Last time, I announced that we would do Exercise V-5.25 of the red book [1], following Hofmann and Lawson [2], in order to show that there exists a space *X* that is core-compact but not locally compact. This is pretty lengthy, but here we go.

We first build a nice space *Y*, out of which we will carve out the stranger space *X*.

## The space *Y*

Let **I** be the closed interval [0,1], with its usual metric topology. Let **J** be the half-open interval [0,1) with the Scott topology of the reverse ordering ≥: its open subsets are [0, *a*) with 0≤*a*≤1. Then, we let *Y* be **I** × **J**, with the product topology.

This is really the space *Y* that is used in Exercise V-5.25 of [1], although you may fail to recognize it. I will make one step in that direction when I talk about lower semicontinuous maps below. You can infer the rest if you do Exercise V-5.23 of [1].

For now, we recognize that *Y* is sober, being a product of sober spaces (Theorem 8.4.8 in the book). Indeed, **I** is Hausdorff, and **J** is a continuous dcpo (where *a* is way-below *b* if and only if *a*>*b*), so both are sober by Proposition 8.2.12 of the book.

*Y* is locally compact, too, because it is a finite product of locally compact spaces (Proposition 4.8.10). **I** is locally compact because it is compact Hausdorff (see Example 4.4.3 and use Proposition 4.8.8), and **J** is locally compact because every continuous dcpo is in its Scott topology (Corollary 5.1.36).

## Lower semicontinuous maps

It will be useful to observe that, given any open subset *W* of *Y*, we can define a map *f* : **I** → [0,1] by *f*(*x*)=sup {*y* | (*x*,*y*) ∈ *W*}, and that this map is lower semicontinuous (i.e., continuous from **I** to [0,1] with the Scott topology of the usual ordering ≤).

Indeed, *f*^{-1}((*a*, 1])={*x* | ∃*y*>*a*, (*x*,*y*) ∈ *W*}. Writing *W* as a union of open rectangles *U _{i}*

**×**

*V*, we see that

_{i}*f*

^{-1}((

*a*, 1]) is equal to the union of all

*U*such that

_{i}*V*contains some

_{i}*y*>

*a*, hence is open in

**I**.

Conversely, given any lower semicontinuous map *f* from **I** to [0,1], we can define an open set *W* as {(*x*,*y*) ∈ **I** **× J** | *y*<*f*(*x*)}. This is open because it is the union over all *a* of *f*^{-1}((*a*, 1]) **×** [0, *a*).

Finally, the two constructions are inverse of each other: starting from an open set *W*, building *f* and going back, we obtain {(*x*,*y*) ∈ **I** **× J** | *y*<sup {*y’* | (*x*,*y’*) ∈ *W*}}, which is just *W*, because (*x*,*y*) ∈ *W* if and only if there is a *y’*>*y* such that (*x*,*y’*) ∈ *W*; and starting from a lower semicontinuous map *f*, building *W* and going back, we obtain a function that maps every *x* to sup {*y* | (*x*,*y*) ∈ {(*x*,*y*) ∈ **I** **× J** | *y*<*f*(*x*)}}=*f*(*x*).

This defines an order isomorphism between **O**(*Y*) and LSC(**I**,[0,1]), the set of lower semicontinuous maps from **I** to [0,1], with the pointwise ordering. This will be useful in the next section.

## The space *X*

We pick a Bernstein subset *A* of **R**. We now define *X* as the subspace of *Y* of points (*x*,*y*) such that:

- either
*y*is rational and*x*is in*A*, - or
*y*is irrational and*x*is not in*A*.

We can expect *X* to inherit much of the pathological character of *A*, but we claim that, despite all odds, *X* is core-compact. This depends on the fact that the set of rationals **Q** is dense, and has dense complement, in **R**. (Any other set with the same properties would have suited our needs. We will need it to be Borel, too, for the later parts of the proof to work.)

In order to show that *X* is core-compact, we show that *X* and *Y* have isomorphic lattices of open subsets.

There is a surjective monotonic map τ : **O**(*Y*) → **O**(*X*) that sends every open subset *V* of *Y* to *V* ∩ *X*. In order to show that it is an order isomorphism, we must realize that every lower semicontinuous map *f* from **I** to [0,1] is entirely characterized, for each *x* in **I**, by:

- the set of all
*rational*points*y*<*f*(*x*) if*x*is in*A*; - the set of all
*irrational*points*y*<*f*(*x*) if*x*is not in*A,*

in other words, *f* is entirely characterized by the set of points (*x*,*y*) in *X* such that *y*<*f*(*x*).

In more detail, imagine we have two open subsets *V* and *V’* such that τ(*V*)⊆τ(*V’*). Through the isomorphism between **O**(*Y*) and LSC(**I**,[0,1]), this means we have two lower semicontinuous maps *f* and *f’* such that for every *x* in *A*, all the rational points *y* strictly below *f*(*x*) are also strictly below *f’*(*x*), and for every *x* in **I** – *A*, all the irrational points *y* strictly below *f*(*x*) are also strictly below *f’*(*x*). This implies that *f*≤*f’*, hence that *V*⊆*V’*.

Having shown that τ(*V*)⊆τ(*V’*) implies *V*⊆*V’*, we now know that τ is an order isomorphism between **O**(*Y*) and **O**(*X*). Since *Y* is locally compact hence core-compact, **O**(*Y*) is a continuous complete lattice. It follows that **O**(*X*) is also a continuous complete lattice, so *X* is core-compact.

One half of the theorem proved—good—only one half to go: we must show that *X* is not locally compact. To this end, we study the shape of the compact subsets of *X*.

## The compact subsets of *X*

In order to show that *X* is not locally compact, we again follow Exercise V-5.25 of [1], and we will show that every compact subset of *X* has empty interior. However, there is nothing to follow, really, here, as [1] recommends to look at [2] for details. Let us do that.

Consider a compact subset *K* of *X*. Imitating the correspondence between open subsets of *Y* and lower semicontinuous maps, *K* allows us to define a map *q* : **I** → [0,1] by *q*(*x*)=sup {*y* | (*x*,*y*) ∈ *K*}. Note that *y* ranges only on rational numbers if *x* is in *A*, and only on irrational numbers if *x* is not in *A*.

**Lemma (7.1 in [2]).** For each *x* in **I**, either there is no *y* ∈ [0,1] such that (*x*,*y*) ∈ *K*, in which case *q*(*x*)=0, or there is a largest one.

In other words, except possibly when *q*(*x*)=0, we have *q*(*x*)=max {*y* ∈ [0,1]* *| (*x*,*y*) ∈ *K*} (a max, not just a sup).

*Proof.* If *q*(*x*)=0, this is obvious, otherwise let *a*=*q*(*x*), and realize that the closed set ({*x*} × [*b*,1]) ∩ *X* intersects *K* for every *b*<*a*. Those closed sets form a filtered family (in fact a chain), so by compactness *K* must also meet their intersection (Proposition 4.4.9 in the book). That intersection is ({*x*} × [*a*,1]) ∩ *X*, and since it meets *K*, (*x*,*a*) is in *K*. ☐

Beware of a tiny mistake in [2]. The authors take *K* saturated, which is OK. However, that means that *K* must be saturated in *X*, but they do as if it were saturated in *Y*: it is claimed that the fact that *K* is saturated means that (*a*,*b*) ∈ *K* implies {*a*} × [0,*b*] ⊆ *K*. This is wrong, since *K* has to be included in *X*.

**Lemma (7.2 in [2]).** The map *q* is upper semicontinuous from **I** to [0,1].

In other words, *q*^{-1}([*a*,1]) is closed for every *a* ∈ [0,1].

*Proof.* This is clear if *a*=0, so let us assume *a*>0. By the previous lemma (and since *a*>0), *q*^{-1}([*a*,1]) is the set of points *x* such that (*x*,*y*) ∈ *K* for some *y* ∈ [*a*,1].

Let us consider the saturation sat(*K*) of *K* in *Y*. I am writing it as sat(*K*) instead of as ↑*K*, as I would usually do, because the specialization ordering on *Y* is heads over heels, and we would only get confused. Explicitly, sat(*K*) is the set of points (*x*,*y*) in *Y* such that there is a *y’* (rational or irrational, depending on *x*) above *y* such that (*x*,*y’*) is in *K*. We now have that *q*^{-1}([*a*,1]) is the set of points *x* such that (*x*,*a*) ∈ sat(*K*).

(Updated February 22nd, 2019, after Xiaodong Jia signaled me a bug in the argument I had previously posted. Begin update:)

I claim that sat(*K*) is closed in **I** × [0,1), where [0,1) has the usual metric topology (I am not writing it **J** in order to avoid confusion, if possible: the topology is finer). The shortest argument is probably the following. Let sat'(*K*) be the saturation of *K* in the slightly larger space **I** × **J**‘, where **J**‘ is the *closed* interval [0, 1], with the Scott topology of ≥. Since *K* is compact in *X*, which is a subspace of *Y* hence of **I** × **J**‘, *K* is compact in **I** × **J**‘. Its saturation sat'(*K*) is therefore compact in **I** × **J**‘ as well (Proposition 4.4.14 in the book). However **I** × **J**‘, being the product of two stably compact spaces, is stably compact (Proposition 9.3.1). Hence sat'(*K*) is compact saturated in **I** × **J**‘, hence closed in its de Groot dual, hence closed in its Nachbin pospace (the space with the patch topology). The latter happens to be **I** × [0, 1], where [0, 1] has its usual metric topology. This is by Proposition 9.3.1 of the book again, which says that the Nachbin space of a product is the product of the Nachbin pospaces, and since the Nachbin pospace of **J**‘ is [0, 1] with its usual metric topology. We observe that sat(*K*)=sat'(*K*) ∩ **I** × [0,1) (remember the ordering is ≥), hence sat(*K*) is closed in **I** × [0,1), as promised.

(End update.)

In order to show that *q*^{-1}([*a*,1]), the set of points *x* such that (*x*,*a*) ∈ sat(*K*), is closed, we show that its complement is open. Let *x* be such that (*x*,*a*) is not in sat(*K*). We have just seen that sat(*K*) is closed, so there is an open rectangle *U* **×** *V* containing (*x*,*a*) and disjoint from sat(*K*). Then *U* is an open neighborhood of *x* disjoint from *q*^{-1}([*a*,1]). ☐

The set **Q _{+}** of (strictly) positive rational numbers is a Borel subset of

**R**, because it is a countable union of one-element sets, and each one-element set is closed. (Closed sets are Borel, since they are complements of open subsets.) It follows that:

**Fact.** *q*^{-1}(**Q _{+}**) is Borel.

This is because every upper semicontinuous (or every lower semicontinuous map) is measurable, but here is a simple proof of what we require in our special case. The set *q*^{-1}(**Q _{+}**) is the union of the countably many sets

*q*

^{-1}({

*a*}),

*a*∈

**Q**, so it suffices to show that each set

_{+}*q*

^{-1}({

*a*}) is Borel. If

*a*is not in [0,1], then that set is empty. If

*a*=1, then

*q*

^{-1}({

*a*})=

*q*

^{-1}([

*a*,1]), which is closed hence Borel. Otherwise, we write {a} as [

*a*,1] minus the union of the countably many closed sets [

*a*+(1–

*a*)/

*n*,1], where

*n*ranges over the non-zero natural numbers, and then

*q*

^{-1}({

*a*}) is equal to

*q*

^{-1}([

*a*,1]) minus the countably many closed sets

*q*

^{-1}([

*a*+(1–

*a*)/

*n*,1]).

## Concluding the proof

Now imagine that *K* contains a non-empty open subset *W*. We wish to reach a contradiction. Here [2] is pretty terse, and I will have to fill in the gaps.

Through the isomorphism between **O**(*X*) and LSC(**I**, [0,1]), we build the associated lower semicontinuous map *f* : *f*(*x*)=sup {*y* | (*x*,*y*) ∈ *W*}, and we realize that *f* is not identically zero (because the zero map is associated with the empty set only), and also that *f*≤*q*. Indeed, for every *x* in **I**, every *y* such that (*x*,*y*) ∈ *W* is also such that (*x*,*y*) ∈ *K*, hence *y*≤*q*(*x*); taking suprema over *y*, *f*(*x*)≤*q*(*x*).

Since *f* is not identically zero, *f*^{-1}((0,1]) is a non-empty open subset *U* of **I**. For every *x* in *U* ∩ *A*, the fact that *x* is in *U* implies *f*(*x*)>0 hence *q*(*x*)>0. By the first of the two lemmata above, (*x*,*q*(*x*)) is in *K*, so, because *x* is in *A*, *q*(*x*) is in **Q**. Since *q*(*x*)>0, it is even in **Q _{+}**. This shows that

*U*∩

*A*is included in

*q*

^{-1}(

**Q**), hence in

_{+}*U*∩

*q*

^{-1}(

**Q**). Conversely, for every

_{+}*x*in

*U*∩

*q*

^{-1}(

**Q**),

_{+}*x*must be in

*A*since (

*x*,

*q*(

*x*)) is in

*K*and

*q*(

*x*)>0. This show that

*U*∩

*A*=

*U*∩

*q*

^{-1}(

**Q**).

_{+}However, *U* ∩ *q*^{-1}(**Q _{+}**) is the intersection of two Borel sets, hence is Borel. But we have seen last time that the intersection of the Bernstein set

*A*with the non-empty open set

*U*cannot be Borel: contradiction.

We conclude that *Q* cannot contain any non-empty open set, namely, it must have empty interior.

Hence we have shown the following, as desired.

**Theorem.** The space *X* is core-compact, but all its compact subsets have empty interior, in particular *X* is not locally compact.

## Further questions

Xiaodong Jia asked the following question: is there any dcpo that is core-compact but not locally compact in its Scott topology? If every dcpo were sober, the answer would be no, but precisely, there are non-sober dcpos, such as Johnstone’s space (Exercise 5.2.15 in the book).

He also asked the following: is every well-filtered core-compact space locally compact? Every sober core-compact space is locally compact, but well-filteredness is a weaker property (except in the presence of… local compactness). Is every well-filtered core-compact dcpo locally compact? (Update: every well-filtered core-compact space is indeed locally compact, see this post.)

Every complete lattice is well-filtered, but not necessarily sober in its Scott topology. Is every core-compact complete lattice locally compact? (Oops, sorry, that is definitely true.)

Then replace ‘dcpo’ by ‘monotone convergence space’: is every core-compact monotone convergence space locally compact?

That makes plenty of open questions… currently, the space *X* of [2] that forms the subject of this post is the only concrete example of a core-compact, non-locally compact space that we know.

- Gerhard Gierz, Karl Heinrich Hofmann, Klaus Keimel, Jimmie D. Lawson, Michael W. Mislove, and Dana S. Scott. Continuous Lattices and Domains. Number 93 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003.
- Karl H. Hofmann and Jimmie D. Lawson. The Spectral Theory of Distributive Continuous Lattices. Transactions of the American Mathematical Society 246 (Dec. 1978), pages 285- 310.

— Jean Goubault-Larrecq (Feb. 20th, 2019)