Given a poset *P*, we can form the space *P*_{σ} obtained by giving *P* the Scott topology. Now we can form two products, either by forming the poset product of *P* with itself, and then taking the Scott topology, yielding (*P* × *P*)_{σ}, or taking the Scott topology first, and then building the topological product *P*_{σ} × *P*_{σ} (namely, with the product topology, generated by open rectangles).

The (Scott) topology of (*P* × *P*)_{σ} is always finer than that of *P*_{σ} × *P*_{σ}, and is often identical, for example if *P* is a continuous poset (Proposition 5.1.54 in the book); but also, more generally, if *P*_{σ} is core-compact; or if *P*_{σ} is first-countable. The latter are points 1(a) and 1(b) from this post, where I discussed the Chen-Kou-Lyu property [1]: *P* has the Chen-Kou-Lyu property if and only if the Scott topology on the product *P ^{n}* of

*P*with itself

*n*times (with

*n*an arbitrary natural number) coincides with the product of the Scott topologies.

I would like to state another result of this kind, due to Hualin Miao, Xiaoyong Xi, Qingguo Li, and Dongsheng Zhao [2, Lemma 3.1]. But this is really an excuse: my initial intention was to talk about a funny dcpo that has the Chen-Kou-Lyu property, but is neither core-compact nor first-countable in its Scott topology; and showing that it has the Chen-Kou-Lyu is best obtained as a consequence of the Miao-Xi-Li-Zhao result.

This funny dcpo is very simple, and I will explore its many properties this month and next month. Doing it in just one post would be too much. I will not say what it is right away: you will have to wait until after the statement and proof of the Miao-Xi-Li-Zhao Lemma! But let me say for now that it seems to have been rediscovered many times. The first occurrence I know of it is due to Xu, Shen, Xi and Zhao [3, Example 4.8] in 2020. Then it was rediscovered independently by Chen, Kou and Lyu [1, Example 4.9] and by Hertling [4, Example 4.7] in 2022.

## The Miao-Xi-Li-Zhao Lemma

The Miao-Xi-Li-Zhao lemma states that, given two posets *P* and *Q* which only have countably many ideals, then (*P* × *Q*)_{σ} = *P*_{σ} × *Q*_{σ}. The proof shows that we can weaken the assumptions somehow. Let us say that a directed family *D* is *trivial* if and only if it has a largest element, and *non-trivial* otherwise. The trivial directed families are uninteresting when studying the Scott topology: a subset *U* of a poset is Scott-open if and only if it is upwards-closed and it intersects every *non-trivial* directed family *D* whose supremum sup *D* exists and is in *U*; indeed, if *D* is a trivial directed family and sup *D* is in *U*, then *D* intersects *U*, whatever *U* may be.

An ideal is a downwards-closed directed subset, and it is non-trivial if and only if it is non-trivial as a directed subset; equivalently, if and only if it is not of the form ↓*x* for any point *x*.

We will not assume that there are countably many ideals, or even countably many ideals that have suprema, but just that the family of *non-trivial* ideals that have suprema is countable. This is a tiny improvement over what Miao, Xi, Li and Zhao stated, but a proper improvement nonetheless: given a poset *P* with countably many non-trivial ideals that have suprema, adding uncountably many pairwise incomparable points, all incomparable to all the points of *P*, to *P*, yields a poset that still only has countably many non-trivial-ideals-that-have-suprema, but uncountably many ideals.

**Theorem A (Miao, Xi, Li, Zhao).** Let *P* and *Q* be two posets. If both *P* and *Q* only have countably many non-trivial ideals that have suprema, then (*P* × *Q*)_{σ} = *P*_{σ} × *Q*_{σ}.

*Proof.* Let me call a (directed) *sup situation* on *P*, for short, any non-trivial ideal *I* that has a supremum; and similarly in *Q*. Let me say that a subset *U* of *P* (resp., *Q*) *respects* a sup situation *I* if and only if sup *I* ∈ *U* implies *I* intersects *U*; in other words, if and only if sup *I* is not in *U* or *I* intersects *U*. These denominations are inspired by very similar notions used in the Rasiowa-Sikorski and Görnemann-Rauszer-Sabalski lemmas, which I have talked about in this post in December 2019. But the sup situations there were not directed, so there are some differences; and we will not need any form of the Baire property to conclude.

I claim that *U* is Scott-open if and only if it is upwards-closed and respects every sup situation. The only if direction is clear. In the if direction, let us consider any directed family *D* with a supremum *x* in *U*. Then ↓*D* is an ideal, with supremum *x*. If *x* is in ↓*D*, then *x*≤*y* for some *y* ∈ *D*, and since *x* is in *U* and *U* is upwards-closed, *y* is in *D* and in *U*, so *D* intersects *U*. Otherwise, ↓*D* is a non-trivial ideal, hence a sup situation. By assumption, *U* respect ↓*D*, so ↓*D* intersects *U*, and since *U* is upwards-closed, *D* must intersect *U* as well.

The key observation we will use to show the theorem is the following:

**Claim.** Let *W* be a Scott-open subset of *P* × *Q*, *I*_{0}, …, *I*_{n–1}, *I*_{n} be finitely many sup situations on *P*, and *A* and *B* be two non-empty finite subsets of *P* and *Q* respectively such that *A* × *B* ⊆ *W*, and such that ↑*A* respects the sup situations *I*_{0}, …, *I*_{n–1}. Then there is a finite subset *A’* of *P* containing *A* such that *A’* × *B* ⊆ *W*, and such that ↑*A’* respects the sup situations *I*_{0}, …, *I*_{n–1} and also *I*_{n}.

*Proof of the claim.* For short, let *x _{k}* denote the supremum of

*I*

_{k}, for each

*k*with 0≤

*k*≤

*n*. If the supremum

*x*of

_{n}*I*

_{n}is not in ↑

*A*, then we simply take

*A’*≝

*A*. Then

*A’*×

*B*⊆

*W*, trivially, ↑

*A’*= ↑

*A*respects the sup situations

*I*

_{0}, …,

*I*

_{n–1}, trivially as well, and also respects the sup situation

*I*because

_{n}*x*is not in ↑

_{n}*A*. Hence, we will assume that

*x*is in ↑

_{n}*A*from now on. The situation is as in the figure below. The dotted, squiggly lines below each point

*x*are meant to denote

_{k}*I*; ↑

_{k}*A*(in orange) respects each of the corresponding sup situations: e.g.,

*x*

_{0}and

*x*

_{1}are not even in ↑

*A*, while

*x*

_{m}_{–1},

*x*, …,

_{m}*x*

_{n}_{–1}and in ↑

*A*, and some points in the dotted lines below them are in ↑

*A*, too. For

*x*, we have assumed that it is in ↑

_{n}*A*, but we don’t know whether any point from the dotted line

*I*

_{n}below it is also in ↑

*A*.

Among *x*_{0}, …, *x*_{n–1}, some elements may be larger than or equal to *x _{n}*, some may not. Up to a permutation of indices, we will assume that

*x*

_{0}, …,

*x*

_{m–1}are those elements that are not larger than or equal to

*x*, while

_{n}*x*

_{m}, …,

*x*

_{n–1}are larger than or equal to

*x*. This is the situation shown above:

_{n}*x*

_{m}, …,

*x*

_{n–1}are the points in the cone of points above

*x*, materialized by the two slanted, dotted lines originating from

_{n}*x*and going up and sideways.

_{n}Let *C* be ↓{*x*_{0}, …, *x*_{m–1}}. This is a Scott-closed subset of *P* which does not contain *x _{n}*. We represent it in blue below.

Since *C* is closed, its complement *U* is a Scott-open neighborhood of *x _{n}*. For each index

*k*with

*m*≤

*k*≤

*n*, we have

*x*

_{k}≥

*x*

_{n}, so

*x*

_{k}is in

*U*as well. (This should be particularly obvious from the picture.) We will build a point

*x’*, one for each index

_{k}*k*with

*m*≤

*k*≤

*n*, as follows.

- The point
*x*_{k}is larger than or equal to some point*a*of*A*. Indeed, when*k*=*n*, this is because*x*is in ↑_{n}*A*, and if*m*≤*k*<*n*, this is because*x*_{k}≥*x*_{n}and*x*is in ↑_{n}*A*. For each one of the finitely many elements*b*of*B*, (*a*,*b*) is then in*A*×*B*, hence in*W*. Therefore (*x*_{k},*b*) is in*W*, since*W*is upwards-closed. We have observed that*x*_{k}is in*U*, so (*x*_{k},*b*) is in*W*∩ (*U*×*Q*), and that intersection is Scott-open. - For each
*b*∈*B*, the pair (*x*_{k},*b*) is the supremum of the directed family of points (*x’*,*b*) where*x’*ranges over*I*_{k}, so since*W*∩ (*U*×*Q*) is Scott-open, we can find a point in*I*_{k}—which we will write as*x’*_{k,b}to show its dependency on both*k*and*b*—such that (*x’*_{k,b},*b*) ∈*W*∩ (*U*×*Q*). Said otherwise,*x’*_{k,b}is in*I*_{k}and in*U*, and (*x’*_{k,b},*b*) ∈*W*. - Since
*I*_{k}is directed and*B*is finite, there is a point*x’*_{k}in*I*_{k}that is larger than or equal to*x’*_{k,b}for every*b*∈*B*. Since*U*and*W*are upwards-closed, we obtain that*x’*_{k}is in*U*, and that (*x’*_{k},*b*) ∈*W*for every*b*∈*B*.

Here is the situation now. We have highlighted the new points *x’*_{m}, …, *x’*_{n}. Since they are all in *U*, we have shown them outside the blue box *C*.

We now define *A’* as *A* ∪ {*x’*_{m}, …, *x’*_{n}}. It is easy to see that *A’* × *B* ⊆ *W*, because *A* × *B* ⊆ *W* and (*x’*_{k}, *b*) ∈ *W* for every *b* ∈ *B*. Let us check that ↑*A’* respects the sup situations *I*_{0}, …, *I*_{n–1} and *I*_{n}. We check that it respects the sup situations *I*_{j} with 0≤*j*<*m* first, then those *I*_{k} with *m*≤*k*≤*n*.

- For every
*j*with 0≤*j*<*m*, if*x*_{j}is in ↑*A’*, then we claim that*x*_{j}is in fact in ↑*A*. Otherwise, we would have*x*_{j}≥*x’*_{k}for some*k*such that*m*≤*k*≤*n*; but*x*_{j}is in the Scott-closed set*C*, by definition of the latter, so*x’*_{k}would be in*C*as well, and that is impossible since*x’*_{k}is in*U*; we recall that*U*is the complement of*C*. Hence*x*_{j}is in ↑*A*. Since ↑*A*respects the sup situation*I*_{j}, some element of*I*_{j}is in ↑*A*, hence in ↑*A’*. Therefore ↑*A’*respects the sup situation*I*_{j}. - For every
*k*with*m*≤*k*≤*n*, in order to show that ↑*A’*respects the sup situation*I*_{k}, we need to show that if*x*_{k}is in ↑*A’*(which happens to be true, but this will not matter), then ↑*A’*contains some element of*I*_{k}. And indeed ↑*A’*contains some element of*I*_{k}, namely*x’*_{k}. (End of proof of Claim.)

By exchanging the roles of *P* and *Q*, we have the following.

**Dual Claim.** Let *W* be a Scott-open subset of *P* × *Q*, *J*_{0}, …, *J*_{n–1}, *J*_{n} be finitely many sup situations on *Q*, and *A* and *B* be two non-empty finite subsets of *P* and *Q* respectively such that *A* × *B* ⊆ *W*, and such that ↑*B* respects the sup situations *J*_{0}, …, *J*_{n–1}. Then there is a finite subset *B’* of *Q* containing *B* such that *A* × *B’* ⊆ *W*, and such that ↑*B’* respects the sup situations *J*_{0}, …, *J*_{n–1} and *J*_{n}.

With the help of the Claim and of the Dual Claim, we can now prove the Theorem.

Every open subset of *P*_{σ} × *Q*_{σ} is Scott-open in *P* × *Q*, namely open in (*P* × *Q*)_{σ}, so we concentrate on the converse statement. Let us fix a point (*x*, *y*) in *P* × *Q*, and *W* be a Scott-open neighborhood of (*x*, *y*) in *P* × *Q*. We wish to show that *W* contains an open rectangle *U* × *V*, where *U* is a Scott-open neighborhood of *x* in *P* and *V* is a Scott-open neighborhood of *y* in *Q*. Once we have managed to do so, and making (*x*, *y*) vary inside *W*, we will have obtained that *W* is an open neighborhood of any of its elements in *P*_{σ} × *Q*_{σ}, hence is open in *P*_{σ} × *Q*_{σ}, and the proof will be complete.

We start with the case where both *P* and *Q* have at least one sup situation. (There is a formal catch when there is no sup situation at all). Then we can enumerate the sup situations of *P* as *I _{n}*,

*n*∈

**N**, and the sup situations of

*Q*as

*J*,

_{n}*n*∈

**N**, possibly repeating some sup situations. For example, if

*P*has just three sup situations

*I*

_{0},

*I*

_{1}and

*I*

_{2}, we can simply decide to define

*I*

_{3},

*I*

_{4}, …, as all being equal to

*I*

_{2}. But we cannot do that if

*P*has no sup situation at all, whence the ‘formal catch’. (I apologize for being so formal.)

Right, anyway, assuming that both *P* and *Q* have at least one sup situation, we define sequences *A*_{0} ⊆ *A*_{1} ⊆ … ⊆ *A _{n}* ⊆ … and

*B*

_{0}⊆

*B*

_{1}⊆ … ⊆

*B*⊆ … of finite subsets of

_{n}*P*and

*Q*respectively in such a way that:

- each
*A*contains_{n}*x*, - each
*B*contains_{n}*y*, - ↑
*A*respects the sup situations_{n}*I*_{0}, …,*I*_{n–1}, - ↑
*B*respects the sup situations_{n}*J*_{0}, …,*J*_{n–1}, and *A*×_{n}*B*⊆_{n}*W*.

We initialize *A*_{0} to {*x*} and *B*_{0} to {*y*}. Then, for every natural number *n*≥1, we apply the Claim to *A* ≝ *A _{n}*

_{–1}and

*B*≝

*B*

_{n}_{–1}; this yields a finite subset

*A’*of

*P*, and we define

*A*≝

_{n}*A’*. Then we apply the Dual Claim that

*A*≝

*A*and

_{n}*B*≝

*B*

_{n}_{–1}; this yields a finite subset

*B’*of

*Q*, and we define

*B*≝

_{n}*B’*.

If instead *P* has at least one sup situation and *Q* has none, then we only use the Claim to define *A _{n}* ≝

*A’*as above, and we let

*B*≝

_{n}*B*

_{n}_{–1}at each step; in that case, we guarantee conditions 1, 2, 3 and 5, but not 4, which would be meaningless. The situation is symmetric if

*Q*has at least one sup situation and

*P*has none. If neither

*P*nor

*Q*has a sup situation, we simply define

*A*as {

_{n}*x*} and

*B*as {

_{n}*y*} for every natural number

*n*.

In any case, once the sequences *A*_{0} ⊆ *A*_{1} ⊆ … ⊆ *A _{n}* ⊆ … and

*B*

_{0}⊆

*B*

_{1}⊆ … ⊆

*B*⊆ … have been built, we define

_{n}*A*as the union of the sets

*A*,

_{n}*n*∈

**N**, and

*U*as ↑

*A*, while we define

*B*as the union of the sets

*B*,

_{n}*n*∈

**N**, and

*V*as ↑

*B*. By construction,

*U*respects all the sup situations of

*P*,

*V*respects all the sup situations of

*Q*, so they are Scott-open in

*P*and in

*Q*respectively. Condition 5 above guarantees that

*U*×

*V*⊆

*W*, while conditions 1 and 2 ensure that

*U*×

*V*contains (

*x*,

*y*).

And that’s it: we have found the two Scott-open neighborhoods *U* of *x* and *V* of *y* such that *U* × *V* ⊆ *W*, as promised. We have argued earlier why this was enough to conclude that *W* is open in *P*_{σ} × *Q*_{σ}. ☐

## The complete lattice *L*_{fan}

I started this month’s post by saying that I would present an example of a dcpo *L*_{fan} that has the Chen-Kou-Lyu property, because of the just mentioned Miao-Xi-Li-Zhao result, although (*L*_{fan})_{σ} is neither core-compact nor first-countable. Here it is at last.

*L*_{fan} is simply **N** × **N**, with equality as ordering on the first component and the usual ordering on the second component, plus an extra bottom element ⊥, plus an extra top element ⊤:

Formally, the elements of *L*_{fan} are ⊥, ⊤ and all pairs (*m*, *n*) of natural numbers, ordered by *x*≤*y* if and only if *x*=⊥, or *y*=⊤, or *x* is a pair (*m*, *n*) and *y* is a pair (*m’*, *n’*) with *m*=*m’* and *n*≤*n’*. In the sequel, I will call the set of pairs whose first element is equal to *m* a *column*, and precisely, *column number m*.

As I have already said, the first occurrence I know of it is due to Xu, Shen, Xi and Zhao [3, Example 4.8] in 2020. Then it was rediscovered independently by Chen, Kou and Lyu [1, Example 4.9] and by Hertling [4, Example 4.7] in 2022.

I have decided to call it *L*_{fan} because what remains when you remove the bottom element ⊥ is very close to the *sequential fan*; just like it, it will be countable but not first-countable.

*L*_{fan} has the Chen-Kou-Lyu property

Hertling showed that *L*_{fan} has the property that the product of the Scott topology is the Scott topology of the product, namely (*L*_{fan} × *L*_{fan})_{σ} = (*L*_{fan})_{σ} × (*L*_{fan})_{σ} [4, Lemma 4.12]. Hertling’s proof is a bit long and technical… but this is an immediate consequence of the Miao-Xi-Li-Zhao lemma. Indeed, *L*_{fan} only has countably many non-trivial ideals, and those are its columns (with ⊥ added). Since *L*_{fan} is countable, it even has countably many ideals, trivial or not.

A poset *P* has the Chen-Kou-Lyu property if and only if, for every natural number *n*, the *n*-fold product of *P* with itself is the same whether with the Scott topology of the poset product, or with the product topology of the Scott topologies. In order to see that *L*_{fan} has the Chen-Kou-Lyu property, we need to observe the following: the ideals of any finite product of posets are the products of ideals. You can check it by hand if you will. Here is a hammer of an argument:

- The irreducible closed subsets of a product of spaces are the products of irreducible closed subsets of each space (Proposition 8.4.7 in the book).
- But the ideals of a poset
*P*are exactly the irreducible closed subsets of*P*, where_{a}*P*is_{a}*P*with the Alexandroff topology (Fact 8.2.49 in the book), and the Alexandroff topology of a finite product of posets is the product topology of the Alexandroff topologies (Exercise 4.5.19 in the book).

Knowing this, the *n*th power (*L*_{fan})* ^{n}* of

*L*

_{fan}(as a poset) has only countably many ideals, since it has exactly as many as there are

*n*-tuples of ideals of

*L*

_{fan}, and

*L*

_{fan}only has countably many ideals. Therefore, by Theorem A (the Miao-Xi-Li-Zhao lemma) and an easy induction on

*n*, ((

*L*

_{fan})

*)*

^{n}_{σ}= ((

*L*

_{fan})

_{σ})

*: this is clear if*

^{n}*n*=0, and otherwise ((

*L*

_{fan})

*)*

^{n}_{σ}= (

*L*

_{fan})

_{σ}× ((

*L*

_{fan})

^{n}^{–1})

_{σ}(by Theorem L, since both

*L*

_{fan}and (

*L*

_{fan})

^{n}^{–1}have countably many ideals) = (

*L*

_{fan})

_{σ}× ((

*L*

_{fan})

_{σ})

^{n}^{–1}(by induction hypothesis) = ((

*L*

_{fan})

_{σ})

*. Therefore, as promised:*

^{n}**Proposition B.** *L*_{fan} has the Chen-Kou-Lyu property.

As a consequence, and in passing, we can then use the argument in Section ‘The Chen-Kou-Lyu property’ of this post (March 2022):

**Corollary C.** The Scott topology coincides with the lower Vietoris topology on **H**((*L*_{fan})_{σ}).

## Basic properties of *L*_{fan}

Let me list some of the basic properties of *L*_{fan}; we will see the others next month.

**Lemma D.** *L*_{fan} is a complete lattice.

*Proof.* The supremum of any subset *A* that contains ⊤, or two elements from different columns, or elements (*m*, *n*) from the same column *m* but with arbitrarily large *n*, is ⊤. The supremum of the empty set, the supremum of {⊥}, is ⊥. All other subsets of *L*_{fan} consist of pairs (*m*, *n*) with the same *m*, with *n* bounded, and at least one such pair (plus possibly ⊥), and their supremum is the point (*m*, *n*) with the largest value of *n*. Infima can be described similarly. ☐

Let us investigate the Scott topology on *L*_{fan}. For every function *f* : **N** → **N**, let *U _{f}* be the subset of those pairs (

*m*,

*n*) of

**N**×

**N**such that

*n*≥

*f*(

*m*), plus ⊤.

**Lemma E.** The Scott-open subsets of *L*_{fan} are exactly the sets *U _{f}*, where

*f*ranges over the functions from

**N**to

**N**, plus the empty set, plus the whole space

*L*

_{fan}.

*Proof.* A Scott-open subset *U* is any upwards-closed subset such that every *non-trivial* directed family *D* whose supremum is in *U* already contains an element that is in *U*. A non-trivial directed family *D* must be infinite. In particular, in *L*_{fan}, every non-trivial directed family *D* must contain infinitely many elements different from ⊥ and from ⊤. If any two (say *x* and *y*) are in two different columns, then *D* must contain a larger element than both *x* and *y*, and that can only be ⊤, which would entail that *D* is trivial. Hence *D* (minus ⊥, whereas ⊤ cannot be in *D*) must be entirely included in a single column, say column *m*. Additionally, since *D* is non-trivial, the elements in that column must be equal to (*m*, *n*) with *n* arbitrarily high. Therefore sup *D*=⊤.

We use this to show that *U _{f}* is Scott-open. It is clearly upwards-closed. We consider any non-trivial directed family

*D*as above. We do not need to require that sup

*D*be in

*, because sup*

*U*_{f}*D*=⊤, which is in

*U*anyway.

_{f}*D*contains pairs (

*m*,

*n*) where

*m*is fixed and

*n*can be made as large as we wish; in particular, such that

*n*≥

*f*(

*m*), and then the pair (

*m*,

*n*) is in

*D*and in

*U*.

_{f}Conversely, let *U* be a non-empty, proper Scott-open set. (“Proper” means different from the whole space.) Since *U* is non-empty, and upwards-closed, it contains ⊤. For every natural number *m*, the directed family {(*m*, *n*) | *n* ∈ **N**} has ⊤ as supremum, which is in *U*, so (*m*, *n*) is in *U* for some *n* ∈ **N**. We pick the smallest possible *n*, and we call it *f*(*m*). This defines a function *f* from **N** to **N**, and since *U* is upwards-closed, the elements of *U* are exactly ⊤ plus all the pairs (*m*, *n*) with *m* arbitrary and *n*≥*f*(*m*). Therefore *U*=*U _{f}*. ☐

The following is easy to see, too:

**Fact F.** For any two functions *f*, *g* : **N** → **N**, *U _{f}* ⊆

*U*if and only if

_{g}*f*≥

*g*, namely if and only if

*f*(

*m*)≥

*g*(

*m*) for every

*m*∈

**N**.

Hence:

**Proposition G.** The lattice **O**(*L*_{fan}) of Scott-open subsets of *L*_{fan} is order-isomorphic to the lattice *M* obtained by adding a top and a bottom element to the poset (**N** → **N**)^{op} of functions from **N** to **N**, ordered with the *opposite* ordering ≥ (not ≤).

*M* is the letter used by Chen, Kou and Lyu [1, Example 4.13].

## Some nasty things about *L*_{fan}

We had promised that *L*_{fan} would neither be first-countable nor core-compact in its Scott topology, and this is now pretty easy.

**Proposition H.** *L*_{fan} is not first-countable in its Scott topology.

*Proof.* It suffices to show that ⊤ does not have a countable base of Scott-open neighborhoods. If it has, by Lemma E, there would be countably many functions *f _{k}*,

*k*∈

**N**, such that every open neighborhood of ⊤ of the form

*U*contains

_{f}*U*for some

_{fk}*k*∈

**N**. In other words, using Fact F, every function

*f*from

**N**to

**N**would have to be smaller than or equal to some

*f*. But this impossible: the function

_{k}*f*that maps every

*k*∈

**N**to

*f*(

_{k}*k*)+1 cannot be smaller than or equal to any

*f*, since that would imply

_{k}*f*(

*k*)=

*f*(

_{k}*k*)+1 ≤

*f*(

_{k}*k*). This is a classical instance of a diagonal argument. ☐

One can see that *L*_{fan} is not a continuous dcpo: no pair (*m*, *n*) is way-below any element, since {(*m*+1, *j*) | *j* ∈ **N**} is a directed family whose supremum (⊤) is above (*m*, *n*), but no element (*m*+1, *j*) is above (*m*, *n*). The top element ⊤ is not way-below anything either, and in fact the way-below relation on *L*_{fan} is given by *x* ≪ *y* if and only if *x*=⊥.

The situation is much worse than that: in its Scott topology, *L*_{fan} is not locally compact, and even more generally, we have the following.

**Proposition I.** *L*_{fan} is not core-compact in its Scott topology.

*Proof.* We use the definition (Definition 5.2.3 in the book), and we show that the lattice of Scott-open subsets **O**(*L*_{fan}) of *L*_{fan} is not a continuous dcpo. By Proposition G, it suffices to show that *M* = (**N** → **N**)^{op} ∪ {⊥, ⊤} is not continuous.

Let *f*, *g* be any element of (**N** → **N**)^{op}, and let us imagine that *f* is way-below *g* in *M*. In particular, *f* is below *g*, namely *f*≥*g* (remember: we use the opposite ordering ≥). For every *n* ∈ **N**, let *f _{n}* :

**N**→

**N**map every

*m*<

*n*to

*g*(

*m*), and every

*m*≥

*n*to

*f*(

*m*)+1. The supremum of the maps

*f*as

_{n}*n*varies, namely its pointwise

*infimum*, is

*g*. But no

*f*is above

_{n}*f*, namely there is no natural number

*n*such that

*f*

_{n}≤

*f*; indeed, applying both sides to

*n*, we have

*f*(

_{n}*n*)=

*f*(

*n*)+1 ≰

*f*(

*n*).

That is enough to conclude that *M* is not continuous: the only element way below an element *g* of (**N** → **N**)^{op} in *M* is ⊥, and the supremum of that lone element is not *g*. ☐

Hertling proves the same thing by showing that (∈) ≝ {(*x*, *U*) | *x* ∈ *U* ∈ **O**(*L*_{fan})} is not open in the product topology [4, Lemma 4.10]. This is equivalent, because (∈) is open in the product topology if and only if the space is core-compact (Exercise 5.2.7 in the book), but I believe the proof above is simpler.

Let me stop here for this month. We will see additional nice and nasty properties that *L*_{fan} enjoys next month.

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— Jean Goubault-Larrecq (May 20th, 2024)