2.1 Given two elements $x,y$ in a partially ordered set, we say that x is way-below y, and write $x\ll y$ , if for every increasing sequence $(z_n)_n$ whose supremum exists and is greater than or equal to y there exists $n\in \mathbb {N}$ such that $x\leq z_n$ .

In [Reference Coward, Elliott and Ivanescu9], it was shown that the Cuntz semigroup S of every $\mathrm {C}^*$ -algebra always satisfies the following properties:

1. (O1) Every increasing sequence in S has a supremum.

2. (O2) Every element in S can be written as the supremum of a $\ll $ -increasing sequence.

3. (O3) For every $x'\ll x$ and $y'\ll y$ , we have $x'+y'\ll x+y$ .

4. (O4) For every pair of increasing sequences $(x_n)_n$ and $(y_n)_n$ , we have $\sup _n x_n +\sup _n y_n = \sup _n (x_n+y_n)_n$ .

2.2 In a more abstract setting, we say that a positively ordered monoid is a $\mathrm {Cu}$ -semigroup if it satisfies (O1)–(O4).

We also say that a map between two $\mathrm {Cu}$ -semigroups is a $\mathrm {Cu}$ -morphism if it is a positively ordered monoid morphism that preserves suprema of increasing sequences and the way-below relation. As proved in [Reference Coward, Elliott and Ivanescu9], every *-homomorphism $\varphi $ between $\mathrm {C}^*$ -algebras induces a $\mathrm {Cu}$ -morphism $\operatorname {\mathrm {Cu}} (\varphi )$ between their Cuntz semigroups.

Thus, one can consider the subcategory $\operatorname {\mathrm {Cu}}$ of the category of positively ordered monoid as the category with $\mathrm {Cu}$ -semigroups and $\mathrm {Cu}$ -morphisms as its objects and morphisms, respectively. By the results from [Reference Coward, Elliott and Ivanescu9], the assignment $\operatorname {\mathrm {Cu}}\colon C^*\to \operatorname {\mathrm {Cu}}$ is functorial.

Moreover, we know from [Reference Antoine, Perera and Thiel3, Corollary 3.2.9] that the category $\operatorname {\mathrm {Cu}}$ has arbitrary inductive limits and that the functor $\operatorname {\mathrm {Cu}}$ is arbitrarily continuous (see also [Reference Coward, Elliott and Ivanescu9]).

2.3 A $\mathrm {Cu}$ -semigroup is said to have weak cancelation if $x\ll y$ whenever $x+z\ll y+z$ for some element z. Given a compact element z, that is, an element such that $z\ll z$ , weak cancelation implies that $x\leq y$ whenever $x+z\leq y+z$ . Stable rank one $\mathrm {C}^*$ -algebras have weakly cancellative Cuntz semigroups by [Reference Rørdam and Winter22, Theorem 4.3].

It was also proved in [Reference Rørdam and Winter22] that the Cuntz semigroup of a $\mathrm {C}^*$ -algebra always satisfies the following property: for every $x'\ll x\leq z$ , there exists c such that

$$\begin{align*}x'+c\leq z\leq x+c. \end{align*}$$

Moreover, it was shown in [Reference Antoine, Perera and Thiel3, Proposition 4.6] that a stronger property termed (O5) was also satisfied in the Cuntz semigroup of any $\mathrm {C}^*$ -algebra:

1. (O5) Given $x'\ll x$ , $y'\ll y$ and $x+y\leq z$ , there exists an element c such that $y'\ll c$ and $x'+c\leq z\leq x+c$ .

Under the assumption of weak cancelation, (O5) and the weaker property from [Reference Rørdam and Winter22] coincide.

We will say that a $\mathrm {Cu}$ -semigroup is countably based if there exists a countable subset such that every element in the semigroup can be written as the supremum of an increasing sequence of elements in the subset. It follows from [Reference Antoine, Perera and Santiago2, Lemma 1.3] that Cuntz semigroups of separable $\mathrm {C}^*$ -algebras are countably based.

2.4 Recall that a $\mathrm {C}^*$ -algebra A is an AI-algebra if it is *-isomorphic to an inductive limit of the form $\lim _n C[0,1]\otimes F_n$ with $F_n$ finite-dimensional for every n.

By [Reference Thomsen24], every unital commutative AI-algebra is *-isomorphic to an inductive limit of the form

with $n_{i}\leq n_{i+1}$ for each i and where all the homomorphisms are in standard form. That is to say, for every i, we have

$$ \begin{align*} \pi_{j,i+1}\varphi_{i+1,i}(f_{1},\ldots ,f_{n_{i}}) = f_{r}\sigma_{j,i}, \end{align*} $$

where $\pi _{j,i+1}\colon C([0,1])^{n_{i+1}}\rightarrow C([0,1])$ is the jth projection map, $\sigma _{j,i}\colon [0,1]\rightarrow [0,1]$ is a continuous function, and $r\leq n_{i}$ .

Thus, every unital commutative AI-algebra is isomorphic to $C(X)$ , where X is an inverse limit of (possibly increasing) finite disjoint unions of unit intervals. Conversely, it is easy to see that $C(X)$ is an AI-algebra for such an inverse limit X.

Since the space X above will always have dimension at most one, we know by [Reference Robert20, Theorem 1.1] that $\operatorname {\mathrm {Cu}} (C(X))\cong \operatorname {\mathrm {Lsc}} (X,\overline {\mathbb {N}})$ .

Moreover, note that, with the above notation, if there exists some i such that $n_{i}=n_{j}$ for every $j\geq i$ , we can write $A\cong \oplus _{k=1}^{n_{i}} C(X_{k})$ with $X_{k}$ an inverse limit of unit intervals.

3.1 Recall that a compactum is a compact metric space, and that a continuum is a connected compactum. As in [Reference Nadler17, Chapter 12], a chain in a continuum X will be a finite nonempty indexed collection $C=\{C_{1},\ldots , C_{k}\}$ of open subsets of X such that

$$\begin{align*}C_{i}\cap C_{j}\neq\emptyset \text{ if and only if }\vert i-j\vert\leq 1. \end{align*}$$

The mesh of a chain $C=\{C_{1},\ldots , C_{k}\}$ , in symbols $\operatorname {\mathrm {mesh}} (C)$ , is defined as

$$\begin{align*}\operatorname{\mathrm{mesh}} (C)=\max\{ \operatorname{\mathrm{diam}} (C_{i})\}. \end{align*}$$

A chain of mesh less than $\epsilon $ is called an $\epsilon $ -chain.

Proof Let X be the inverse limit of unit intervals. Since the unit interval is compact and connected, X is connected and, consequently, a continuum.

If X is degenerate (i.e., a point), it is clearly chainable, so we may assume otherwise.

If X is nondegenerate, then [Reference Nadler17, Theorems 12.11 and 12.19] and the comments following Theorem 12.19 of [Reference Nadler17] imply that X is chainable.

Conversely, if X is chainable, it can either be degenerate (in which case we are done) or nondegenerate. By [Reference Nadler17, Theorem 12.11], a nondegenerate chainable continuum is an inverse limit of unit intervals, so the result follows.

Proof This is a generalization of the equality $x+y=(x\vee y) + (x\wedge y)$ and is proved by induction.

Notation 4.4 Given a $\mathrm {Cu}$ -semigroup S and an element $y\in S$ , we write $\infty y:=\sup _n ny$ . Furthermore, if S has a greatest element, we denote it by $\infty $ .

4.7 Note that an $\operatorname {\mathrm {Lsc}}$ -like $\mathrm {Cu}$ -semigroup S has no maximal idempotent elements $x<\infty $ if and only if $S= \{ 0\}$ . Indeed, if S has no maximal idempotents, then every idempotent z satisfies $z\leq 0$ by condition (C1). This implies that every idempotent element is zero and, since for every $s\in S$ the element $\infty s$ is idempotent and $\infty s\geq s$ , we get $s=0$ for every s, as desired. The converse is trivial.

Similarly, an element $s\in S$ satisfies $\infty s=\infty $ if and only if there are no maximal idempotents $x<\infty $ with $\infty s\leq x$ .

Proof Write y as $y=\sup _{r} y_{r}$ , where each $y_{r}$ is a finite ordered sum of elements in $\downarrow e$ (see Lemma 4.2 and Remark 4.3).

As $y_{r}\leq ne$ for each r and $\downarrow e$ is topological, each $y_{r}$ has at most n nonzero summands.

Furthermore, since $y_{r}\leq y_{r+1}$ , the kth summands of $(y_{r})_{r}$ form an increasing sequence for every k.

Taking their suprema, this shows that y is an ordered sum of at most n elements in $\downarrow e$ .

Proof Take any $s'\ll s$ , and let $y^{\prime }_{i}\ll y_{i}$ be such that $y^{\prime }_{1}+\cdots +y^{\prime }_{n}\geq s'$ . Since $\infty e\geq y_{1}+\cdots + y_{n}\gg y^{\prime }_{1}+\cdots +y^{\prime }_{n}$ , there exists some $k\in \mathbb {N}$ such that $ke\geq y^{\prime }_{1}+\cdots +y^{\prime }_{n}$ .

By Lemma 4.8, each $y^{\prime }_{i}$ can be written as an ordered sum of at most k elements below e. That is, we can write $y^{\prime }_{i}=\sum _{j=1}^{k} y_{j,i}$ with $e\geq y_{j,i}\geq y_{j+1,i}$ for each $j<k$ .

Applying Lemma 4.2 iteratively, we have that $y^{\prime }_{1}+\cdots +y^{\prime }_{n}$ can be written as a finite ordered sum of elements below e, and that the greatest summand of the sum is $y_{1,1}\vee \cdots \vee y_{1,n}$ .

Since the order in $\downarrow e$ is topological, we get $y_{1,1}\vee \cdots \vee y_{1,n}\geq s'$ . It follows that

$$\begin{align*}s'\leq y_{1,1}\vee\cdots \vee y_{1,n}\leq y^{\prime}_{1}\vee\cdots\vee y^{\prime}_{n}\leq y_{1}\vee\cdots \vee y_{n}. \end{align*}$$

Since this holds for every $s'\ll s$ , we have $s\leq y_{1}\vee \cdots \vee y_{n}$ as required.

Proof By Lemma 4.8 and condition (C2) in Definition 4.3, x can be written as the supremum of finite ordered sums of elements in $\downarrow e$ . Moreover, using that the order in $\downarrow e$ is topological, it follows that the sequence formed by the greatest element of each ordered sum is increasing. Let $x'$ be the supremum of such a sequence.

Then, it is clear that $\infty x=\infty x'$ and $x\wedge e=x'$ . Therefore, we have

$$\begin{align*}\infty x=\infty x' = \infty (x\wedge e), \,\,\,\text{ and }\,\,\, (\infty x)\wedge e = (\infty x')\wedge e = x'=x\wedge e, \end{align*}$$

as desired.

Proof If there are no maximal idempotent elements $x<\infty $ , we know that $S=\{ 0\}$ , so we may assume otherwise.

We claim that the maximal idempotent elements $x<\infty $ are precisely the elements $\infty s$ with $s<e$ maximal.

To see this, let $x<\infty $ be a maximal idempotent and take $s=x\wedge e$ . By Lemma 4.10, we have $\infty s= \infty (x\wedge e)=\infty x=x$ . Let us check that s is maximal, and thus take $t\in S$ with $s\leq t\leq e$ .

By maximality of x, we either have $\infty t=\infty $ or $\infty t=x$ . Using Lemma 4.10 once again, it follows that we have $t= (\infty t)\wedge e=\infty \wedge e = e$ or $t=x\wedge e=s$ , as required.

Conversely, let $s<e$ be maximal and consider the element $\infty s=\sup _{n} ns$ . Let $(s_k)_k$ be a $\ll $ -increasing sequence with supremum s.

Given an idempotent element x such that $\infty s\leq x\leq \infty $ , we know by Lemma 4.8 that there exists a $\ll $ -increasing sequence $(x_{m})_m$ with supremum x such that each $x_m$ can be written as a finite increasing sum of elements in $\downarrow e$ .

Let $(m_k)_k$ be an increasing sequence of integers such that $ks_{k}\leq x_{m_{k}}$ for every k.

Using that $\downarrow e$ is topological, each $s_{k}$ is less than or equal to each of the first k summands of $x_{m_{k}}$ . As in the proof of Lemma 4.10, note that the largest summands of each sum form an increasing sequence of elements below e. Letting $x'$ be the supremum of this sequence, we get $s\leq x'\leq e$ .

By maximality of s, we either have $s=x'$ or $x'=e$ . Thus, it follows from Lemma 4.10 that $\infty s= x$ or $x=\infty x'=\infty $ . This proves that $\infty s$ is a maximal idempotent, as desired.

A similar argument shows that for any $y,z\leq e$ we have $y\leq z\leq e$ if and only if $\infty y\leq \infty z$ . Consequently, if $z\leq s$ for every maximal element $s<e$ such that $y\leq s$ , we know that $\infty z\leq \infty y$ . Taking the infimum with e, one gets $z=(\infty z)\wedge e\leq (\infty y)\wedge e=y$ , as required.

Proof By definition, we have to see that for every subset $T\subseteq \downarrow e$ , one has $s\wedge (\vee _{t\in T} t)=\vee _{t\in T}(s\wedge t)$ for every $s\in \downarrow e$ . Thus, let $x<e$ be maximal with $x\geq \vee _{t\in T}(s\wedge t)$ , which happens if and only if $x\geq s\wedge t$ for every $t\in T$ . Since $x\neq e$ and $x\vee x\geq x\vee (s\wedge t)=(x\vee s)\wedge (x\vee t)$ , we either have $x\vee s=x$ or $x\vee t=x$ (since otherwise both of these unions would be equal to e and then $x\geq e$ , a contradiction).

If $x= x\vee s \geq s$ , we have $x\geq s\wedge (\vee _{t\in T} t)$ . Else, $x\geq t$ for each $t\in T$ , so we also get $x\geq s\wedge (\vee _{t\in T} t)$ .

By Lemma 4.11, this shows that $s\wedge (\vee _{t\in T} t)\leq \vee _{t\in T}(s\wedge t)$ .

The other inequality holds in any lattice.

Proof We will construct our almost complement $y\setminus z$ in three steps.

Step 1. Let e be the least order unit of S and assume that $y\leq z\leq e$ . Then, consider the subset $T:=\{ x\in S\mid y+x\leq z \}$ and, since arbitrary suprema exist in S, we define

$$\begin{align*}y\setminus z :=\vee \{ x\in S\mid y+x\leq z \} = \vee_{x\in T}x. \end{align*}$$

Furthermore, note that, for $x\in T$ , $y+x=(y\vee x)+(y\wedge x)$ and, as $z\leq e$ and $\downarrow e$ is topological, we have $y\vee x\leq z$ and $y\wedge x=0$ .

Using that $\downarrow e$ is a complete Heyting algebra, we get $(\vee _{x\in T}x)\vee y=\vee _{x\in T} (x\vee y)\leq z$ and $(\vee _{x\in T}x)\wedge y=\vee _{x\in T}(x\wedge y)=0$ . Consequently, we get

$$\begin{align*}y+(y\setminus z)=y\vee (y\setminus z)+y\wedge (y\setminus z)=y\vee (y\setminus z)\leq z. \end{align*}$$

This shows that $x\leq y\setminus z$ if and only if $y+x\leq z$ .

Step 2. Now, given any pair of elements $y\leq z\leq ne$ for some $n\in \mathbb {N}$ , write them as finite ordered sums of elements in $\downarrow e$ , say $y=\sum _{i=1}^{n} y_{i}$ and $z=\sum _{j=1}^{n} z_{j}$ .

Since $\downarrow e$ is topological, $y_{i}\leq z_{i}\leq e$ for every i, so we can define $y\setminus z:=\sum _{i=1}^{n} y_{i}\setminus z_{i}$ .

Note that, given any element x such that $y+x\leq z\leq ne$ , we have $x\leq ne$ and so, by Lemma 4.8, x can be written as a finite ordered sum $\sum _{i=1}^{n} x_{i}$ with $x_i\leq e$ for every i. Moreover, we also have

$$\begin{align*}\sum_{i=1}^{n} y_{i}\vee x_{i}\leq y+x\leq \sum_{i=1}^{n} z_{i} \end{align*}$$

and so $y_{i}\vee x_{i}\leq z_{i}$ , which happens if and only if $x_{i}\leq y_{i}\setminus z_{i}$ .

This shows that $x+y\leq z\leq ne$ for some n if and only if $x\leq y\setminus z$ .

Step 3. Given $y\leq z\wedge ne$ for some n, we have that $y\leq z\wedge me$ for every $m\geq n$ .

Thus, we can consider the element $y\setminus (z\wedge me)$ . Furthermore, it is easy to check that $y\setminus (z\wedge me)\leq y\setminus (z\wedge (m+1)e)$ for every m. We define $y\setminus z:=\sup _{m} y\setminus (z\wedge me)$ , which has the required property.

Finally, to see that $x+ (y\vee z)=(x+y)\vee (x+z)$ for any given $x,y,z$ , note that “ $\geq $ ” is clear. To prove “ $\leq $ ,” let $x'\ll x$ and let s be such that $x'+(y\vee z)\leq s$ .

Since $x'\ll x\leq \infty e$ , there exists some n with $x'\leq ne$ , so we can consider the element $x'\setminus s$ .

Thus, we know that $x'+(y\vee z)\leq s$ holds if and only if $y\vee z\leq x'\setminus s$ , which in turn holds if and only if $x'+y,x'+z\leq s$ . Consequently, we get

$$\begin{align*}x'+(y\vee z)=(x'+y) \vee (x'+z). \end{align*}$$

Since the equality holds for every $x'\ll x$ , it also holds for x.

Proof First, note that $y\leq x\vee y\leq x\vee z$ . Indeed, since $x,y,z\leq e$ and we know that $x+y=(x\vee y)+(x\wedge y)$ and $x+z=(x\vee z) +(x\wedge z)$ , we have that $(x\vee y)+ (x\wedge y)\leq (x\vee z) +(x\wedge z)$ .

Since the right- and left-hand sides of the previous inequality are ordered sums of elements below e, we can use that the order in $\downarrow e$ is topological to get $x\vee y\leq x\vee z$ , as desired.

Therefore, the sum $(z\vee x)+y$ is ordered, as we have $y\leq z\vee x$ .

Using Lemma 4.14 at the first and third steps, the inequality $x+y\leq x+z$ at the second step, and that S is distributively lattice ordered at the last step, one obtains

$$\begin{align*}\begin{aligned} (z\vee x) +y &= (y+z)\vee (y+x) \leq (y+z)\vee (z+x) = z+(y\vee x)\\ &=(z\vee y\vee x) + z\wedge (y\vee x). \end{aligned} \end{align*}$$

Using once again that the order in $\downarrow e$ is topological, it follows that $y\leq z\wedge (y\vee x)\leq z$ , as required.

Proof Write $y'=\sup _{n} y^{\prime }_{n}$ and $z'=\sup _{n} z^{\prime }_{n}$ with $(y^{\prime }_{n})_{n}$ and $(z^{\prime }_{n})_{n}$ $\ll $ -increasing. Since $y+z\ll y' +z'$ , we have $y+z\leq y^{\prime }_{n}+z^{\prime }_{n}$ for some n.

Moreover, as we have that $z^{\prime }_{n}\ll z'\leq y'$ , there exists a k such that $z^{\prime }_{n}\leq y^{\prime }_{k}$ . This implies that

$$\begin{align*}y+z\leq y^{\prime}_{n}+z^{\prime}_{n}\leq y^{\prime}_{\max\{n,k\} }+z^{\prime}_{n}, \end{align*}$$

and, since the order is topological, we obtain $y\leq y^{\prime }_{\max \{n,k\} }\ll y'$ and $z\leq z^{\prime }_{n}\ll z'$ .

Proof First, let us assume that $f\ll g$ . Fix $n\in \mathbb {N}$ and consider an increasing sequence $(h_{k})_{k}$ such that

$$\begin{align*}\chi_{\{ g\geq n\}}\leq \sup_{k} h_{k}, \end{align*}$$

which happens if and only if $\chi _{\{ g\geq n\}}\leq \chi _{\cup _{k}\text {supp}(h_{k})}$ .

Define the increasing sequence of functions

$$\begin{align*}G_{k}:= (n-1) + \chi_{\text{supp}(h_{k})} \sum_{r=0}^{\infty }\chi_{\{ g\geq n+r\}}, \end{align*}$$

and note that $g\leq \sup _{k}G_{k}$ .

Since $f\ll g$ , we get that $f\leq G_{k}$ for some k and, consequently,

$$\begin{align*}\{f\geq n\} \subseteq \{G_{k}\geq n\}=\text{supp}(h_{k})\cap \{g\geq n\}\subseteq \text{supp}(h_{k}). \end{align*}$$

This in turn implies $\chi _{\{f\geq n\}}\leq \chi _{\text {supp}(h_{k})}\leq h_{k}$ , so it follows that $\chi _{\{f\geq n\}}\ll \chi _{\{g\geq n\}}$ . That $\sup (f)<\infty $ is clear.

Conversely, if $\sup (f)<\infty $ , we know that $f=\sum _{i=1}^{m} \chi _{\{ f\geq i \}}$ for some $m<\infty $ . Furthermore, given an increasing sequence $(h_{k})_{k}$ with $g\leq \sup _{k}h_{k}$ , it follows that

$$\begin{align*}\{g\geq n\}\subseteq \bigcup_{k} \{h_{k}\geq n\}, \end{align*}$$

so $\chi _{\{ g\geq n\}}\leq \sup _{k}\chi _{\{h_{k}\geq n\}}$ . Since $\chi _{\{ f\geq n\}}\ll \chi _{\{ g\geq n\}}$ for every n, we get that for each i there exists an integer $k_{i}$ with

$$\begin{align*}\chi_{\{ f\geq i\}}\leq\chi_{\{ h_{k_{i}}\geq i\}}. \end{align*}$$

Taking $k=\max _{i=1,\ldots ,m}\{k_{i}\}$ , we have

$$\begin{align*}f\leq \sum_{i=1}^{m} \chi_{\{ h_{k}\geq i\}}\leq h_{k}, \end{align*}$$

as desired.

Proof Axioms (O1) and (O4) are always satisfied in $\operatorname {\mathrm {Lsc}} (X,\overline {\mathbb {N}} )$ , so we only need to prove (O2) and (O3).

Let $U,V$ be open subsets of X, and note that $\chi _U\ll \chi _V$ if and only if U is compactly contained in V. Indeed, if $\chi _U\ll \chi _V$ , we can write V as a countable increasing union of open sets $V_n$ such that $V_n$ is compactly contained in $V_{n+1}$ for every n. Thus, one gets $\chi _U\ll \sup _n \chi _{V_n}$ , which implies that U is contained in $V_n$ for some n. Conversely, if $\overline {U}\subseteq V$ and $(W_n)_n$ is an increasing sequence of open sets with $V=\cup _n W_n$ , it is clear that $U\subseteq W_n$ for some n. This shows $\chi _U\ll \chi _V$ , as required.

In particular, it follows that every indicator function can be written as the supremum of a $\ll $ -increasing sequence. Since every element in $S=\operatorname {\mathrm {Lsc}} (X,\overline {\mathbb {N}} )$ is the supremum of finite sums of indicator functions, one can check that S satisfies (O2).

Now, let $f'\ll f$ and $g'\ll g$ in S, which by Lemma 4.18 implies that $\sup (f),\sup (g)<m\leq \infty $ and $\{ f'\geq n \}, \{ g'\geq n\}$ are compactly contained in $\{ f\geq n \}, \{ g\geq n\}$ , respectively. Thus, we have

$$\begin{align*}\bigcup_{k=0}^{m}\overline{(\{ f'\geq k \}\cap\{ g'\geq n-k \})} \subseteq \bigcup_{k=0}^{m}(\{ f\geq k \}\cap\{ g\geq n-k \}) \end{align*}$$

for every $n\leq \sup (f)+\sup (g)$ , where note that the left-hand side is equal to $\overline {\{ f'+g'\geq n\}}$ and the right-hand side is contained in $\{f+g\geq n\}$ . By Lemma 4.18, we have $f'+g'\ll f+g$ , which shows that S satisfies (O3).

5.1 Let S be an $\operatorname {\mathrm {Lsc}}$ -like $\mathrm {Cu}$ -semigroup with least order unit e. The topological space $X_S$ of S is defined as

$$\begin{align*}X_{S}:=\{ x\in S\mid x<e \text{ maximal} \}, \end{align*}$$

with closed subsets

$$\begin{align*}C_{y}:=\{ x\in X_{S}\mid x\geq y\}, \quad y\leq e. \end{align*}$$

Proof First, note that $C_{0}=X_S$ and that $C_{e}=\emptyset $ . Moreover, $C_{x}=\{ x\}$ for every $x\in X_S$ . Thus, our topology is $T_1$ .

To see that arbitrary intersections of $C_{y}$ ’s are of the form $C_{z}$ for some $z\leq e$ , simply note that

$$\begin{align*}\bigcap_{i} C_{y_{i}} = C_{\vee_{i}y_{i}}. \end{align*}$$

Furthermore, one also has that

$$\begin{align*}\bigcup_{i=1}^{n} C_{y_{i}} = C_{\wedge_{i=1}^{n}y_{i}}. \end{align*}$$

Indeed, given $x\in X_S$ with $x\geq \wedge _{i=1}^{n}y_{i}$ , we have

$$\begin{align*}x= x\vee x \geq x\vee (\wedge_{i=1}^{n}y_{i})=(x\vee y_{1})\wedge \cdots\wedge (x\vee y_{n}). \end{align*}$$

Since x is maximal, for each i, we either have $x\vee y_{i}=x$ or $x\vee y_{i}=e$ .

However, note that the previous inequality implies that we cannot have $x\vee y_{i}=e$ for every i, so $x=x\vee y_{j}\geq y_{j}$ for some j.

The other inclusion is clear.

Proof To see (i), recall that, by definition, $C_{y}\subseteq C_{z}$ if and only if $x\geq z$ for every $x\geq y$ with $x<e$ maximal. Using Lemma 4.11, we see that this is equivalent to $y\geq z$ .

For (ii), let $y\leq e$ and take $x<e$ be maximal. Thus, we have $x\leq y\vee x\leq e$ . Since x is maximal, we either have $x=y\vee x\geq y$ (i.e., $x\in C_{y}$ ) or $y\vee x=e$ . Thus, $U_{y}= \{ x\in X_S: y\vee x=e \}$ .

To prove (iii), let us assume, for the sake of contradiction, that $y\wedge z\neq 0$ . Then, $U_{y\wedge z}$ is nonempty and we can consider a maximal element $x\in U_{y\wedge z}$ .

By (ii), we have $x\vee (y\wedge z)=e$ . Thus, one gets $x\vee y=e$ and, consequently, $x\geq z$ from our assumption that $U_y \subseteq C_z$ . However, we also have

$$\begin{align*}e= x\vee (y\wedge z) = (x\vee y)\wedge (x\vee z) = e\wedge x=x, \end{align*}$$

which is a contradiction, as required.

Let us now prove (iv) and, consequently, (v). First, note that, if $y\vee x=e$ , we have

$$\begin{align*}y+x=(y\vee x) + (y\wedge x)\geq e\geq y\setminus e +y. \end{align*}$$

Canceling y (see Lemma 4.15), we have that $x\geq y\setminus e$ . This shows $U_{y}\subseteq C_{y\setminus e}$ .

Conversely, let z be such that $U_{y}\subseteq C_{z}$ . By (iii), we know that this implies $y\wedge z=0$ and, consequently, $y+z\leq e$ . Thus, $z\leq y\setminus e$ or, equivalently, $C_{y\setminus e}\subseteq C_{z}$ .

Finally, for (vi), assume first that $C_{y}\subseteq U_{z}$ . Furthermore, assume for the sake of contradiction that $y\vee z\neq e$ . Then, there exists $x\in X_{S}$ with $x\geq y\vee z$ . This implies $x\geq y$ and, consequently, $x\vee z=e$ from $C_y\subseteq U_z$ . However, we have

$$\begin{align*}x = x\vee y\vee z \geq x\vee z=e, \end{align*}$$

which is a contradiction.

Conversely, if $y\vee z=e$ , take $x\geq y$ , which implies that $x\vee z\geq y\vee z=e$ . In particular, $x\in U_{z}$ .

Proof Let $x,y<e$ be two elements such that $C_{x}\cap C_{y}=C_{e}=\emptyset $ . In terms of the elements in S, this is equivalent to $x+y\geq e\gg e$ . Then, we can take $x'\ll x$ and $y'\ll y$ such that $x'+y'\gg e$ .

Using (O5), there exist $c,d\leq e$ such that

$$\begin{align*}x'+c\leq e\leq x+c, \quad y'+d\leq e\leq y+d. \end{align*}$$

Consequently, we also have $x'+c+y'+d\leq e+e$ with $x'+y'\gg e$ . Since every $\operatorname {\mathrm {Lsc}}$ -like $\mathrm {Cu}$ -semigroup has weak cancelation, it follows that $c+d\leq e$ .

Since our order is topological, we get $c\wedge d=0$ , $x\vee c= e$ , and $y\vee d=e$ .

Using the properties listed in Proposition 5.3, the previous inequalities imply that $C_{x}\subseteq U_{c}$ , $C_{y}\subseteq U_{d}$ , and $U_{c}\cap U_{d}= U_{c\wedge d}=\emptyset $ . Thus, $X_{S}$ is normal.

Now, let $y,z\leq e$ and assume that $\overline {U_{y}}\subseteq U_{z}$ , which by (iv) in Proposition 5.3 happens if and only if $(y\setminus e)\vee z=e$ . Furthermore, since $y\setminus e + z\geq e$ , we have $y\setminus e + z\geq e\gg e\geq y\setminus e + y$ . As elements below e have cancelation (see Lemma 4.15), one gets $z\gg y$ .

Conversely, assume that $e\geq z\gg y$ . Then, by (O5), there exists an element x such that $y+x\leq e$ and $e\leq z+x$ .

From the first inequality, it follows that $x\leq y\setminus e$ , so $e\leq z+x\leq z+y\setminus e$ as required.

Proof Let $e\in S$ be the least order unit of S. To show (i), assume first that S is countably based with a countable basis B, and let $\sum '(\downarrow e)$ denote the set of finite sums of elements in $\downarrow e$ . Naturally, the set $B':=B\cap \sum '(\downarrow e)$ is also a countable basis for S.

Given an open set $U_{y}$ with $y\leq e$ , write $y=\sup _{n} y_{n}$ with $y_{n}\in B'$ . We have $\cup _{n} U_{y_{n}} = U_{y}$ , and so $X_{S}$ is second countable.

Conversely, assume that $X_{S}$ is second countable with basis $C=\{U_{z_{n}}\}_{n}$ . Note that the family $C'$ consisting of all the finite unions of sets in C is also countable.

Then, any open subset $U_{y}$ can be written as the countable union of increasing open subsets $U_{z^{\prime }_{n}}$ of $C'$ . We know that this is equivalent to $y=\sup _{n}z^{\prime }_{n}$ . This implies that the set $\sum ' \{ z^{\prime }_{n}\}_{n}$ of finite sums from $\{ z^{\prime }_n\}_n$ is a countable basis for $\sum ' (\downarrow e)$ and, since $\sum ' (\downarrow e)$ is dense in S, $\sum ' \{ z^{\prime }_{n}\}_{n}$ is a countable basis for S.

To prove (ii), note that it is easy to check that $e\in S$ is compact if and only if $X_{S}$ is countably compact. Therefore, we are left to prove that, if there exists a compact order unit in S, then e must also be compact.

To see this, let p be a compact order unit in S, which implies that $p\leq ne$ for some $n\in \mathbb {N}$ . Since we know that p can be written as a finite ordered sum of elements below e, there exist $m\in \mathbb {N} $ and elements $q_{m}\leq \cdots \leq q_{1}\leq e$ such that $p=q_{1}+\cdots +q_{m}$ .

By weak cancelation applied to $q_1+\cdots +q_m\ll q_1+\cdots +q_m$ , the element $q_{1}$ is compact and satisfies $\infty q_{1}=\infty p=\infty $ . Thus, $q_{1}$ is a compact order unit with $e\geq q_{1}$ . By minimality of e, one gets $e=q_{1}$ compact as required.

Let us now show (iii). First, assume that S is normal, and let z be the associated order unit. Let $C_x , C_y$ be closed subsets of $X_S$ with $x,y\leq e$ , and recall that $C_{x}, C_{y}$ are disjoint if and only if $x\vee y=e$ .

Since e is an order unit, we have $\infty x + \infty y= \infty \geq z$ . Thus, we get $s,t$ such that $\infty x +s\geq z$ , $\infty y +t\geq z$ , and $s\wedge t=0$ . As $z\geq e$ , we know by Corollary 4.9 that $\infty x\vee s\geq e$ and $\infty y\vee t\geq e$ .

Since $x,y\leq e$ , taking the infimum with e and using Lemma 4.10, we have $e=(\infty x\wedge e)\vee (s\wedge e)=x\vee (s\wedge e)$ , $y\vee (t\wedge e)=e$ , and $(s\wedge e)\wedge (t\wedge e)=0$ . By (vi) in Proposition 5.3, it follows that $C_{x}\subseteq U_{s\wedge e}$ , $C_{y}\subseteq U_{t\wedge e}$ , and $U_{s\wedge e}\cap U_{t\wedge e}=\emptyset $ . This implies that $X_{S}$ is normal.

Conversely, if $X_{S}$ is normal, it is easy to see that S is normal by setting $z=e$ in the definition of normality.

To prove (iv), we have that $X_{S}$ is second countable and normal by (i) and (iii), and that $C_{x}=\{x \}$ for any $x\in X_{S}$ . Thus, points are closed in our topology, so $X_{S}$ is Hausdorff. We can now use Urysohn’s metrization theorem to conclude that $X_{S}$ is metric (see, e.g., [Reference Munkres16, Theorem 34.1]).

For (v), note that $e\ll e$ . Thus, Lemma 5.5 implies that $X_S$ is normal. Following the arguments above, one gets that $X_S$ is metric. Moreover, we also know that $X_S$ is second countable and countably compact by (i) and (ii) above. Thus, $X_S$ is compact.

Proof Assume $y\ll z$ , and let $(f_{n})_{n}$ be an increasing sequence in $\operatorname {\mathrm {Lsc}} (X_{S},\overline {\mathbb {N}} )$ such that $\chi _{U_{z}}\leq \sup _{n} f_{n}$ . In particular, note that this holds if and only if $\chi _{U_{z}}\leq \chi _{\cup _{n} \text {supp}(f_{n})}$ or, equivalently, if

$$\begin{align*}\bigcap_{n}(X_S\setminus\text{supp}(f_{n})) \subseteq C_{z}.\\[-9pt] \end{align*}$$

Denote by $z_{n}$ the elements in $\downarrow e$ with $C_{z_{n}}=X_S\setminus \text {supp}(f_{n})$ . Given that $\text {supp}(f_{n})\subseteq \text {supp}(f_{n+1})$ , (i) in Proposition 5.3 implies that $z_{n}$ is increasing.

Using the proof of Lemma 5.2 in the first step, we can rewrite the previous inclusion as

$$\begin{align*}C_{\sup_{n}(z_{n})} = \bigcap_{n} C_{z_n}\subseteq C_{z}.\\[-9pt] \end{align*}$$

Applying (i) in Proposition 5.3 once again, one gets $z\leq \sup _{n}(z_{n})$ and, consequently, $y\leq z_{n}$ for some n. This implies that $U_{y}\subseteq U_{z_{n}}$ or, equivalently, that $\chi _{U_{y}}\leq \chi _{U_{z_{n}}}= \chi _{\text {supp}(f_{n})}\leq f_{n}$ . This shows $\chi _{U_{y}}\ll \chi _{U_{z}}$ .

Now, let $y,z\leq e$ be such that $\chi _{U_{y}}\ll \chi _{U_{z}}$ , and consider an increasing sequence $(h_{n})_{n}$ in S such that $z\leq \sup _{n}(h_{n})$ . Note that, by taking $z\wedge h_{n}$ instead of $h_{n}$ , we can assume $h_{n}\leq e$ for every n.

Applying again (the proof of) Lemma 5.2, one gets

$$\begin{align*}\bigcap_{n} C_{h_{n}}=C_{\sup_{n}(h_{n})}\subseteq C_{z}, \end{align*}$$

and, consequently, we have $\sup _{n}\chi _{U_{h_{n}}}\geq \chi _{U_{z}}$ since $\chi _{U_{y}}\ll \chi _{U_{z}}$ , there exists some n with $\chi _{U_{y}}\leq \chi _{U_{h_{n}}}$ ; i.e., $C_{h_n}\subseteq C_y$ .

Using (i) in Proposition 5.3 one last time, one sees that $y\leq h_{n}$ as required.

Proof Take $f\in \operatorname {\mathrm {Lsc}} (X_{S},\overline {\mathbb {N}} )$ , and let $(y_{i})_{i}$ be the sequence on $\downarrow e$ such that

$$\begin{align*}\{f\geq i\} = U_{y_{i}}, \end{align*}$$

where recall that the sequence is decreasing as a consequence of (i) in Proposition 5.3.

Since S satisfies (O2), we have $y_{i}=\sup _{n} y_{i,n}$ with $y_{i,n}\ll y_{i,n+1}$ for every n.

For every fixed k, we have

$$\begin{align*}y_{1}\geq\cdots\geq y_{k}\gg y_{k,n} \text{ for all }n. \end{align*}$$

Thus, for every i, one can choose inductively $n_{i,k}$ with $k\geq i$ such that

$$\begin{align*}y_{i,n_{i,k}}\ll y_{i,n_{i,k+1}}, \,\,\,\text{ and }\,\,\, y_{1,n_{1,k}}\geq\cdots\geq y_{k,n_{k,k}}. \end{align*}$$

Indeed, we begin by setting $n_{1,1}=1$ (i.e., $y_{1,n_{1,1}}=y_{1,1}$ ). Then, assuming that we have defined $n_{i,k}$ for every $i,k\leq m-1$ (and $k\geq i$ ) for some fixed m, we set $n_{m,m}=1$ , that is, $y_{m,n_{m,m}}=y_{m,1}$ . We then set $n_{m-1,m}$ large enough so that $y_{m-1,n_{m-1,m}}\geq y_{m,n_{m,m}}$ and $n_{m-1,m}\geq n_{m-1,m-1}$ . Similarly, we set $n_{m-2,m}\geq n_{m-2,m-1}$ such that $y_{m-2,n_{m-2,m}}\geq y_{m-1,n_{m-1,m}}$ and define $n_{i,m}$ for every $i\leq m-2$ in the same fashion.

Now, consider the sums $f_{k}=\sum _{i=1}^{k} \chi _{U_{y_{i,n_{i,k}}}}$ , which are ordered by construction. Thus, one has $U_{y_{i,n_{i,k}}}=\{ f_k\geq i\}$ for every i. Since $y_{i,n_{i,k}}\ll y_{i,n_{i,k+1}}$ for every i, it follows from Lemmas 4.18 and 5.7 that

$$\begin{align*}f_{k}=\sum_{i=1}^{k} \chi_{U_{y_{i,n_{i,k}}}}\ll \sum_{i=1}^{k} \chi_{U_{y_{i,n_{i,k+1}}}} \leq f_{k+1}. \end{align*}$$

It is now easy to check that $\sup _{k} f_{k}=f$ .

Proof Note that the semigroup $\operatorname {\mathrm {Lsc}} (X_S ,\overline {\mathbb {N}} )$ always satisfies (O1) and (O4). Moreover, we already know that (O2) is also satisfied by Proposition 5.8. Thus, we are left to prove (O3).

Let $f\ll f'$ and $g\ll g'$ . By Lemma 4.18, this implies that

$$\begin{align*}\chi_{\{ f\geq i\}} \ll \chi_{\{ f'\geq i\}}\,\,\,\text{ and }\,\,\, \chi_{\{ g\geq i\}} \ll \chi_{\{ g'\geq i\}} \end{align*}$$

for every i. We also know that there exists $m<\infty $ such that $\sup (f),\sup (g)\leq m$ .

Let $y_{i},y^{\prime }_{i},z_{i}$ , and $z^{\prime }_{i}$ be elements in $\downarrow e$ such that

$$\begin{align*}U_{y_{i}}= \{ f\geq i\},\quad U_{y^{\prime}_{i}}= \{ f'\geq i\},\quad U_{z_{i}}= \{ g\geq i\},\,\,\,\text{ and }\,\,\, U_{z^{\prime}_{i}}= \{ g'\geq i\}. \end{align*}$$

By Lemma 5.7, we have $y_{i}\ll y^{\prime }_{i}$ and $z_{i}\ll z^{\prime }_{i}$ for every i and, since S satisfies (O3), one gets

$$\begin{align*}\sum_{i=1}^{m} (y_{i}+z_{i}) \ll \sum_{i=1}^{m} (y^{\prime}_{i}+z^{\prime}_{i}). \end{align*}$$

By Lemma 4.2, these sums can be rewritten as

$$\begin{align*}\sum_{i=1}^{2m} \vee_{j=0}^{m}(y_{j}\wedge z_{i-j}) \ll \sum_{i=1}^{2m} \vee_{j=0}^{m}(y^{\prime}_{j}\wedge z^{\prime}_{i-j}) \end{align*}$$

and, since both the right- and left-hand sides of the previous inequality are ordered, we can use Lemma 4.17 to obtain

$$\begin{align*}\vee_{j=0}^{m}(y_{j}\wedge z_{i-j}) \ll \vee_{j=0}^{m}(y^{\prime}_{j}\wedge z^{\prime}_{i-j}) \end{align*}$$

for every i.

Note that

$$\begin{align*}\{ f+g\geq i \} = \bigcup_{j=0}^{m}(\{ f\geq j \}\cap\{ g\geq i-j \}), \end{align*}$$

so, by the equalities in the proof of Lemma 5.2, one gets

$$\begin{align*}X_S\setminus\{ f+g\geq i \} &= \bigcap_{j=0}^{m} ((X_S\setminus \{ f\geq j \})\cup (X_S\setminus\{ g\geq i-j \})) = \bigcap_{j=0}^{m} (C_{y_{j}\wedge z_{i-j}})\\ &= C_{\vee_{j=0}^{m}(y_{j}\wedge z_{i-j})} \end{align*}$$

and, consequently, $\chi _{\{ f+g\geq i \}} = \chi _{U_{\vee _{j=0}^{m}(y_{j}\wedge z_{i-j})}}$ .

The same argument also shows that $\chi _{\{ f'+g'\geq i \}}\geq \chi _{ U_{\vee _{j=0}^{m}(y^{\prime }_{j}\wedge z^{\prime }_{i-j})}}$ .

By using Lemma 5.7, we get $\chi _{\{ f+g\geq i \}}\ll \chi _{\{ f'+g'\geq i \}}$ for every i. Lemma 4.18 then implies that $f+g\ll f'+g'$ , as desired.

Proof It is easy to see that $\varphi $ is a global order embedding.

To prove surjectivity, let $h\in H$ . Since $\varphi (S)$ is a basis for H, we can write $h=\sup _{n} \varphi (s_{n})$ for some $s_{n}\in S$ . Furthermore, as we know that $\varphi $ is an order embedding, the sequence $(s_{n})_{n}$ is increasing in S, so $\sup _{n} \varphi (s_{n})=\varphi (s)$ for $s=\sup _{n} s_{n}$ .

Proof Let $f=\sum _{i=1}^{m} \chi _{U_{z_{i}}}$ as above and take $g=\sum _{j=1}^{n} \chi _{U_{y_{j}}}$ . We will first prove by induction on m that $\varphi ' (f+g)=\varphi '(f)+\varphi '(g)$ .

For $m=1$ , f is simply $\chi _{U_{z}}$ for some open subset $U_{z}$ . Since $\operatorname {\mathrm {Lsc}} (X_{S},\overline {\mathbb {N}} )$ is distributively lattice ordered and $0\leq 0\leq \cdots \leq 0\leq f$ is an increasing sequence, we can apply Lemma 4.2 to get

$$\begin{align*}f + \sum_{j=1}^{n} \chi_{U_{y_{j}}} = \chi_{U_{z}\cup U_{y_{1}}} + \chi_{(U_{z}\cap U_{y_{1}})\cup U_{y_{2}} } + \cdots + \chi_{U_{z}\cap (\cap_{j} U_{y_{j}})}.\\[-9pt] \end{align*}$$

Applying $\varphi '$ and the equalities in the proof of Lemma 5.2 at the first step, and Lemma 4.2 at the second step, one gets

$$\begin{align*}\begin{aligned} \varphi'(f+g) &= (z\vee y_{1}) + ((z\wedge y_{1})\vee y_{2}) + \cdots + (z\wedge y_{1}\wedge\cdots\wedge y_{n})\\ &=z+(y_{1}+\cdots +y_{n}) = \varphi'(f)+ \varphi'(g). \end{aligned} \end{align*}$$

Now, fix any finite m and assume that the result has been proved for any $k\leq m-1$ . Then, using the induction hypothesis at the second step, and the case $m=1$ at the third and fourth steps, we have

$$\begin{align*}\begin{aligned} \varphi '(f+g) &= \varphi' ( (f-\chi_{U_{z_{m}}}) + (g+\chi_{U_{z_{m}}}) ) = \varphi' ( f-\chi_{U_{z_{m}}}) + \varphi' (g+\chi_{U_{z_{m}}})\\ &= \varphi' ( f-\chi_{U_{z_{m}}}) + \varphi' (g) + \varphi' (\chi_{U_{z_{m}}})= \varphi' ( f-\chi_{U_{z_{m}}}+\chi_{U_{z_{m}}}) + \varphi' (g)\\ &= \varphi' (f) + \varphi' (g), \end{aligned} \end{align*}$$

as desired.

Note that this could have also been proved using Lemma 4.2.

To see that $\varphi '$ is an order embedding, let $f\leq g$ in $\operatorname {\mathrm {Lsc}} (X_{S},\overline {\mathbb {N}} )_{fs}$ and note that $f\leq g$ if and only if $\{ f\geq i\}\subseteq \{ g\geq i\}$ for every i. Let $z_i, y_i\in S$ be such that $U_{z_i}=\{ f\geq i\}$ and $U_{y_i}=\{ g\geq i\}$

By (i) in Proposition 5.3, $U_{z_i}\subseteq U_{y_i}$ if and only if $z_{i}\leq y_{i}$ for every i. Furthermore, note that the sequences $(z_{i})_{i=1}^{m},(y_{i})_{i=1}^{m}$ are both decreasing. Since we have a topological order, $z_{i}\leq y_{i}$ for every i if and only if

$$\begin{align*}\sum_{i=1}^{m} z_{i}\leq \sum_{i=1}^{m} y_{i}, \end{align*}$$

where note that the right- and left-hand sides correspond to $\varphi '(f)$ and $\varphi '(g)$ , respectively.