This is exercise 4.7.7 in Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.

Consider a dcpo $(X,\leq)$ and equip it with the Scott topology where the opens are the upward-closed sets $U$ such that for any directed family $(x_i)_{i \in I}$ with $\sup_{i \in I}x_i \in U$ we can find some $i \in I$ such that $x_i \in U$.

Now let $(x_i)_{i \in I}$ be a directed family, and let $N = (x_i)_{i \in I, \sqsubseteq}$ be the net given by $i \sqsubseteq j$ if and only if $x_i \leq x_j$. I want to show that

a point $y$ is a limit of the net $N$ $\iff$ $y \leq \sup_{i \in I} x_i$

(this supremum exists because we assumed $X$ was a dcpo). I have been able to show that if $y \leq \sup_{i \in I} x_i$, then $y$ is a limit of the net $N$. However, the other direction is causing me trouble.

If we assume that $y$ is a limit point of $N$, then by definition this means that for every neighborhood $U$ of $y$, we can find some $x_i \in U$ and hence it follows that $x_j \in U$ for all $x_j \geq x_i$. Because $U$ is Scott-open, it follows that $\sup_{i \in I} x_i$, so we know that $\sup_{i \in I} x_i$ is in any neighborhood of $y$. However, I don't see how that allows me to conclude anything about the ordering $\leq$ between $y$ and $\sup_{i \in I} x_i$.

Could somebody help me prove this? Do I need a different approach to the problem?


1 Answer 1


That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $y\leqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.

For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset $\{y\mid y\le x\}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)

Thus by the above, $y\leqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $y\le x$ in the original order.


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