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?

Question

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)

Answer 1

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.

Answer 2

We have (only) to prove that for every open set $A\supseteq P$, where $P=\bigcap F$, we have $A\in F$ (for the whole notation you can see the question).

The missing step.

(we refer to [3], thanks to the suggestion of მამუკა ჯიბლაძე)

Suppose $P\subseteq A\notin F$ then (since $X\in F$) there must exist $V\in\tau_X$ s.t. $A\subseteq V\notin F$ with $T\in F$ for every open set $T\supseteq V$, i.e. $V$ is the maximal open set containing $A$ with respect to not being in $F$.

So if $B,C\in\tau$ s.t. $B\cap C=V$ then $(B=V) \vee (C=V)$ or $V\in F$ because $F$ is a filter.

Then $X\setminus V$ is a closed (obviously) and irreducible set, indeed if $C_1,C_2\subsetneq(X\setminus V)$ are two closed sets s.t. $C_1\cup C_2=(X\setminus V)$ then $(X\setminus C_1)\cap(X\setminus C_2)=V$ but $(X\setminus C_1)\supsetneq V \wedge (X\setminus C_2)\supsetneq V$, absurd.

Thanks to sobriety, there exists $p\in X$ s.t. $\overline{\{p\}}=X\setminus V$.

Now, for all $G\in F$ we have $p\in G$, because if $p\notin G$ then $\overline{\{p\}}\cap G=(X \setminus V)\cap G=\emptyset$ and so $G\subseteq V$ and $V\in F$, because $F$ is a filter; abusurd.

So, for all $G\in F$ we have $p\in G$ but this means that $p\in P$ and then $p\in A\subseteq V=(X\setminus\overline{\{p\}})$, absurd.

So $A\in F.\quad\square$

The existence of V

Let $D=\{E\in(\tau_X\setminus F)\mid A\subseteq E\}$ ordered by inclusion. So D is a poset. Let $M$ the lower upper bound of a generical chain in $D$, i.e. $M$ is the union of the chain. If $M\in F$ so, thanks to the Scott openness of $F$, there must be an element of the chain which lies in $F$, absurd.

So, the lower upper bound of every chain in $D$ lies in $D$. By Zorn's lemma there is a maximal element in $D$. Call one of them $V$.

Reference.

[3]: G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove and D. S. Scott. Continuous lattices and domains (Lemma II-1.19, page 146).