we see that indeed $s\in\LL\_D$.


Note also that for any $x\in\LL\_D$ we have $L\_x\subseteq\LL\_D$. Therefore and because $s\in\LL\_D$, we see that $L\_s\subseteq\LL\_D$.


Now, in view of \eqref{1}, we conclude that
\begin{equation\*}
 \LL\_D=L\_s=L\_{\sup D},
\end{equation\*}
as claimed.


