3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.