2.1 (The Cuntz semigroup).

Let A be a $\mathrm {C}^*$ -algebra. Given $a, b \in A_+$ , one says that a is Cuntz subequivalent to b, written $a\precsim b$ if there exists a sequence $(v_n)_n$ in A such that $a=\lim _n v_n b v_n^*$ . Further, one says that a is Cuntz equivalent to b, written $a\sim b$ , if $a\precsim b$ and $b \precsim a$ .

The Cuntz semigroup of A, denoted by $ {\mathrm {Cu}} (A)$ , is the positively ordered monoid defined as the quotient $(A\otimes \mathcal {K})_+/{\sim }$ equipped with the order induced by $\precsim $ and the addition induced by addition of orthogonal elements. For further details, we refer to [Reference Antoine, Perera and Thiel4, Reference Ara, Perera and Toms8, Reference Gardella and Perera24].

2.2 ( $\mathrm {Cu}$ -semigroups).

Let $(P, \leq )$ be a partially ordered set. Suppose that every increasing sequence in P has a supremum. Given two elements $x,y$ in P, one says that x is way below y, denoted $x\ll y$ , if for every increasing sequence $(z_n)_n$ in P satisfying $y\leq \sup _n z_n$ , there exists some $m\in {\mathbb {N}}$ such that $x\leq z_m$ .

As defined in [Reference Coward, Elliott and Ivanescu17], a $\mathrm {Cu}$ -semigroup is a positively ordered monoid S satisfying two domain-type conditions and two compatibility conditions:

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

2. (O2) For every element x in S, there exists a sequence $(x_n)_{n}$ in S such that $x_0 \ll x_1 \ll x_2 \ll \cdots $ and $x = \sup _n x_n$ .

3. (O3) The addition is compatible with the way-below relation, that is, for every $x', x, y', y \in S$ satisfying $x' \ll x$ and $y' \ll y$ , we have $x' + y' \ll x + y$ .

4. (O4) The addition is compatible with suprema of increasing sequences, that is, for every increasing sequences $(x_n)_n$ and $(y_n)_n$ in S, we have

   $$\begin{align*}\sup _n (x_n+y_n) = \sup _nx_n + \sup _ny_n.\end{align*}$$

It follows from [Reference Coward, Elliott and Ivanescu17] that the Cuntz semigroup of any $\mathrm {C}^*$ -algebra always satisfies (O1)–(O4). Specifically, the Cuntz semigroup of any $\mathrm {C}^*$ -algebra is a $\mathrm {Cu}$ -semigroup.

Given a monoid morphism $\varphi $ between two $\mathrm {Cu}$ -semigroups, we say that $\varphi $ is a $\mathrm {Cu}$ -morphism if it preserves the order, suprema of increasing sequences and the way-below relation. A generalized $\mathrm {Cu}$ -morphism is a monoid map that preserves order and suprema of increasing sequences (but not necessarily the way-below relation).

The following properties, which will often be considered throughout the paper, are also satisfied in the Cuntz semigroup of any $\mathrm {C}^*$ -algebra; see [Reference Antoine, Perera and Thiel4, Proposition 4.6] and its precursor [Reference Rørdam and Winter35, Lemma 7.1] for (O5), [Reference Robert32, Proposition 5.1.1] for (O6) and [Reference Antoine, Perera, Robert and Thiel1, Proposition 2.2] for (O7).

1. (O5) For every $x, y, x', y', z\in S$ satisfying $x+y\leq z$ and $x'\ll x$ and $y'\ll y$ , there exists $c \in S$ such that $y'\ll c$ and $x'+c\leq z \leq x+c$ .

   This property is often applied with $y'=y=0$ .

2. (O6) For every $x, x', y, z\in S$ satisfying $x'\ll x\ll y+z$ , there exist $v, w \in S$ such that

   $$\begin{align*}v \leq x,y, \quad w \leq x,z, \,\,\,\text{ and }\,\,\, x'\leq v+w. \end{align*}$$

3. (O7) For every $x, x', y, y', w \in S$ satisfying $x'\ll x\leq w$ and $y'\ll y\leq w$ , there exists $z \in S$ such that $x',y'\ll z\leq w,x+y$ .

2.3 (The Global Glimm Property and nowhere scatteredness).

A $\mathrm {C}^*$ -algebra A is said to be nowhere scattered if no hereditary sub- $\mathrm {C}^*$ -algebra of A has a nonzero one-dimensional representation. Equivalently, A is nowhere scattered if and only if A has no nonzero elementary ideal quotients; see [Reference Thiel and Vilalta39, Definition A] and [Reference Thiel and Vilalta39, Theorem 3.1].

We say that A has the global Glimm property (in the sense of [Reference Kirchberg and Rørdam29, Definition 4.12]) if, for every $a\in A_+$ and $\varepsilon>0$ , there exists a square-zero element $r\in \overline {aAa}$ such that $(a-\varepsilon )_+\in \overline {\mathrm {span}}ArA$ ; see [Reference Thiel and Vilalta43, Section 3].

A $\mathrm {C}^*$ -algebra satisfying the global Glimm property is always nowhere scattered. The converse remains open and is known as the global Glimm problem. The problem has been answered affirmatively under the additional assumption of real rank zero ([Reference Elliott and Rørdam21]) or stable rank one ([Reference Antoine, Perera, Robert and Thiel2]).

A $\mathrm {Cu}$ -semigroup is said to be $(2,\omega )$ -divisible if, for every pair $x',x\in S$ with $x'\ll x$ , there exists $y\in S$ such that $2y\leq x$ and $x'\leq \infty y$ ; see [Reference Robert and Rørdam33, Definition 5.1].

For a detailed study of the global Glimm problem and its relation with the Cuntz semigroup, we refer to [Reference Thiel and Vilalta43]; see also [Reference Vilalta48]. Among other results, it follows from [Reference Thiel and Vilalta43, Theorem 3.6] that a $\mathrm {C}^*$ -algebra A has the global Glimm property if and only if $ {\mathrm {Cu}} (A)$ is $(2,\omega )$ -divisible.

Proof. By [Reference Thiel and Vilalta40, Proposition 3.6], a positive element c in a $\mathrm {C}^*$ -algebra is soft if and only if for every $\varepsilon>0$ there exists $r \in (\overline {cAc})_+$ such that r is orthogonal to $(c-\varepsilon )_+$ and such that $c \lhd r$ . Using this characterization for b, we show that it is satisfied for $a+b$ .

To verify that $a+b$ is soft, let $\varepsilon>0$ . Using that b is soft, we obtain $r \in (\overline {bAb})_+$ such that r is orthogonal to $(b-\varepsilon )_+$ and such that $b \lhd r$ . Since a and b are orthogonal, we have

$$\begin{align*}((a+b)-\varepsilon)_+ = (a-\varepsilon)_+ + (b-\varepsilon)_+. \end{align*}$$

Since r belongs to $\overline {bAb}$ , it is also orthogonal to a, and thus also orthogonal to $((a+b)-\varepsilon )_+$ . Further, we have $a+b \lhd b \lhd r$ , as desired.

Proof. Choose $x"\in {\mathrm {Cu}} (A)$ such that $x'\ll x"\ll x$ . Using that x is strongly soft, we know that there exists $t\in {\mathrm {Cu}} (A)$ such that $x"\ll \infty t$ and $x"+t\ll x$ . Choose orthogonal positive elements $c,d \in A\otimes {\mathbb {K}}$ and $\varepsilon>0$ such that

$$\begin{align*}x" = [c], \quad t = [d], \quad x' \ll [(c-\varepsilon)_+], \,\,\,\text{ and }\,\,\, x" \ll \infty[(d-\varepsilon)_+]. \end{align*}$$

Using that $c+d \precsim a$ , we can apply Rørdam’s lemma (see, for example, [Reference Thiel36, Theorem 2.30]) to obtain $x \in A\otimes {\mathbb {K}}$ such that

$$\begin{align*}((c+d)-\varepsilon)_+ = xx^*, \,\,\,\text{ and }\,\,\, x^*x \in \overline{aAa}. \end{align*}$$

Set

$$\begin{align*}c' := x^*(c-\varepsilon)_+x, \,\,\,\text{ and }\,\,\, d' := x^*(d-\varepsilon)_+x. \end{align*}$$

Then $c',d' \in \overline {aAa}$ . Since c and d are orthogonal, we have

$$\begin{align*}((c+d)-\varepsilon)_+ = (c-\varepsilon)_+ + (d-\varepsilon)_+. \end{align*}$$

It follows that $c'$ and $d'$ are orthogonal and that $c' \sim (c-\varepsilon )_+$ and $d' \sim (d-\varepsilon )_+$ .

In particular, we have $x" \ll \infty [(d-\varepsilon )_+] = \infty [d']$ , and we obtain $\delta>0$ such that $x" \ll \infty [(d'-\delta )_+]$ . Applying that A has an abundance of soft elements for $d'$ and $\delta $ , we obtain a soft element $e \in (\overline {d'Ad'})_+$ such that $(d'-\delta )_+ \lhd e$ . Since $c'$ and $d'$ are orthogonal, and e belongs to $\overline {d'Ad'}$ , it follows that $c'$ and e are orthogonal.

Using that positive elements $g,h$ in a $\mathrm {C}^*$ -algebra satisfy $g \lhd h$ if and only if $[g] \leq \infty [h]$ , we have

$$\begin{align*}[c'] = [(c-\varepsilon)_+] \leq [c] = x" \leq \infty[(d'-\delta)_+] \leq \infty[e] \end{align*}$$

and thus $c' \lhd e$ . By Proposition 3.1, $c'+e$ is soft.

Note that $c'$ and e belong to $\overline {aAa}$ . In particular, $c'+e$ belongs to $A_+$ , and we can apply [Reference Thiel and Vilalta40, Theorem 6.9] to obtain a completely soft element $f \in A_+$ such that $\overline {fAf} = \overline {(c'+e)A(c'+e)} \subseteq \overline {aAa}$ . Then $f \in \overline {aAa}$ , and therefore $[f] \leq [a] = x$ . Further, we have

$$\begin{align*}x' \ll [(c-\varepsilon)_+] = [c'] \leq [c'+e] = [f]. \end{align*}$$

Choose $\delta>0$ such that

$$\begin{align*}x' \ll [(f-\delta)_+], \end{align*}$$

and set $b := (f-\delta )_+$ . Since cut downs of $(f-\delta )_+$ are also cut downs of f, we see that b is completely soft. Further, we have

$$\begin{align*}x' \ll [b] = [(f-\delta)_+] \ll [f] \leq x, \end{align*}$$

which shows that b has the desired properties.

Proof. Choose a $\ll $ -increasing sequence $(x_n)_n$ in $ {\mathrm {Cu}} (A)$ with supremum $[a]$ . We will inductively choose completely soft elements $a_n \in (\overline {aAa})_+$ such that

$$\begin{align*}x_n \ll [a_n] \ll [a], \,\,\,\text{ and }\,\,\, [a_n] \ll [a_{n+1}] \end{align*}$$

for $n\in {\mathbb {N}}$ . To start, apply Lemma 3.2 for $x_0 \ll [a]$ to obtain a completely soft element $a_0 \in (\overline {aAa})_+$ such that $x_0 \ll [a_0] \ll [a]$ . Assuming we have chosen $a_0,\ldots ,a_n$ , find $x_n' \in {\mathrm {Cu}}(A)$ such that $[a_n],x_n \ll x_n' \ll [a]$ . Applying Lemma 3.2 for $x_n' \ll [a]$ , we obtain a completely soft element $a_{n+1} \in (\overline {aAa})_+$ such that $x_n' \ll [a_{n+1}] \ll [a]$ . Proceeding inductively, we obtain the desired sequence $(a_n)_n$ .

Next, assume that A has weak stable rank one. By [Reference Thiel and Vilalta40, Proposition 4.16], soft operators have strongly soft Cuntz classes. Conversely, assuming that $[a]$ is strongly soft, we will show that $[a] = [b]$ for some completely soft element $b\in A_+$ .

Let $(a_n)_n$ be as above. We will show that $\sup _n [a_n]$ (which is $[a]$ ), has a soft representative. Given $c,d \in A_+$ , we will write $c \sim _u d$ if there exists a unitary $u\in \tilde {A}$ such that $c=udu^*$ , and we write $c \subseteq d$ if $\overline {cAc} \subseteq \overline {dAd}$ .

Using [Reference Thiel36, §2.5], one can find a sequence $(\delta _n)_n$ in $(0,\infty )$ and a sequence of contractive elements $(b_n)_n$ in $A_+$ such that

$$\begin{align*}\begin{matrix} a_1 & \precsim & a_2 & \precsim & a_3 & \precsim & \ldots\\{\leq} & & {\leq} & &{\leq} & &\\(a_1-\delta_1)_+ & & (a_2-\delta_2)_+ & & (a_3-\delta_3)_+ & & \ldots \\{\sim_u} & & {\sim_u} & & {\sim_u} & &\\b_1 & \subseteq & b_2 & \subseteq & b_3 & \subseteq & \ldots \end{matrix} \end{align*}$$

and, setting $b_\infty :=\sum _n \frac {1}{2^n \Vert b_n \Vert } b_n$ , such that $[b_\infty ]=\sup _n [a_n]$ .

For each $n\in {\mathbb {N}}$ , since $a_n$ is completely soft, so is the element $(a_n-\delta _n )_+$ . Since $(a_n-\delta _n )_+$ and $b_n$ are unitarily equivalent, they generate $\ast $ -isomorphic hereditary sub- $\mathrm {C}^*$ -algebras of A, and it follows that $b_n$ is completely soft as well.

Further, since $b_0\subseteq b_1\subseteq \ldots $ and $b_\infty =\sum _n \frac {1}{2^n \Vert b_n \Vert } b_n$ , the sequence of hereditary sub- $\mathrm {C}^*$ -algebras $\overline {b_{n}Ab_{n}}$ is increasing with $\overline {b_\infty A b_\infty } = \overline {\bigcup _n \overline {b_n A b_n}}$ . Since each $\overline {b_n A b_n}$ has no nonzero unital quotients, it follows from [Reference Thiel and Vilalta40, Proposition 2.17] that neither does $\overline {b_\infty A b_\infty }$ . This proves that $b_\infty $ is soft.

Note that $b_\infty $ belongs to $A_+$ . Applying [Reference Thiel and Vilalta40, Theorem 6.9], we obtain a completely soft element $b \in A_+$ such that $\overline {bAb} = \overline {b_\infty A b_\infty }$ . Then $[b]=[b_\infty ]=[a]$ , as desired.

Proof. It follows from [Reference Thiel and Vilalta43, Theorem 3.6] that $A\otimes {\mathbb {K}}$ has the global Glimm property. Hence, $A\otimes {\mathbb {K}}$ has an abundance of soft elements by [Reference Thiel and Vilalta43, Proposition 7.7]. Further, $A\otimes {\mathbb {K}}$ has weak stable rank one by [Reference Blackadar, Robert, Tikuisis, Toms and Winter13, Lemma 4.3.2]. Now, the result follows from Theorem 3.3.

3.5 (The strongly soft subsemigroup).

Given a $\mathrm {Cu}$ -semigroup S, we let $S_{\mathrm {{soft}}}$ denote the set of strongly soft elements in S. By Corollary 3.4, given a $\mathrm {C}^*$ -algebra A with the global Glimm property, we have

$$\begin{align*}{\mathrm{Cu}}(A)_{\mathrm{{soft}}} = \big\{ [a] : a \in (A\otimes{\mathbb{K}})_+ \text{ soft} \big\}. \end{align*}$$

In particular, if A is stably finite, simple and unital, it follows from [Reference Thiel and Vilalta40, Proposition 4.16] that the subset $ {\mathrm {Cu}} (A)_{\mathrm {{soft}}}\setminus \{ 0\}$ coincides with $ {\mathrm {Cu}}_{+} (A)$ , the set of Cuntz classes of purely positive elements as introduced in [Reference Perera and Toms30, Definition 2.1]; see also [Reference Asadi-Vasfi, Golestani and Phillips11, Definition 3.8].

Proof. By [Reference Thiel and Vilalta40, Proposition 7.7], if a $\mathrm {Cu}$ -semigroup is $(2,\omega )$ -divisible and satisfies (O5), then it has an abundance of soft elements, which then by [Reference Thiel and Vilalta40, Proposition 5.6] implies that its strongly soft elements form a sub- $\mathrm {Cu}$ -semigroup. Thus, $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup.

Let us verify that $S_{\mathrm {{soft}}}$ satisfies (O5). By [Reference Antoine, Perera and Thiel4, Theorem 4.4(1)] it suffices to show that for all $x',x,y',y,z',z \in S_{\mathrm {{soft}}}$ satisfying

(1) $$ \begin{align} x' \ll x, \quad y' \ll y, \,\,\,\text{ and }\,\,\, x+y \ll z' \ll z, \end{align} $$

there exist $c',c \in S_{\mathrm {{soft}}}$ such that

(2) $$ \begin{align} x'+c \ll z, \quad z' \ll x+c', \,\,\,\text{ and }\,\,\, y' \ll c' \ll c. \end{align} $$

So let $x',x,y',y,z',z \in S_{\mathrm {{soft}}}$ satisfy Equation (1). Choose $v',v \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}z' \ll v' \ll v \ll z.\end{align*}$$

Applying (O5), we obtain $b \in S$ such that

$$\begin{align*}x'+b \leq v' \leq x+b, \,\,\,\text{ and }\,\,\, y' \ll b. \end{align*}$$

Using that $v' \ll v$ and that v is strongly soft, we apply [Reference Thiel and Vilalta40, Proposition 4.13] to find $t \in S_{\mathrm {{soft}}}$ such that $v'+t \leq v \leq \infty t$ . Set $c := b+t$ . Since $b \leq v' \leq v \leq \infty t$ and t is strongly soft, we have $c \in S_{\mathrm {{soft}}}$ by [Reference Thiel and Vilalta40, Theorem 4.14(2)]. Thus, one gets

$$\begin{align*}x'+c = x'+b+t \leq v'+t \leq v \ll z, \end{align*}$$

and

$$\begin{align*}z' \ll v' \leq x+b \leq x+c, \,\,\,\text{ and }\,\,\, y' \ll b \leq c. \end{align*}$$

Using also that $S_{\mathrm {{soft}}}$ is a $\mathrm {Cu}$ -semigroup and $c \in S_{\mathrm {{soft}}}$ , we can find $c' \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}c' \ll c, \quad z' \ll x+c', \,\,\,\text{ and }\,\,\, y' \ll c'. \end{align*}$$

This shows that $c'$ and c satisfy Equation (2), as desired.

That $S_{\mathrm {{soft}}}$ satisfies (O6) (respectively (O7)) whenever S does is proven analogously.

4.1 (Cuntz semigroups of stable rank one $\mathrm {C}^*$ -algebras).

Let A be a stable rank one $\mathrm {C}^*$ -algebra. As shown in [Reference Rørdam and Winter35, Theorem 4.3], the Cuntz semigroup $ {\mathrm {Cu}} (A)$ satisfies a cancellation property termed weak cancellation: If $x,y,z \in {\mathrm {Cu}}(A)$ satisfy $x+z\ll y+z$ , then $x\ll y$ .

If A is also separable, then $ {\mathrm {Cu}}(A)$ is inf-semilattice ordered, that is, for every pair of elements $x,y \in {\mathrm {Cu}}(A)$ their infimum $x \wedge y$ exists, and for every $x,y,z \in {\mathrm {Cu}}(A)$ one has $(x+z) \wedge (y+z) = (x \wedge y) +z$ ; see [Reference Antoine, Perera, Robert and Thiel2, Theorem 3.8].

Proof. It follows directly from the definitions that weak cancellation is stronger than left-soft separativitiy. By [Reference Rørdam and Winter35, Theorem 4.3], the Cuntz semigroup of a stable rank one $\mathrm {C}^*$ -algebra is weakly cancellative.

Proof. The backwards implication is straightforward to verify and even holds for general $\mathrm {Cu}$ -semigroups. To show the forward implication, assume that S is left-soft separative, and let $x,y,t',t\in S$ as in the statement. By Proposition 3.6, we know that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup. In particular, x can be written as the supremum of a $\ll $ -increasing sequence of strongly soft elements.

Take $x'\in S_{\mathrm {{soft}}}$ such that $x'\ll x$ . We have

$$\begin{align*}x'+t'\ll x+t\leq y+t',\quad t' \ll \infty x,\,\,\,\text{ and }\,\,\, t' \ll \infty y. \end{align*}$$

By left-soft separativity, we deduce $x'\ll y$ . Since x is the supremum of such $x'$ , one gets $x\leq y$ , as required.

Proof. Take $t"\in S$ such that $t'\ll t"\ll t$ . Using that t is strongly soft, one finds $s\in S_{\mathrm {{soft}}}$ such that $t"+s\leq t\leq \infty s$ ; see [Reference Thiel and Vilalta40, Proposition 4.13]. Note that, since x and s are strongly soft, so is $x+s$ by [Reference Thiel and Vilalta40, Theorem 4.14]. We get

$$\begin{align*}(x+s)+t" = x+(s+t") \leq x+t \leq y+t'. \end{align*}$$

Further, we have $t'\ll \infty y$ and $t'\ll t"\leq \infty s\leq \infty (x+s)$ .

An application of Lemma 4.4 shows that $x+s\leq y$ and, therefore, that $x\leq y$ .

Proof. In general, (1) implies (2), which implies (3); see [Reference Antoine, Perera and Thiel4, Proposition 5.6.3]. To verify that (3) implies (1), let $x,y\in S_{\mathrm {{soft}}}$ and $n\geq 1$ satisfy $nx\leq ny$ . Then $\widehat {x} \leq \widehat {y}$ ; see Paragraph 5.1. By [Reference Thiel and Vilalta40, Proposition 4.5], x is functionally soft. Thus, we deduce from [Reference Antoine, Perera and Thiel4, Theorem 5.3.12] that $x \leq y$ , as desired.

Proof. Let S be an almost unperforated $\mathrm {Cu}$ -semigroup satisfying (O5). To verify that S is left-soft separative, let $y,t\in S$ and $x\in S_{\mathrm {{soft}}}$ satisfy $x+t\ll y+t$ and $t\ll \infty x,\infty y$ . Choose $y'\in S$ such that

$$\begin{align*}x+t\ll y'+t, \quad t\ll\infty y', \,\,\,\text{ and }\,\,\, y'\ll y. \end{align*}$$

Then $x \leq _p y'$ by [Reference Antoine, Perera and Thiel4, Proposition 5.6.8(ii)]. In particular, there exists $k\in {\mathbb {N}}$ such that $kx\leq ky'$ , and thus $\widehat {x} \leq \widehat {y'}$ ; see Paragraph 5.1. By [Reference Thiel and Vilalta40, Proposition 4.5], x is functionally soft. Using that S is almost unperforated, we obtain that $x \leq y' \ll y$ , by [Reference Antoine, Perera and Thiel4, Theorem 5.3.12].

Proof. A $\mathrm {C}^*$ -algebra has strict comparison of positive elements if and only if its Cuntz semigroup is almost unperforated; see [Reference Elliott, Robert and Santiago20, Proposition 6.2]. Since every Cuntz semigroup satisfies (O5), the result follows from Lemma 4.7.

5.1 (Functionals and ranks).

Given a $\mathrm {Cu}$ -semigroup S, we will denote by $F(S)$ the set of its functionals, that is to say, the set of monoid morphisms $S\to [0,\infty ]$ that preserve the order and suprema of increasing sequences. If S satisfies (O5), then $F(S)$ becomes a compact, Hausdorff space – and even an algebraically ordered compact cone [Reference Antoine, Perera, Robert and Thiel1, Section 3] – when equipped with a natural topology [Reference Elliott, Robert and Santiago20, Reference Keimel28, Reference Robert32].

Given a $\mathrm {C}^*$ -algebra, the cone $ {\mathrm {QT}}(A)$ of lower-semicontinuous 2-quasitraces on A is naturally isomorphic to $F( {\mathrm {Cu}}(A))$ , as shown in [Reference Elliott, Robert and Santiago20, Theorem 4.4].

We let $ {\mathrm {LAff}}(F(S))$ denote the monoid of lower-semicontinuous, affine functions $F(S)\to (-\infty ,\infty ]$ , equipped with pointwise order and addition. For $x\in S$ , the rank of x is defined as the map $\widehat {x} \colon F(S) \to [0,\infty ]$ given by

$$\begin{align*}\widehat{x}(\lambda) := \lambda(x) \end{align*}$$

for $\lambda \in F(S)$ . The function $\widehat {x}$ belongs to $ {\mathrm {LAff}}(F(S))$ and the rank problem of determining which functions in $ {\mathrm {LAff}}(F(S))$ arise this way has been studied extensively in [Reference Thiel37] and [Reference Antoine, Perera, Robert and Thiel2].

Sending an element $x \in S$ to its rank $\widehat {x}$ defines a monoid morphism from S to $ {\mathrm {LAff}} (F(S))$ which preserves both the order and suprema of increasing sequences.

Proof. Let $u"\in S$ be such that $u'\ll u"\ll u$ . By [Reference Thiel and Vilalta40, Proposition 4.13], there exists $s\in S$ satisfying

$$\begin{align*}u"+s\leq u\leq \infty s. \end{align*}$$

Since $u"\ll u\leq \infty s$ , there exists $s' \in S$ such that

$$\begin{align*}s' \ll s, \,\,\,\text{ and }\,\,\, u"\ll \infty s'. \end{align*}$$

We have

$$\begin{align*}u"+s \leq x, \quad u' \ll u", \,\,\,\text{ and }\,\,\, s'\ll s. \end{align*}$$

Applying (O5), we obtain $d\in S$ such that $u'+d\leq x\leq u"+d$ with $s'\leq d$ . Since $u"\leq \infty s'$ , it follows that $x\leq \infty d$ . Finally, apply [Reference Thiel and Vilalta40, Proposition 7.7] to d in order to obtain $c\in S_{\mathrm {{soft}}}$ such that $2c\leq d\leq \infty c$ . This element satisfies the required conditions.

Proof. Let $u',v'\in L_x$ , let $x' \in S$ satisfy $x' \ll x$ , and let $k \in {\mathbb {N}}$ . By definition, there exist $u,v\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}u'\ll u\ll x,\,\,\,\text{ and }\,\,\, v'\ll v\ll x. \end{align*}$$

Choose $y',y\in S$ such that

$$\begin{align*}x' \ll y' \ll y \ll x, \quad v \ll y', \,\,\,\text{ and }\,\,\, u \ll y'. \end{align*}$$

Using that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup by Proposition 3.6, we can choose elements $u",u"',v" \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}u' \ll u" \ll u"' \ll u, \,\,\,\text{ and }\,\,\, v' \ll v" \ll v. \end{align*}$$

Applying Lemma 5.2 for $u"'\ll u\ll y$ and $v"\ll v\ll y$ , we obtain $c,d \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}u"'+c\leq y\leq \infty c,\,\,\,\text{ and }\,\,\, v"+2d\leq y\leq \infty d. \end{align*}$$

Then, applying [Reference Thiel and Vilalta43, Proposition 4.10] for $y'\ll y\leq \infty c,\infty d$ , we get $e\in S$ such that

$$\begin{align*}y' \ll \infty e, \,\,\,\text{ and }\,\,\, e\ll c,d. \end{align*}$$

By [Reference Thiel and Vilalta40, Proposition 7.7], there exists a strongly soft element $e_0$ such that $e_0 \leq e \leq \infty e_0$ . Replacing e by $e_0$ , we may assume that $e \in S_{\mathrm {{soft}}}$ . Using again that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup, we can find $e',e"\in S_{\mathrm {{soft}}}$ satisfying

$$\begin{align*}y' \ll \infty e', \,\,\,\text{ and }\,\,\, e'\ll e"\ll e. \end{align*}$$

By [Reference Thiel and Vilalta40, Proposition 4.13], there exists $r\in S$ such that

$$\begin{align*}e"+r\leq e\leq \infty r. \end{align*}$$

Since $e"\ll e$ , we can find $r' \in S$ such that

$$\begin{align*}r' \ll r, \,\,\,\text{ and }\,\,\, e"\leq \infty r'. \end{align*}$$

Thus, one has

$$\begin{align*}e"+(r+u"')\leq e+u"'\leq c+u"'\leq y, \quad e' \ll e", \,\,\,\text{ and }\,\,\, r'+u" \ll r+u"'. \end{align*}$$

Applying (O5), we obtain $z\in S$ such that

$$\begin{align*}e'+z \leq y \leq e"+z, \,\,\,\text{ and }\,\,\, r'+u"\leq z. \end{align*}$$

Using again that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup, choose $d' \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}e \ll d' \ll d. \end{align*}$$

We have

(3) $$ \begin{align} (v"+d)+d=v"+2d\leq y \leq z+e" \leq z+d', \end{align} $$

with $v"+d\in S_{\mathrm {{soft}}}$ . Note that

$$\begin{align*}d'\ll d\leq \infty (v"+d),\,\,\,\text{ and }\,\,\, d'\ll d\leq y\leq z+e"\leq z+\infty r'\leq \infty z. \end{align*}$$

In particular, since $d'\ll \infty z$ , there exists $M\in {\mathbb {N}}$ such that $d'\leq Mz$ . Set

$$\begin{align*}l:= \infty (u"+v"), \,\,\,\text{ and }\,\,\, w:=e'+ (z\wedge l), \end{align*}$$

where $z\wedge l$ exists because l is idempotent, and S is countably based and satisfies (O7); see [Reference Antoine, Perera, Robert and Thiel1, Theorem 2.4].

Note that, since $l\leq \infty y'\leq \infty e'$ and $e'\in S_{\mathrm {{soft}}}$ , it follows from [Reference Thiel and Vilalta40, Theorem 4.14] that $w\in S_{\mathrm {{soft}}}$ . We get

$$\begin{align*}w \leq e'+z\leq y\ll x,\quad x'\ll y'\leq \infty e'\leq \infty w,\,\,\,\text{ and }\,\,\, u'\ll u"\leq z\wedge l\leq w. \end{align*}$$

By [Reference Antoine, Perera, Robert and Thiel1, Theorem 2.5], the map $S \to S$ , $s \mapsto s\wedge l$ , is additive. Using this at the second and fourth step, we get

$$ \begin{align*} v" + 2(d'\wedge l) &= (v" \wedge l) + 2(d'\wedge l) = (v"+2d')\wedge l \\ &\leq (z+d')\wedge l = (z \wedge l) + (d'\wedge l) \leq w + (d'\wedge l). \end{align*} $$

We also have $d'\wedge l\leq (Mz)\wedge l = M(z\wedge l)\leq Mw$ , and this implies that

$$\begin{align*}\widehat{v"}\leq \widehat{w}. \end{align*}$$

Now, since $v'\ll v"$ and $\frac {k}{k+1}<1$ , we can apply [Reference Robert32, Lemma 2.2.5] to obtain

$$\begin{align*}\frac{k}{k+1} \widehat{v'} \ll \widehat{v"}\leq \widehat{w}. \end{align*}$$

Since w is strongly soft and $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup, there exists a $\ll $ -increasing sequence of soft elements with supremum w. Using that the rank map $x\mapsto \widehat {x}$ preserves suprema of increasing sequences, we can find $w' \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}w'\ll w, \quad \frac{k}{k+1} \widehat{v'} \leq \widehat{w'}, \quad x'\ll \infty w',\,\,\,\text{ and }\,\,\, u'\ll w'. \end{align*}$$

Further, we have $w'\ll w\ll x$ . This shows that $w'$ is a strongly soft element in $L_x$ , as desired.

If, additionally, S is left-soft separative, we can apply Lemma 4.4 on (3) to obtain that $v"+d \leq z$ , and so $v" \leq z$ . We also have $v" \leq l$ and thus

$$\begin{align*}v' \ll v" \leq z \wedge l \leq w. \end{align*}$$

We also have $u' \ll u" \leq w$ and $x' \ll \infty w$ . Using that w is strongly soft and that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup, we can find $w' \in S_{\mathrm {{soft}}}$ such that $u',v' \ll w' \ll w$ and $x' \ll \infty w'$ . Then $w'$ has the desired properties.

Proof. We will prove the result by induction on $\vert C\vert $ , the size of C. If $\vert C\vert =1$ , the result follows from Lemma 5.3.

Thus, fix $n\in {\mathbb {N}}$ with $n \geq 2$ , and assume that the result holds for any finite subset of $n-1$ elements. Given $C\subseteq L_x$ with $\vert C\vert =n$ , pick some $v_0\in C$ . Applying the induction hypothesis, we get an element $w"\in L_x$ such that

$$\begin{align*}u' \ll w",\quad x' \ll \infty w",\,\,\,\text{ and }\,\,\, \frac{k}{k+1}\widehat{v'} \leq \widehat{w"} \end{align*}$$

for every $v'\in C \setminus \{v_0\}$ .

Now, applying Lemma 5.3 to $x'$ , $w"$ and $v_0$ , we get a strongly soft element $w' \in L_x$ such that

$$\begin{align*}w" \ll w',\quad x' \ll \infty w',\,\,\,\text{ and }\,\,\, \frac{k}{k+1}\widehat{v_0} \leq \widehat{w'}. \end{align*}$$

Then $\widehat {w"}\leq \widehat {w'}$ , which shows that $w'$ satisfies the required conditions.

Proof. By definition of $L_x$ , we obtain $u \in S_{\mathrm {{soft}}}$ such that $u' \ll u \ll x$ . Let $(x_n)_n$ be a $\ll $ -increasing sequence with supremum x and such that $u \ll x_0$ . Note that the sets $L_{x_n}$ form an increasing sequence of subsets of S with $L_x = \bigcup _n L_{x_n}$ .

Let B be a countable basis for S. Then

$$\begin{align*}B \cap L_x = \bigcup_n (B \cap L_{x_n}), \end{align*}$$

and we can choose a $\subseteq $ -increasing sequence $(C_n)_n$ of finite subsets of $B \cap L_x$ such that

$$\begin{align*}B \cap L_x = \bigcup_n C_n, \,\,\,\text{ and }\,\,\, C_n \subseteq B \cap L_{x_n} \text{ for each } n. \end{align*}$$

We have $u'\in L_{x_0} \subseteq L_{x_1}$ . Apply Proposition 5.5 to $k=1,(0 \ll x_1),u'$ , and $C_1$ to obtain a strongly soft element $w_1'\in L_{x_1}$ such that

$$\begin{align*}u' \ll w_1',\quad 0 \ll \infty w_1', \,\,\,\text{ and }\,\,\, \frac{1}{2}\widehat{v'} \leq \widehat{w_1'} \end{align*}$$

for every $v' \in C_1$ .

We have $w_1' \in L_{x_2}$ . Applying Proposition 5.5 again to $k=2,(x_1 \ll x_2),w_1'$ and $C_2$ , we obtain a strongly soft element $w_2' \in L_{x_2}$ such that

$$\begin{align*}w_1' \ll w_2',\quad x_1 \ll \infty w_2',\,\,\,\text{ and }\,\,\, \frac{2}{3} \widehat{v'}\leq \widehat{w_2'} \end{align*}$$

for every $v'\in C_2$ .

Proceeding inductively, we get a $\ll $ -increasing sequence of strongly soft elements $(w_n')_n$ such that

$$\begin{align*}w_n'\in L_{x_n},\quad x_{n-1}\ll \infty w_{n}',\,\,\,\text{ and }\,\,\, \frac{n}{n+1}\widehat{v'}\leq \widehat{w_n'} \end{align*}$$

for every $v'\in C_n$ and $n \geq 2$ .

Set $w:=\sup _n w_n'$ , which is strongly soft by [Reference Thiel and Vilalta40, Theorem 4.14]. Note that we get $u' \ll w_1' \leq w \leq x$ by construction. Further, since $x_n\leq \infty w_{n+1}'\leq \infty w$ for each $n \geq 2$ , we deduce that $x\leq \infty w$ .

Now, take $\lambda \in F(S)$ . Given $v'\in B \cap L_x$ , choose $n_0 \geq 2$ such that $v' \in C_{n_0}$ . We have

$$\begin{align*}\frac{n}{n+1} \lambda(v') \leq \lambda(w_n') \leq \lambda(w) \end{align*}$$

for every $n\geq n_0$ . Thus, it follows that $\lambda (v') \leq \lambda (w)$ for every $v' \in B \cap L_x$ .

Since $L_x$ is downward hereditary, every element in $L_x$ is the supremum of an increasing sequence from $B \cap L_x$ . Using also that functionals preserve suprema of increasing sequences, we obtain

$$\begin{align*}\sup_{v'\in L_x} \lambda (v') \leq \sup_{v'\in B\cap L_x} \lambda (v') \leq \lambda (w) = \sup_n \lambda (w_n')\leq \sup_{v'\in L_x} \lambda (v'), \end{align*}$$

which shows that w has the desired properties.

Proof. Choose $x" \in S$ such that $x'\ll x"\ll x$ . Applying [Reference Thiel and Vilalta43, Proposition 4.10] to

$$\begin{align*}x"\ll x\leq\infty x,\infty t, \end{align*}$$

we get $s\in S$ such that

$$\begin{align*}x" \ll \infty s, \,\,\,\text{ and }\,\,\, s\ll x,t. \end{align*}$$

By [Reference Thiel and Vilalta40, Proposition 7.7], we can choose $s'\in S_{\mathrm {{soft}}} $ such that

$$\begin{align*}x" \leq \infty s', \,\,\,\text{ and }\,\,\, s'\ll s. \end{align*}$$

Then $x" \ll \infty s'$ . Applying (O5) to $s'\ll s\leq x$ , we obtain $v\in S$ satisfying

$$\begin{align*}v+s'\leq x\leq v+s. \end{align*}$$

In particular, one has $x"\ll v+s$ . Applying (O6) to $x' \ll x" \leq v+s$ , we find $u\in S$ such that

$$\begin{align*}x'\ll u+s,\,\,\,\text{ and }\,\,\, u\ll x" , v. \end{align*}$$

Since $u\ll x"\leq \infty s'$ , it follows from [Reference Thiel and Vilalta40, Theorem 4.14] that $u+s'$ is soft. Further, we get

$$\begin{align*}x' \ll u+s \leq u+t \leq (u+s')+t, \,\,\,\text{ and }\,\,\, u+s'\leq v+s'\leq x. \end{align*}$$

Using that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup by Proposition 3.6, we can find $u'\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}x'\ll u'+t,\,\,\,\text{ and }\,\,\, u'\ll u+s'\leq x. \end{align*}$$

Then $u'\in L_x$ , which shows that $u'$ has the desired properties.

Proof. By [Reference Thiel and Vilalta40, Proposition 4.13], there exists $s\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}t'+s\leq t\leq \infty s. \end{align*}$$

Applying Lemma 5.7 to $x'\ll x\leq \infty s$ , we obtain a strongly soft element $v'\in L_x$ satisfying $x'\leq v'+s$ . Consequently, we obtain

$$\begin{align*}x'+t'\leq v'+s+t'\leq v'+t.\\[-35pt] \end{align*}$$

Proof. Strong softness is preserved under $\mathrm {Cu}$ -morphisms, and the inclusion map of a sub- $\mathrm {Cu}$ -semigroup is a $\mathrm {Cu}$ -morphism. Hence, given sub- $\mathrm {Cu}$ -semigroups $T_1 \subseteq T_2 \subseteq S$ containing u, if u is strongly soft in $T_1$ , then it is also strongly soft in $T_2$ . This implies in particular that $\mathcal {R}$ is $\sigma $ -complete.

To show that $\mathcal {R}$ is cofinal, let $T_0 \subseteq S$ be a countably based sub- $\mathrm {Cu}$ -semigroup, and let $B_0 \subseteq T_0$ be a countable basis, that is, a countable subset such that every element in $T_0$ is the supremum of an increasing sequence from $B_0$ .

Let $(u_n)_n$ be a $\ll $ -increasing sequence in S with supremum u. Since u is strongly soft in S, for each n we obtain $t_n \in S$ such that

$$\begin{align*}u_n + t_n \ll u, \,\,\,\text{ and }\,\,\, u_n \ll \infty t_n. \end{align*}$$

By [Reference Thiel and Vilalta38, Lemma 5.1], there exists a countably based sub- $\mathrm {Cu}$ -semigroup $T \subseteq S$ containing

$$\begin{align*}B_0 \cup \{u_0,u_1,\ldots\} \cup \{t_0,t_1,\ldots\}. \end{align*}$$

One checks that $T_0 \subseteq T$ , and that u is strongly soft in T.

Proof. We first prove the result under the additional assumption that S is countably based. Use Proposition 5.6 to obtain $w\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}u' \ll w \leq x \leq \infty w, \,\,\,\text{ and }\,\,\, \lambda (w)=\sup_{v'\in L_x} \lambda (v'), \end{align*}$$

for every $\lambda \in F(S)$ . Since $w\leq x$ , we have $\widehat {w} \leq \widehat {x}$ . To show the reverse inequality, let $\lambda \in F(S)$ . We need to prove that $\lambda (x) \leq \lambda (w)$ .

Take $x',w'\in S$ such that $x' \ll x$ and $w'\ll w$ . Applying Lemma 5.8, we obtain an element $v'\in L_x$ such that

$$\begin{align*}x'+w'\leq v'+w. \end{align*}$$

Since $v'$ belongs to $L_x$ , we have $\lambda (v')\leq \lambda (w)$ . This implies

$$\begin{align*}\lambda (x')+\lambda (w')\leq \lambda (v')+\lambda (w) \leq 2\lambda (w). \end{align*}$$

Passing to the supremum over all $x'$ way below x, and all $w'$ way below w, we get

$$\begin{align*}\lambda (x)+\lambda (w) \leq 2\lambda (w). \end{align*}$$

This proves $\lambda (x)\leq \lambda (w)$ . Indeed, if $\lambda (w)=\infty $ , then there is nothing to prove. If $\lambda (w)\neq \infty $ , we can cancel $\lambda (w)$ from the previous inequality.

We now consider the case that S is not countably based. Choose $u \in S_{\mathrm {{soft}}}$ such that $u' \ll u \ll x$ . Since $(2,\omega )$ -divisibility and (O5)–(O7) each satisfy the Löwenheim-Skolem condition, and using also Lemma 5.9, we can use the technique from [Reference Thiel and Vilalta38, Section 5] to deduce that there exists a countably based, $(2,\omega )$ -divisible sub- $\mathrm {Cu}$ -semigroup $H \subseteq S$ satisfying (O5)–(O7), containing x, u and $u'$ , and such that u is strongly soft in H.

Applying the first part of the proof to H, we find $w\in H_{\mathrm {{soft}}}$ such that

$$\begin{align*}u'\ll w\leq x\leq \infty w,\,\,\,\text{ and }\,\,\, \lambda (x)=\lambda (w) \end{align*}$$

for every $\lambda \in F(H)$ .

Since the inclusion $\iota \colon H\to S$ is a $\mathrm {Cu}$ -morphism, it follows that w is strongly soft in S. Further, any functional $\lambda $ on S induces the functional $\lambda \iota $ on H. This shows that w satisfies the required conditions.

Proof. Let $a\in A_+$ . Since A has the global Glimm property, it follows from [Reference Thiel and Vilalta43, Theorem 3.6] that $ {\mathrm {Cu}} (A)$ is $(2,\omega )$ -divisible. Using Theorem 5.10, find $w\in {\mathrm {Cu}} (A)_{\mathrm {{soft}}}$ such that $w\leq [a]$ and $\lambda (w)=\lambda ([a])$ for every $\lambda \in F( {\mathrm {Cu}} (A))$ .

By Theorem 3.3, there exists a soft element $b\in A_+$ such that $w=[b]$ . The result now follows from the fact that the map

$$\begin{align*}\tau\mapsto \left([a]\mapsto d_\tau(a)\right) \end{align*}$$

is a natural bijection from $\mathrm {QT} (A)$ to $F( {\mathrm {Cu}} (A))$ ; see [Reference Elliott, Robert and Santiago20, Theorem 4.4].

Proof. Given $v' \in L_x$ , there exists $v \in S_{\mathrm {{soft}}}$ with $v' \leq v \leq x$ , which shows the inequality ‘ $\geq $ ’.

Conversely, let $v \in S_{\mathrm {{soft}}}$ with $v \leq x$ . Since $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup by Proposition 3.6, there exists a $\ll $ -increasing sequence $(v_n')_n$ in $S_{\mathrm {{soft}}}$ with supremum v. Each $v_n'$ belongs to $L_x$ , and one gets

$$\begin{align*}\lambda(v) = \sup_n \lambda(v_n') \leq \sup_{v'\in L_x} \lambda (v'). \end{align*}$$

This shows the the inequality ‘ $\leq $ ’.

Proof. It is easy to see that $\lambda _{\mathrm {{soft}}}$ preserves order. Further, given an increasing sequence $(x_n)_n$ with supremum x in S, we have that for every $v'\in L_x$ there exists $n\in {\mathbb {N}}$ with $v'\in L_{x_n}$ . Thus, using Lemma 5.12, we get

$$\begin{align*}\lambda_{\mathrm{{soft}}} (x) = \sup_{v'\in L_x}\lambda (v') \leq \sup_n \left(\sup_{v'\in L_{x_n}}\lambda (v') \right) =\sup_n \lambda_{\mathrm{{soft}}} (x_n). \end{align*}$$

Since $\lambda _{\mathrm {{soft}}}$ is order preserving, we also have $\sup _n \lambda _{\mathrm {{soft}}} (x_n)\leq \lambda _{\mathrm {{soft}}} (x)$ , which shows that $\lambda _{\mathrm {{soft}}}$ preserves suprema of increasing sequences.

Given $x,y\in S$ and $u,v\in S_{\mathrm {{soft}}}$ such that $u\leq x$ and $v\leq y$ , we have $u+v\in S_{\mathrm {{soft}}}$ and $u+v\leq x+y$ . This implies that

$$\begin{align*}\lambda_{\mathrm{{soft}}} (x)+\lambda_{\mathrm{{soft}}} (y)\leq \lambda_{\mathrm{{soft}}} (x+y). \end{align*}$$

Thus, $\lambda _{\mathrm {{soft}}}$ is subadditive.

Finally, we show that $\lambda _{\mathrm {{soft}}}$ is superadditive. Given $x,y\in S$ and $w'\in L_{x+y}$ , take $x',x",y',y"\in S$ such that

$$\begin{align*}x'\ll x"\ll x,\quad y'\ll y"\ll y,\,\,\,\text{ and }\,\,\, w'\ll x'+y'. \end{align*}$$

By [Reference Thiel and Vilalta40, Proposition 7.7], there exist $s,t\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}s\leq x"\leq \infty s, \,\,\,\text{ and }\,\,\, t\leq y"\leq \infty t. \end{align*}$$

Take $s',t'\in S$ such that $s'\ll s$ and $t'\ll t$ . Using Lemma 5.8, we find $u'\in L_x$ and $v'\in L_y$ such that

$$\begin{align*}x'+s'\leq u'+s,\,\,\,\text{ and }\,\,\, y'+t'\leq v'+t. \end{align*}$$

Consequently, one has

$$\begin{align*}w'+s'+t' \leq x'+y'+s'+t' \leq u'+s+v'+t. \end{align*}$$

Applying Theorem 5.10, find $u,v\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}u'\ll u\leq x\leq \infty u,\,\,\,\text{ and }\,\,\, v'\ll v\leq y\leq \infty v. \end{align*}$$

This implies

$$\begin{align*}w'+s'+t' \leq u+s+v+t \end{align*}$$

and, therefore,

$$\begin{align*}\lambda (w')+\lambda (s' + t') \leq \lambda (u)+\lambda (v)+\lambda (s+t). \end{align*}$$

Passing to the suprema over all $s'$ way below s, and all $t'$ way below t, we deduce that

$$\begin{align*}\lambda (w')+\lambda (s + t) \leq \lambda (u)+\lambda (v)+\lambda (s+t). \end{align*}$$

Note that $s+t\leq x"+y" \ll x+y \leq \infty (u+v)$ . This allows us to cancel $\lambda (s + t)$ , and we obtain

$$\begin{align*}\lambda (w') \leq \lambda (u)+\lambda (v) \leq \lambda_{\mathrm{{soft}}} (x)+\lambda_{\mathrm{{soft}}} (y). \end{align*}$$

Since this holds for every $w'\in L_{x+y}$ , we can apply Lemma 5.12 to get

$$\begin{align*}\lambda_{\mathrm{{soft}}}(x+y) = \sup_{\{ w\in S_{\mathrm{{soft}}} : w \leq x+y \}}\lambda(w) = \sup_{w'\in L_{x+y}} \lambda(w') \leq \lambda_{\mathrm{{soft}}} (x)+\lambda_{\mathrm{{soft}}} (y). \end{align*}$$

This show that $\lambda _{\mathrm {{soft}}}$ is superadditive and thus a functional.

Proof. Given $\lambda \in F(S_{\mathrm {{soft}}})$ , let $\lambda _{\mathrm {{soft}}} \in F(S)$ be defined as in Proposition 5.13. This defines a map $\phi \colon F(S_{\mathrm {{soft}}} )\to F(S)$ by $\phi (\lambda ) :=\lambda _{\mathrm {{soft}}}$ . We verify that $\iota ^*\phi ={\operatorname{id}}_{F(S_{\mathrm {{soft}}} )}$ and $\phi \iota ^*={\operatorname {id}}_{F(S)}$ .

Given $\lambda \in F(S_{\mathrm {{soft}}} )$ and $w\in S_{\mathrm {{soft}}}$ , we have

$$\begin{align*}\iota^*\phi(\lambda )(w) = \iota^*\lambda_{\mathrm{{soft}}} (w) = \lambda_{\mathrm{{soft}}} (\iota (w)) = \sup_{\{ v\in S_{\mathrm{{soft}}} : v\leq w \}} \lambda (v) = \lambda (w), \end{align*}$$

which shows $\iota ^*\phi ={\operatorname {id}}_{F(S_{\mathrm {{soft}}} )}$ .

Conversely, if $\lambda \in F(S)$ and $x\in S$ , we can use Theorem 5.10 at the last step to obtain

$$\begin{align*}\phi\iota^*(\lambda)(x) = \phi (\lambda \iota )(x) = \sup_{\{ v\in S_{\mathrm{{soft}}} : v\leq x \}}\lambda (v) = \lambda (x). \end{align*}$$

This shows that $\iota ^*$ is a bijective, continuous map. Since $F(S)$ and $F(S)_{\mathrm {{soft}}}$ are both compact, Hausdorff spaces, it follows that $\iota ^*$ is a homeomorphism.

Proof. The result follows from Theorem 5.14 and the fact that $ {\mathrm {QT}}(A)$ is naturally homeomorphic to $F( {\mathrm {Cu}} (A))$ ; see [Reference Elliott, Robert and Santiago20, Theorem 4.4].

Proof. To verify (1), note that the members of $L_x$ are bounded by x, and consequently $\sigma (x) \leq x$ . To see that $\sigma (x)$ is strongly soft, let $s\in S$ be such that $s\ll \sigma (x)$ . We will find $t \in S$ such that $s + t \ll \sigma (x)$ and $s \ll \infty t$ .

Since $\sigma (x)=\sup L_x$ , there exists $u' \in L_x$ such that $s \ll u' \leq \sigma (x)$ . Using that $u'\in L_x$ , we find $u\in S_{\mathrm {{soft}}}$ with $u'\ll u\ll x$ . By Proposition 3.6, $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup, and we obtain $u" \in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}s \ll u' \ll u" \ll u \ll x. \end{align*}$$

Then $s \ll u" \in S_{\mathrm {{soft}}}$ and by the definition of strong softness we obtain $t \in S$ such that $s+t\ll u"$ and $s \ll \infty t$ . We have $u" \in L_x$ and therefore $u" \leq \sigma (x)$ , which shows that t has the desired properties.

Thus, $\sigma (x)$ is a strongly soft element dominated by x. To show that it is the largest element with these properties, let $w \in S_{\mathrm {{soft}}}$ satisfy $w\leq x$ . We can use once again that $S_{\mathrm {{soft}}}$ is a sub- $\mathrm {Cu}$ -semigroup to find a $\ll $ -increasing sequence $(w_n)_n$ of strongly soft elements with supremum w. Then $w_n \in L_x$ for each n, and consequently

$$\begin{align*}w = \sup_n w_n \leq \sup\ L_x = \sigma (x). \end{align*}$$

This also shows that $x=\sigma (x)$ if and only if x is strongly soft. We have proved (1) and (3).

To verify (2), we first note that $\infty \sigma (x) \leq \infty x$ since $\sigma (x) \leq x$ . For the converse inequality, use Proposition 5.6 to obtain $w\in S_{\mathrm {{soft}}}$ with $w\leq x\leq \infty w$ . By (1), we have $w\leq \sigma (x)$ , and we get

$$\begin{align*}\infty x= \infty w\leq \infty \sigma(x). \end{align*}$$

Finally, to prove (4), let $t \in S$ satisfy $x \leq \infty t$ . Let $x' \in S$ satisfy $x' \ll x$ . Applying Lemma 5.7, we obtain $u' \in L_x$ such that $x' \ll u'+t$ . Then

$$\begin{align*}x' \ll u'+t \leq \sigma(x)+t. \end{align*}$$

Passing to the supremum over all $x'$ way below x, we get $x \leq \sigma (x)+t$ , as desired.

Proof. Using that $\sigma (x)\leq x$ , we have $2\sigma (x)\leq x+\sigma (x)$ . To show the reverse inequality, let $w \in S$ satisfy $w\ll \sigma (x)$ . Since $\sigma (x)$ is strongly soft, it follows from [Reference Thiel and Vilalta40, Proposition 4.13] that there exists $t\in S$ with $w+t\leq \sigma (x)\leq \infty t$ .

We have $x \leq \infty \sigma (x)$ by Proposition 6.3 (2), and thus $x \leq \infty t$ . Therefore, $x\leq \sigma (x)+t$ by Proposition 6.3 (4). Thus, we have

$$\begin{align*}x+w\leq \sigma (x) + t +w \leq 2\sigma (x). \end{align*}$$

Passing to the supremum over all w way below $\sigma (x)$ , we get $x+\sigma (x)\leq 2\sigma (x)$ .

Proof. To show that $\sigma $ is order preserving, let $x,y\in S$ satisfy $x\leq y$ . Then $L_x \subseteq L_y$ , and thus

$$\begin{align*}\sigma(x) = \sup L_x \leq \sup L_y = \sigma(y). \end{align*}$$

To show that $\sigma $ preserves suprema of increasing sequences, let $(x_n)_n$ be an increasing sequence in S with supremum x. Since $\sigma $ is order-preserving, one gets $\sup _n\sigma (x_n) \leq \sigma (x)$ . Conversely, given $u' \in L_x$ , choose $u \in S_{\mathrm {{soft}}}$ with $u' \ll u \ll x$ . Then there exists $n\in {\mathbb {N}}$ such that $u \ll x_n$ , and thus $u' \in L_{x_n}$ . We deduce that

$$\begin{align*}u' \leq \sup L_{x_n} = \sigma(x_n) \leq \sup_n \sigma(x_n). \end{align*}$$

Hence, $\sigma (x) = \sup L_x \leq \sup _n \sigma (x_n)$ , as desired.

To see that $\sigma $ is superadditive, let $x,y\in S$ . Note that $\sigma (x)+\sigma (y)$ is a strongly soft element bounded by $x+y$ . Using Proposition 6.3 (1), we get $\sigma (x)+\sigma (y)\leq \sigma (x+y)$ .

Next, given $x,y \in S$ , let us show that $2\sigma (x+y) \leq 2\sigma (x) + 2\sigma (y)$ . To prove this, let $w \in S$ satisfy $w \ll \sigma (x+y)$ . By [Reference Thiel and Vilalta40, Proposition 4.13], there exists $s\in S$ satisfying

$$\begin{align*}w + s \leq \sigma(x+y) \leq \infty s. \end{align*}$$

Applying [Reference Thiel and Vilalta40, Proposition 7.7], we find $t \in S$ such that $2t \leq s\leq \infty t$ . Using also Proposition 6.3 (2), we deduce that

$$\begin{align*}w+2t \leq w+s \leq \sigma(x+y), \,\,\,\text{ and }\,\,\, x,y \leq \infty (x+y) = \infty \sigma (x+y) \leq \infty s \leq \infty t. \end{align*}$$

Using Proposition 6.3 (4) at the second step and Lemma 6.5 at last step, we get

$$ \begin{align*} \sigma (x+y) + w &\leq x+y+w\leq \sigma (x)+\sigma (y)+w+2t\leq \sigma (x)+\sigma (y) + \sigma (x+y)\\ &\leq \sigma (x)+\sigma (y) + x + y = 2\sigma (x) + 2\sigma (y). \end{align*} $$

Passing to the supremum over all elements w way below $\sigma (x+y)$ , we obtain

$$\begin{align*}2\sigma (x+y) \leq 2\sigma (x) + 2\sigma (y). \end{align*}$$

Next, given $x,y \in S$ , using the above inequality together with the established superadditivity of $\sigma $ , we get

$$\begin{align*}2\sigma (x+y) \leq 2\sigma (x) + 2\sigma (y) \leq \sigma (x+y) + \big( \sigma (x) + \sigma (y) \big) \leq 2\sigma (x+y), \end{align*}$$

as desired.

Proof. By Theorem 6.6, we only need to check that $\sigma $ is subadditive.

(i): If S is almost unperforated, then it follows from Proposition 4.6 that $S_{\mathrm {{soft}}}$ is unperforated. Given any pair $x,y\in S$ , we know from Theorem 6.6 that

$$\begin{align*}2\sigma (x+y) = 2\big(\sigma (x) + \sigma (y)\big). \end{align*}$$

Since this equality is in $S_{\mathrm {{soft}}}$ , it follows that $\sigma (x+y)=\sigma (x) + \sigma (y)$ .

For (ii) and (iii), note that it is enough to prove that $\sigma (x+y)\leq x+\sigma (y)$ for all $x,y \in S$ . Indeed, if this inequality holds, one can use it at the second and last steps to get

$$\begin{align*}\sigma (x+y) = \sigma (\sigma (x+y))\leq \sigma (x+\sigma (y)) = \sigma (\sigma (y) +x)\leq \sigma (y)+\sigma (x), \end{align*}$$

as required.

Given $x,y \in S$ , we proceed to verify that $\sigma (x+y)\leq x+\sigma (y)$ . Let $w\in S$ satisfy $w \ll \sigma (x+y)$ . Choose $y'\in S$ such that

$$\begin{align*}y'\ll y, \,\,\,\text{ and }\,\,\, w \ll x+y'. \end{align*}$$

Since $\sigma (x+y)$ is strongly soft, it follows from [Reference Thiel and Vilalta40, Proposition 4.13] that there exists $r\in S_{\mathrm {{soft}}}$ such that

$$\begin{align*}w+r \leq \sigma (x+y) \leq \infty r. \end{align*}$$

Applying Proposition 6.3 (2), one gets

$$\begin{align*}y' \ll y \leq \infty \sigma (x+y)\leq \infty r. \end{align*}$$

Applying [Reference Thiel and Vilalta43, Proposition 4.7], we obtain $t',t\in S$ such that

$$\begin{align*}y'\leq\infty t', \,\,\,\text{ and }\,\,\, t'\ll t\ll r,y. \end{align*}$$

Using that S is $(2,\omega )$ -divisible, it follows from [Reference Thiel and Vilalta40, Proposition 5.6] that we may assume both $t'$ and t to be strongly soft. Thus, as in the proof of Lemma 5.2, we can apply (O5) to obtain an element b satisfying

$$\begin{align*}t'+b\leq y\leq t+b,\,\,\,\text{ and }\,\,\, y\leq \infty b, \end{align*}$$

which implies

$$\begin{align*}w+r\leq\sigma(x+y)\leq x+y \leq x+t+b \end{align*}$$

with $t\ll r \leq \infty (x+y)=\infty (x+b)$ .

Thus, since both w and r are strongly soft, left-soft separativity (in the form of Lemma 4.5) implies that $w\leq x+b$ . Since S is countably based and satisfies (O7), the infimum $(b\wedge \infty t')$ exists. Note that $(b\wedge \infty t') + t'$ is soft because $(b\wedge \infty t') \leq \infty t'$ ; see [Reference Thiel and Vilalta40, Theorem 4.14]. Then

$$\begin{align*}(b\wedge\infty t') + t' \leq b+t' \leq y, \end{align*}$$

and thus $b\wedge \infty t' \leq (b\wedge \infty t') + t' \leq \sigma (y)$ by Proposition 6.3 (1).

(ii): Assuming that S is inf-semilattice ordered, it now follows that

$$\begin{align*}w\leq (x+b)\wedge(x+\infty t') = x+(b\wedge\infty t') \leq x+\sigma(y). \end{align*}$$

Passing to the supremum over all w way below $\sigma (x+y)$ , we get $\sigma (x+y)\leq x+\sigma (y)$ , as desired. This proves the case (ii).

(iii): Let us additionally assume that $y\ll \infty y$ . Then, given w and r as before, we have that $y\ll \infty y\leq \infty r$ . This implies that there exists $r'\in S$ such that $r'\ll r$ and $y\leq \infty r'$ . Using Proposition 6.3 at the last step, one gets

$$\begin{align*}w+r\leq \sigma (x+y)\leq x+y\leq x+\sigma (y)+r' \end{align*}$$

with $r'\ll r \leq \infty (x+y)=\infty (x+\sigma (y))$ .

Therefore, we can use Lemma 4.4 to deduce that $w\leq x+\sigma (y)$ . Since this holds for every w way below $\sigma (x+y)$ , it follows that $\sigma (x+y)\leq x+\sigma (y)$ whenever $y\ll \infty y$ .

If $S\otimes \{ 0,\infty \}$ is algebraic, then by [Reference Thiel and Vilalta43, Lemma 4.16] every $y \in S$ is the supremum of an increasing sequence $(y_n)_n$ of elements $y_n \in S$ such that $y_n \ll \infty y_n$ . Using the above for each $y_n$ and using that $\sigma $ preserves suprema of increasing sequences, we get

$$\begin{align*}\sigma(x+y) = \sup_n \sigma(x+y_n) \leq \sup_n \big( x+\sigma(y_n) \big) = x + \sigma(y), \end{align*}$$

as desired.

Proof. The Cuntz semigroup $ {\mathrm {Cu}}(A)$ is countably based and satisfies (O5)–(O7). Since A has the global Glimm property, it follows from [Reference Thiel and Vilalta43, Theorem 3.6] that $ {\mathrm {Cu}} (A)$ is $(2,\omega )$ -divisible. We check that the additional conditions of Proposition 6.8 are satisfied:

(i): Assume that A has has strict comparison of positive elements. Then $ {\mathrm {Cu}}(A)$ is almost unperforated by [Reference Elliott, Robert and Santiago20, Proposition 6.2] and left-soft separative by Proposition 4.8. This verifies Proposition 6.8 (i).

(ii): Assume that A has stable rank one. Then $ {\mathrm {Cu}}(A)$ is inf-semilattice ordered by [Reference Antoine, Perera, Robert and Thiel2, Theorem 3.8], and left-soft separative by Proposition 4.3. This verifies Proposition 6.8 (ii).

(iii): Assume that A has topological dimension zero, and $ {\mathrm {Cu}} (A)$ is left-soft separative. Then $ {\mathrm {Cu}} (A)\otimes \{ 0,\infty \}$ is algebraic by [Reference Thiel and Vilalta43, Proposition 4.18]. This verifies Proposition 6.8 (iii).

Question 6.10. Let S be a countably based, weakly cancellative, $(2,\omega )$ -divisible $\mathrm {Cu}$ -semigroup satisfying (O5)–(O7). Is the map $\sigma \colon S\to S_{\mathrm {{soft}}}$ subadditive?

Question 6.11. Let S be the Cuntz semigroup of a $\mathrm {C}^*$ -algebra. Let $x,y,z,w\in S$ satisfy

$$\begin{align*}w=2w, \quad x\leq y+z, \,\,\,\text{ and }\,\,\, x\leq y+w. \end{align*}$$

We know that $z\wedge w$ exists. Does it follow that $x\leq y+(z\wedge w)$ ?

7.1 (Dimension of $\mathrm {Cu}$ -semigroups).

Recall from [Reference Thiel and Vilalta42, Definition 3.1] that, given a $\mathrm {Cu}$ -semigroup S and $n\in {\mathbb {N}}$ , we say that S has dimension n, in symbols $\dim (S)=n$ , if n is the least integer such that, for any $r\in {\mathbb {N}}$ , any pair $x',x \in S$ , and any tuple $y_1,\ldots ,y_r\in S$ with $x'\ll x\ll y_1+\ldots +y_r$ , there exist elements $z_{j,k}\in S$ with $j=1,\ldots ,r$ and $k=0,\ldots ,n$ such that:

1. (i) $z_{j,k}\ll y_j$ for every j and k;

2. (ii) $x'\ll \sum _{j,k}z_{j,k}$ ;

3. (iii) $\sum _{j}z_{j,k}\ll x$ for each k.

If no such n exists, we say that S has dimension $\infty $ , in symbols $\dim (S)=\infty $ .