# 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**)