Definitions and notations.
==========================


Let $\mathcal{P}(X)$ the **power set** of $X$.


Let $\tau\_X\subseteq\mathcal{P}(X)$ a **topology** on X.


We call $A$ **irreducible** if every time $A=B\cup C$ with $B,C$ closed set then $(B=A)\vee(C=A)$.


We call $X$ **sober** if every non empty irreducible closed set is the closure of a (single one) point.


We Call $K$ **compact** if every open covering $(U\_i)\_{i\in I}\subseteq\tau\_X$ of $K$ (i.e. $K\subseteq\bigcup\_{i\in I}U\_i$) admits a finite subcovering of $K$ (i.e. there is a finite $J\subseteq I$ s.t. $K\subseteq\bigcup\_{j\in J}U\_j$). Note that $(X,\tau\_X)$ is not required to be T$\_2$.


We call $A$ **relatively compact in** $B$ if $A\subseteq B$ and every open covering of $B$ admits a finite subcovering of $A$. Write $A\ll B$ if $A$ is relatively compact in $B$ (note: by definitions $A$ is compact iff $A\ll A$).


We say that $F$ has the **relatively compactness property** if for all $A\in F$ exist $B\in F$ s.t. $B\ll A$.


We call $D\subseteq\mathcal{P}(X)$ **direct** if $D\neq\emptyset$ and for all $A,B\in D$ exist $C\in D$ s.t. $A\cup B\subseteq C$. In such case we call $C$ an **upper bound** of $\{A,B\}$. In other words $D$ is directed if it is non empty and closed by upper bounds of his finite subsets.


We call **supremum** of $A\subseteq\mathcal{P}(X)$ the lower upper bound (by inclusion) of $A$, i.e. $S$ is a supremum of $A$ if for all $B\in A$ we have $B\subseteq S$ and $S$ is a subset of all other sets with the same property.
(note: if it exists, there is at most one supremum).


We call $S\subseteq \mathcal{P}(X)$ **scott open** if $S$ is upward closed and every time it contains the supremum of a direct set $D$ then $S\cap D\neq\emptyset$.


We call $F\subset\mathcal{P}(X)$ a **filter** if $\emptyset\notin F$, it is an upward set (i.e. if $A\in F$ and $A\subseteq B$ then $B\in F$) and it is closed by finite intersections.


We call $\mathcal{Ofilt}(X)$ the space of the scott open filters on $X$.


We call $A\subseteq X$ **saturated** if $A=\bigcap\{U\in\tau\_X\mid A\subseteq U\}$.


We call $\mathcal{Q}(X)\subseteq\mathcal{P}(X)$ the set of all saturated and compact subset of $X$.


The claim.
==========


Let $(X,\tau\_X)$ a sober (and second countable) space. Then


$\begin{align}
f\colon\mathcal{Q}(X)&\to\mathcal{Ofilt}(\tau\_X)\\
Q&\mapsto f(Q)=\{U\in\tau\_X\mid Q\subset U\},
\end{align}$


is a bijective function whose inverse is the map which associates to a scott open filter in $\tau\_X$ the intersection of the filter.


Note: we’ve put between brackets the assumption for X to be second countable because, for our purpose, we have it. In any case the proposition seems to be true without that assumption, as is shown in the Theorem 2.16 of [1]


My question, some explanations and some requests.
=================================================


I'm able to prove that the function $f$ is well defined and injective. On the other hand the proof that the intersection of such a filter is compact (it is obviously a saturated set) is really an hard problem for me.


If it is possible I’m looking for a self-cotained (maybe direct) proof: I lost myself in cross-references from an article to another in which the authors refer.


What follows is my steps (without the final one).


Beginning of (my) proof.
------------------------


Note: I'm supposing that X is second countable.


Let $F$ be a scott open filter of $\tau\_X$ and let $P=\bigcap F$.


Let $(V\_n)\_{n\in\omega}$ be a arbitrary open covering of $P$ (eventually with repetitions). We have to prove that it has a finite subcovering of $P$ (we can suppose the covering to be countable because we have supposed $X$ is second countable).


Let $W\_k=\bigcap\_{n≤k}V\_n$, so for any $k\in\omega$ we have $W\_k\subseteq W\_{k+1}$ and $P\subset\bigcup\_{k\in\omega}W\_k$. We note that $\{W\_k\mid k\in\omega\}$ form a direct set and that $\bigcup\_{k\in\omega}W\_k$, his supremum, is open. So if we prove that $\bigcup\_{k\in\omega}W\_k\in F$ we can conclude thanks to the scott openness and because each $W\_k$ is a finite union, covering $P$, of some sets in $(V\_n)\_{n\in\omega}$ sequence.


On the other hand, if we suppose that we have proved the statement, then the intersection of the filter (i.e. $P$) is in $\mathcal{Q}(X)$ and by $f$ it would be mapped back again to $F$ (thanks to the injectivity). So, if the statement is true, $F$ contains all open set containing $P$.


If we are able to prove that a general open set containing $P$ is in $F$ (that we know is consistent and "true"...), then we'll conclude that $\bigcup\_{k\in\omega}W\_k\in F$ (because it is open) and so we'll conclude the proof.


So, let $A\in\tau\_X$ an open set of $X$ containing $P$.


First of all if $P\in F$ then $A\in F$ (because $F$ is a filter).


So we suppose $P\notin F$. Then (only by second countability) we can take a decrescent (by inclusion) sequence $(U\_n)\_{n\in\omega}\subseteq F$ s.t. $\bigcap\_{n\in\omega}U\_n=P$.


On the first hand if exist $m\in\omega$ s.t. $U\_m\subseteq A$ then $A\in F$ (because $F$ is a filter) so we can suppose that for all $n\in\omega$ we have $U\_n\setminus A\neq\emptyset$.


So let $C\_n=U\_n\setminus A\neq\emptyset$ and let $C=\bigcap\_{n\in\omega}C\_n$.
There are only two cases: $C\neq\emptyset$ or $C=\emptyset$.


If $C\neq\emptyset$ then (because of $C\subseteq\bigcap\_{n\in\omega}U\_n=P$) we conclude an absurd, i.e. $C\subseteq P\subseteq A$ when $C$ does not contain any points of $A$; so it must be $C=\emptyset$ and so...


Step-conclusion.
----------------


With regard to the proof of the statement I'm not able to go on from what I did in the last section; but I'm sure that without the assumption of scott openness the theorem is false.


we give a counterexample: let $X=\mathbb{R}$, $\tau\_\mathbb{R}="\text{standard topology generated by the open intervals}"$, $P=\mathbb{Z}$, $F=\{A\in\tau\_\mathbb{R}\mid \mathbb{Z}\subseteq A\}$; then let $\{\bigcup\_{z\in\mathbb{Z}}(a\_z,b\_z)\mid\forall z\in\mathbb{Z}\; a\_z,b\_z\in\mathbb{Q}\wedge z\in(a\_z,b\_z)\}\subseteq F$ be the sequence of "$(U\_n)\_{n\in\omega}$", decrescent by inclusion, whose intersection is $P$.


So,


$\mathbb{R}$ is sober and second countable;


$F$ is the filter of all open set containing $P$ (but F is not scott open, e.g. $((-n,n))\_{n\in\omega}$ is clearly a direct sequence whose union is $\mathbb{R}\in F$ but for all $n\in\omega$ we have $(-n,n)\notin F$);


the intersection of $F$ is $P$ but $P$ is clearly not compact, e.g. $\{(z-\frac{1}{2},z+\frac{1}{2})\mid z\in\mathbb{Z}\}$ is clearly a covering of $\mathbb{Z}$ without a finite sub covering.


Moreover.
---------


I'm aware that the guideline of the proof that I followed cannot be applied in the general setting (without second countably hypothesis on $X$)...but I'm not in the general case and I'm looking for a "simple", direct and clear proof (if it exist).


In a little more specific setting (which is my case) we don't require directly that $F$ is scott open but that it respects the relatively compactness property, which implies the scott openness for $F$.


Indeed, if $D$ is a direct subset of $\mathcal{P}(\tau\_X)$ whose supremum $S$ (i.e. $S=\bigcup D$) lies on $F$ (so $D$ is a covering of $S$), then there exists $A\in F$ with $A\ll S$, i.e. there must be a finite subset of $D$ that covers a set $A$ who lies on $F$ and so the union of $D$ must lie on $F$ too (because $F$ is a filter). To conclude we have only to note that a finite union of elements of a direct set lies on the direct set (by definition of direct set) and so our union is in $D$. Then $D\cap F\neq\emptyset$ and so $F$ is scott open.


Note that in my counterexample, obviously, $F$ fails this property too...


In any case if you use the relatively compactness instead of the scott openness to find a (simpler) proof... it will be fine for me!


Thank you all,


Corrado.


References.
===========


[1]: Karl Heinrich Hofmann and Michael William Mislove. *Local compactness and continuous lattices*. In Bernhard Banaschewski and Rudolf-Eberhard Hoffmann, editors, Continuous Lattices, volume 871 of Lecture Notes in Mathematics, pages 209–248. Springer Berlin Heidelberg, 1981 (**Theorem 2.16 pag. 226**)


see also (if interested on my specific setting)


[2]: Matthew de Brecht. *Quasi-polish spaces*. Annals of Pure and Applied Logic, (164):356–381, 2013. (last three lines of the proof of the **Theorem 44 pag. 369**)



Why the intersection of a scott open (or \w the relatively compactness property) filter on a topology of a sober (and 2nd countable) space is compact?
A self-contained proof is in the book "Continuous lattices and domains" by G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove and D. S. Scott.


The particular place you need is Lemma II-1.19 on page 146.


