Proof. (i) follows directly from the definition of $\operatorname {div}_m ([a])$.

To see (ii), let $n_j=\operatorname {div}_m ([b_j])$ for $j=1,\,\ldots,\,r$. We may assume that all quantities are finite, since otherwise we are done.

Take $x\ll [a]$. Since $[a]\leq N[b_1]+\ldots +N[b_r]$, we can find elements $x_j$ such that $x_j\ll [b_j]$ and $x\leq Nx_1+\ldots +Nx_r$. The element $[b_j]$ is weakly $(m,\,n_j)$-divisible, so we can find elements $y_{1,j},\,\ldots,\,y_{n_j,j}$ such that

\[ my_{i,j} \leq [b_j],\text{ and } x_{j}\leq y_{1,j}+ \ldots +y_{n_j,j} \]

for each $i$ and $j$.

Using that $[b_j]\leq [a]$, one has $my_{i,j}\leq [a]$. Further, since we also have $x\leq N\sum _j x_j$, we deduce that $x\leq N\sum _{i,j} y_{i,j}$. It follows that $\operatorname {Div}_m ([a])\leq N(n_1+\ldots +n_r)$, as desired.

Note that, given $a,\,b\in (A\otimes \mathcal {K})_+$, we have $[a+b]\leq [a]+[b]$ and $[a],\,[b]\leq [a+b]$. Thus, (iii) follows directly from (ii).

For (iv), let $x,\,y\in \operatorname {Cu} (A)$ be as stated. As before, we may assume that $\operatorname {div}_m (x)=n$ for some $n\in {\mathbb {N}}$. Take $y'\ll y$, and apply the weak $(m,\,n)$-divisibility of $x$ to obtain elements $y_1,\,\ldots,\,y_n$ satisfying the properties from paragraph 2.3 for $y'$ and $x$.

Since $my_j\leq x$ for every $j$, and $x\wedge \infty y=y$, one obtains $m(y_j\wedge \infty y)\leq x\wedge \infty y=y$. To see that $y'\leq (y_1\wedge \infty y)+\ldots +(y_n\wedge \infty y)$, note that $y'$ is bounded by $y$ (and thus $\infty y$) and by $y_1+\ldots +y_n$. Using that the map $t\mapsto t\wedge \infty y$ preserves addition [Reference Antoine, Perera, Robert and Thiel1, Theorem 2.5(i)], one has

\[ y'\leq (y_1+\ldots +y_n)\wedge\infty y = (y_1\wedge\infty y)+\ldots +(y_n\wedge\infty y), \]

which shows that the elements $y_j\wedge \infty y$ satisfy the conditions from paragraph 2.3 for $y'$ and $y$, as required.

To prove (v), assume that the supremum $\sup _k \operatorname {div}_m (x_k)$ is finite, since otherwise there is nothing to prove. Let $n\in {\mathbb {N}}$ be such that $\sup _k \operatorname {div}_m (x_k)\leq n$, and take $x'\in \operatorname {Cu} (A)$ such that $x'\ll x$. Since $x=\sup _k x_k$, it follows that $x'\ll x_k$ for some $k\in {\mathbb {N}}$. Thus, using that $\operatorname {div}_m (x_k)\leq n$, we obtain elements $y_1,\,\ldots,\,y_n$ with $x'\leq y_1+\ldots +y_n$ and $my_j\leq x_k\leq x$ for each $j$. We get that $\operatorname {div}_m (x)\leq n$, as desired.

Proof. Let $n=\operatorname {div}_m ([a])$, which we may assume to be finite, and take $x\ll k[a]$. Then, there exists $\varepsilon >0$ such that $x\ll k[(a-\varepsilon )_+]$. Since $[(a-\varepsilon )_+]\ll [a]$, there exist $y_1,\,\ldots,\,y_n\in \operatorname {Cu} (A)$ such that

\[ [(a-\varepsilon )_+]\leq y_1+\ldots +y_n,\text{ and } my_j\leq [a] \]

for each $j$.

In particular, one gets

\[ x\ll k[(a-\varepsilon )_+]\leq ky_1+\ldots +ky_n,\text{ and } m(ky_j)\leq k[a], \]

which implies $\operatorname {div}_{km}(k[a])\leq n$, as desired.

Now assume that $A$ has strict comparison of positive elements, and let $n=\operatorname {div}_{k(m+1)}(k[a])$. As before, we may assume $n$ to be finite. Let $x\ll [a]$, which implies $kx\ll k[a]$. Then, there exist $z_1,\,\ldots,\,z_n$ such that $x\leq kx\leq z_1+\ldots +z_n$ and $k(m+1)z_j\leq k[a]$. Thus, one has $(km+1)mz_j\leq km(m+1)z_j\leq (km)[a]$.

Using [Reference Elliott, Robert and Santiago17, Proposition 6.2], we get $mz_j\leq [a]$. This implies $\operatorname {div}_{m} ([a])\leq \operatorname {div}_{k(m+1)}(k[a])$.

Proof. Let $m\in {\mathbb {N}}$. If $A$ is simple and non-elementary, it follows from [Reference Robert38, Proposition 5.2.1] that $\operatorname {Cu} (A)$ is $(m,\,\omega )$-divisible.

Let $x\in \operatorname {Cu} (A)$ be such that $x\ll \infty$. Then, using that $x$ is $(m,\,\omega )$-divisible, there exists a nonzero element $y\in \operatorname {Cu} (A)$ such that $my\leq x$. Using that $x\ll \infty = \infty y$, we find $n\in {\mathbb {N}}$ such that $x\leq ny$. This implies that $\operatorname {div}_m (x)\leq n$. Since the subset of elements compactly contained in $\infty$ is sup-dense in $\operatorname {Cu} (A)$, the result follows.

Now assume that $A$ is not stably finite. Then, the element $\infty$ in $\operatorname {Cu} (A)$ is compact; see, for example, [Reference Thiel46, Lemma 6.21] or the proof of [Reference Bosa7, Theorem 2.6].

It follows that every element in $\operatorname {Cu} (A)$ is compactly contained in $\infty$. Using the argument above, we see that $\operatorname {Cu} (A)_{\operatorname {div}}=\operatorname {Cu} (A)$.

Proof. Take $y\in \operatorname {Cu} (A)$, and assume that there exists $y'\in \operatorname {Cu} (A)$ such that $y'\ll y\ll \infty y'$. Let $M\in {\mathbb {N}}$ be such that $y\leq My'$. Using that $y$ is weakly $(m,\,\omega )$-divisible, one finds $y_1,\,\ldots,\, y_n$ such that $my_i\leq y$ and $y'\leq y_1+\ldots y_n$. Thus, one has $y\leq My_1+\ldots +My_n$, and we get $\operatorname {div}_m (y)\leq Mn$.

Since $A$ is separable and of topological dimension zero, it follows from [Reference Thiel and Vilalta51, Proposition 4.18] that the Cu-semigroup $\operatorname {Cu} (A)\otimes \{ 0,\,\infty \}$ is algebraic. Thus, we know from [Reference Thiel and Vilalta51, Lemma 4.16] that for every pair $x',\,x\in \operatorname {Cu} (A)$ such that $x'\ll x$ there exist $y',\,y\in \operatorname {Cu} (A)$ satisfying $x'\ll y\ll x$ and $y'\ll y\ll \infty y'$. The first part of the proof shows that $\operatorname {div}_m (y)<\infty$ and, consequently, that $\operatorname {Cu} (A)_{\operatorname {div}}$ is dense in $\operatorname {Cu} (A)$.

Proof. If $A$ is nowhere scattered and of nuclear dimension at most $k$, so is $A\otimes \mathcal {K}$.

Let $[a]\in \operatorname {Cu} (A)$. Upon passing to the hereditary $\mathrm {C}^*$-algebra $\overline {a(A\otimes \mathcal {K})a}$, we may assume that $a$ is full in $A\otimes \mathcal {K}$; note that $\overline {a(A\otimes \mathcal {K})a}$ is still nowhere scattered by [Reference Thiel and Vilalta49, Proposition 4.1], and has nuclear dimension at most $k$ by [Reference Winter and Zacharias55, Proposition 2.5].

Recall from [Reference Thiel and Vilalta49, Theorem 3.1] that $A\otimes \mathcal {K}$ is nowhere scattered if and only if it has no finite-dimensional irreducible representations. By [Reference Robert and Tikuisis41, Proposition 3.2], there exists $[b]\in \operatorname {Cu} (A)$ such that

\[ [(a-\varepsilon )_+]\leq 4(k+1)[b],\text{ and } 4(k+1)[b]\leq 2(k+1)[a]. \]

Thus, we obtain that $2(k+1)[(a-\varepsilon )_+]\leq 8(k+1)^2[b]$. Since $2(k+1)[a]$ can be written as the supremum of the elements $2(k+1)[(a-\varepsilon )_+]$ with $\varepsilon \to 0$, it follows that $\operatorname {div}_{4(k+1)} (2(k+1)[a])< 8(k+1)^2$, as desired.

Proof. It follows from [Reference Robert and Rørdam40, Corollary 5.6] that $F(A)$ has no characters if and only if $\operatorname {div}_2 ([1])<\infty$ in $\operatorname {Cu} (F(A))$. Let $n_0=\operatorname {div}_2 ([1])$. [Reference Christensen12, Part II, B, Proposition 2.8] implies that $\operatorname {div}_2 ([a])\leq n_0$ for every $a\in A_+$.

An inductive argument now shows that $\sup _{a\in A_+}\operatorname {div}_m ([a])<\infty$ for every $m\in {\mathbb {N}}$; see, for example, [Reference Thiel and Vilalta51, Lemma 3.4].

Proof. Let $x:=[a]$. By [Reference Thiel and Vilalta48, Proposition 6.1] and [Reference Farah, Hart, Lupini, Robert, Tikuisis, Vignati and Winter20, Proposition 3.8.1], there exists a separable sub-$\mathrm {C}^*$-algebra $B\subseteq A$ that has a surjective rank map, is nowhere scattered, contains $x$ and is such that the induced inclusion map $\operatorname {Cu} (B)\to \operatorname {Cu} (A)$ is an order-embedding.

By definition, every element in $L (F(\operatorname {Cu} (B)))$ can be realized as a rank function. Thus, there exists $y\in \operatorname {Cu} (B)$ such that $\widehat {y}=\frac {1}{3(k+1)}\widehat {x}$. This implies, in particular, that

\[ 2(k+1)\widehat{y}\leq \frac{2}{3}\widehat{x},\quad \text{ and }\quad \widehat{x}\leq 3(k+1)\widehat{y}= \frac{3}{4}(4(k+1))\widehat{y}. \]

Applying $k$-comparison, one obtains $2(k+1)y\leq (k+1)x$ and $x\leq 4(k+1)^2 y$ in $\operatorname {Cu} (B)$. Using that the map $\operatorname {Cu} (B)\to \operatorname {Cu} (A)$ is an order-embedding, the same is true in $\operatorname {Cu} (A)$. Thus, we have

\[ 2(k+1) y\leq (k+1)x\leq 4(k+1)^3 y, \]

as desired.

Proof. It follows from [Reference Ciuperca, Robert and Santiago13] that $\operatorname {Cu} (I)$ can be identified with an ideal of $\operatorname {Cu} (A)$, and that $\operatorname {Cu} (A/I)$ is naturally isomorphic to the quotient $\operatorname {Cu} (A)/\operatorname {Cu} (I)$. We will make these identifications without explicitly writing the isomorphisms.

Let $c\in A_+$, and set $x:=[c]$. Denote by $\operatorname {Cu} (\pi )\colon \operatorname {Cu} (A)\to \operatorname {Cu} (A/I)$ the induced quotient map. Given $x',\,x''$ in $\operatorname {Cu} (A)$ such that $x'\ll x''\ll x$, we can apply the bounded divisibility of $\operatorname {Cu} (A/I)$ to obtain elements $y_j'',\,y_j',\,y_j\in \operatorname {Cu} (A)$ for $j=1,\,\ldots,\,n_0$ such that

\[ (2m_0 )\operatorname{Cu} (\pi ) (y_j)\leq \operatorname{Cu} (\pi ) (m_0 x),\quad \operatorname{Cu} (\pi )(m_0 x'')\leq \operatorname{Cu} (\pi ) (y_1'')+\ldots +\operatorname{Cu} (\pi ) (y_{n_0 }''), \]

and $y_j''\ll y_j '\ll y_j$ for each $j$. Note that this is possible because $\operatorname {Cu} (\pi ) (x)=[\pi (c)]$ and $\operatorname {Cu} (\pi ) (x)$ has a representative in $(A/I)_+$.

Take $w\in \operatorname {Cu} (I)$ such that $w=2w$, and

\[ (2m_0 )y_j'\ll (2m_0 )y_j\leq m_0 x+w,\text{ and } m_0 x''\leq y_1''+\ldots + y_{n_0 }' +w. \]

Using [Reference Thiel and Vilalta49, Proposition 7.8], we find elements $z_j',\, z_j\in \operatorname {Cu} (A)$ satisfying

\[ (2m_0 )z_j\ll m_0 x,\quad y_j''\ll z_j'+w,\quad \text{ and } \quad z_j'\ll z_j \]

for every $j=1,\,\ldots,\,n_0$.

In particular, we have

\[ x'\ll x''\leq m_0 x''\leq z_1'+\ldots + z_{n_0 }' +w. \]

Applying (O6), we obtain an element $\tilde {r}\in \operatorname {Cu} (A)$ such that

\[ x'\ll z_1'+\ldots + z_{n_0 }' +\tilde{r},\quad \text{ and }\quad \tilde{r}\leq x'',w. \]

Note that, since $\tilde {r}\leq x''\ll [c]$ with $c\in A_+$, there exists $b\in A_+$ such that $r:=[b]$ satisfies $r\leq \tilde {r}$ and $x'\ll z_1'+\ldots + z_{n_0 }' +r$; see, for example, [Reference Thiel and Vilalta49, Lemma 3.3(1)]. Further, since we also have $[b]\leq w$, one gets $b\in I_+$. Take $r'\ll r$ such that $x'\leq z_1'+\ldots + z_{n_0}' +r'$.

Now, using that $\operatorname {div}_{2m_1}(m_1[b])\leq n_1$, one gets elements $r_1,\,\ldots,\, r_{n_1}$ such that

\[ (2m_1)r_i\ll m_1 r,\quad \text{ and } \quad m_1 r'\ll r_1+\ldots +r_{n_1} \]

for each $i\leq n_1$.

Thus, it follows that

\[ (m_0 m_1 )x'\leq (m_0 m_1 )z_1'+\ldots +(m_0 m_1)z_{n_0}'+m_0 r_1+\ldots +m_0 r_{n_1} \]

and

\[ (2m_0 m_1)z_j'\leq (m_0 m_1)x,\quad (2m_0 m_1)r_i\leq (m_0 m_1)r\leq (m_0 m_1)x \]

for each $i$ and $j$.

This proves that

\[ \operatorname{div}_{2m_0 m_1}((m_0 m_1)x)\leq m_0(m_1 n_0+n_1), \]

as desired.

The converse follows from a standard argument; see [Reference Thiel and Vilalta51, Proposition 3.9].

Proof. Assume first that $I$ and $A/I$ satisfy the stated conditions. Let $c\in A_+$ and take $\varepsilon >0$. Set

\[ x:=[c],\quad x'':=[(c-\varepsilon )_+],\quad \text{ and } \quad x':=[(c-2\varepsilon )_+]. \]

As before, denote by $\pi$ the induced quotient map $\operatorname {Cu} (A)\to \operatorname {Cu} (A/I)$. Applying $\operatorname {Div}_{4m_1} (m_1\pi (x))\leq n_1$ to $\pi (x'')\ll \pi (x)$, we find $y\in \operatorname {Cu} (A)$ such that

\[ 4m_1\pi (y)\leq m_1\pi (x),\text{ and } m_1\pi (x'')\ll n_1\pi (y). \]

Take $y',\,y''\in \operatorname {Cu} (A)$ such that $y'\ll y''\ll y$ and $m_1\pi (x'')\leq n_1\pi (y')$. Find $w\in \operatorname {Cu} (I)$ such that

\[ 4m_1 y''\ll 4m_1y\leq m_1 x+w,\quad m_1 x''\leq n_1 y'+w,\quad \text{ and } \quad w=2w. \]

Applying [Reference Thiel and Vilalta49, Proposition 7.8] at the first inequality and the pair $y'\ll y''$, we find $z\in \operatorname {Cu} (A)$ satisfying

\[ 4m_1 z\ll m_1 x,\quad \text{ and } \quad y'\ll z+w. \]

Choose $z'\in \operatorname {Cu} (A)$ such that $z'\ll z$ and $y'\ll z'+w$. Note that we have

(4.1)\begin{equation} 4m_1 z + m_1 x\leq 2m_1 x. \end{equation}

Further, using that $y'\ll z'+w$, we obtain

\[ x'\ll x''\leq m_1 x''\leq n_1 y' +w\leq n_1 z' + w. \]

Applying (O6), we find $\tilde {r}\in S$ such that

\[ x'\ll n_1 z' + \tilde{r},\quad \text{ and }\quad \tilde{r}\leq x', w. \]

Proceeding as in the previous proof, note that, since $x''=[(c-\varepsilon )_+ ]$ and $\tilde {r}\leq w\in \operatorname {Cu} (I)$, we can find $a\in I_+$ such that $r:=[a]$ satisfies $r\leq \tilde {r}$ and $x'\ll n_1 z' + r$. Take $r'$ with $r'\ll r$ and $x'\ll n_1 z' + r'$.

Applying $\operatorname {Div}_{4m_0} (m_0 r)\leq n_0$ to $m_0 r'\ll m_0 r$, we get $t\in \operatorname {Cu} (I)$ such that

(4.2)\begin{equation} 4m_0 t\leq m_0 r,\quad \text{ and } \quad m_0 r'\leq n_0 t. \end{equation}

Recall that $r\leq x$. Combining (4.1) and (4.2) we obtain

\[ 4m_0m_1 (z+t)\leq 4m_0m_1 z + m_0m_1 x\leq 2m_0m_1 x. \]

One also gets

\[ 2m_0m_1x'\leq 2m_0m_1(n_1z'+r')\leq 2m_1 (m_0 n_1 z' + n_0 t)\leq 2m_0 m_1 n_0 n_1 (z'+t) , \]

which shows that $\operatorname {Div}_{4m_0m_1}(2m_0m_1 x)\leq 2m_0 m_1 n_0n_1$, as desired.

As in the proof of the previous proposition, we note that the converse follows from similar arguments to those in [Reference Thiel and Vilalta51, Proposition 3.9].

Proof. We know from [Reference Thiel and Vilalta49, Theorem 8.9] that $A$ is nowhere scattered if and only if every element in $\operatorname {Cu} (A)$ is weakly $(2,\,\omega )$-divisible. Using [Reference Thiel and Vilalta49, Theorem 8.9], it follows that every element is $(2n,\,\omega )$-divisible for every $n\in {\mathbb {N}}$. This proves that (i) implies (ii), and it is trivial that (ii) implies (iii).

To see that (iii) implies (i), assume for the sake of contradiction that $A$ is not nowhere scattered. Then, we know from [Reference Thiel and Vilalta49, Theorem 3.1] that there exist ideals $I\subseteq J\subseteq A$ such that $J/I$ is elementary, that is, $\operatorname {Cu} (I/J)\cong \operatorname {Cu} (\mathbb {C})\cong \overline {{\mathbb {N}}}$; see [Reference Engbers19, Theorem 4.4.4].

Note that, since we are assuming that $A$ satisfies (iii), both $I$ and $I/J$ satisfy (iii) as well. Let $\phi$ be an isomorphism from $\operatorname {Cu} (I/J)$ to $\overline {{\mathbb {N}}}$. Now, it follows from [Reference Thiel and Vilalta51, Lemma 3.3] that for every positive element $b\in (I/J)\otimes \mathcal {K}$ there exists $a\in (I/J)_+$ such that $[b]\leq \infty [a]$ in $\operatorname {Cu} (I/J)$. Thus, let $a\in (I/J)_+$ be such that $\phi ([a])\neq 0$. We get $1\ll 1\leq \phi ([a])$ and, using Rørdam's lemma (see e.g. [Reference Thiel46, Theorem 2.30]), there exists $a'\in (I/J)_+$ such that $1=\phi ([a'])$.

Take $n\in {\mathbb {N}}$ such that $[a'\otimes 1_n]$ is weakly $(n+1,\,\omega )$-divisible. Since $\phi$ is an isomorphism, we must have that $\phi ([a'\otimes 1_n])=n$ is also weakly $(n+1,\,\omega )$-divisible. However, the only element $x\in \overline {{\mathbb {N}}}$ such that $(n+1)x\leq n$ is zero. This contradicts the weak $(n+1,\,\omega )$-divisibility of $[a'\otimes 1_n]$.

Thus, $A$ has no nonzero elementary ideal-quotients, as desired.

Proof. Let $n_a,\,n_b\in {\mathbb {N}}$ be such that $\operatorname {div}_{2n_a}(n_a[a]),\,\operatorname {div}_{2n_b}(n_b[b])<\infty$.

Using that $n_a n_b[a+b]=[a\otimes 1_{n_a n_b} + b\otimes 1_{n_a n_b}]$, we can apply lemma 3.3 (iii) and get

\[ \operatorname{div}_{2n_a n_b}(n_an_b[a+b]) \leq \operatorname{div}_{2n_a n_b}(n_an_b[a]) + \operatorname{div}_{2n_a n_b}(n_an_b[b]). \]

Thus, applying lemma 3.4 to both summands, we obtain

\[ \operatorname{div}_{2n_a n_b}(n_an_b[a+b]) \leq \operatorname{div}_{2n_a}(n_a[a]) + \operatorname{div}_{2 n_b}(n_b[b])<\infty , \]

as desired.

Proof. Let $\varepsilon _0>0$. Assume that $n:=\sup _i\sup _{\varepsilon \leq \varepsilon _0} \operatorname {div}_{2m}(m[(a_i-\varepsilon )_+])$ is finite, since otherwise there is nothing to prove. Fix $\varepsilon \leq \varepsilon _0$ positive. By assumption, $(a_i-\tfrac {\varepsilon }{4})_+\otimes 1_m$ is weakly $(2m,\,n)$-divisible for each $i$. Thus, we can find elements $b_{i,j}\in A\otimes \mathcal {K}$ for $j=1,\,\ldots,\, n$ such that

\[ m[(a_i-\varepsilon/3 )_+]\leq [b_{i,1}]+\ldots +[b_{i,n}] ,\quad \text{ and } \quad 2m[b_{i,j}]\leq m[(a_i-\varepsilon/4 )_+] \]

for each $i$ and $j$.

It follows from [Reference Robert and Rørdam40, Lemma 2.3(i)] that there exist elements $c_{i,j}\in M_m(A)_+$ with $c_{i,j}\precsim b_{i,j}$ such that

\[ (a_i - 2\varepsilon/3 )_+{\otimes} 1_m = \sum_{j=1}^n c_{i,j}. \]

Set $d_{i,j}:= ((a_i - 2\varepsilon /3 )_+\otimes 1_m) c_{i,j} ((a_i - 2\varepsilon /3)_+\otimes 1_m)\in M_m (A)$. Note that every entry in $d_{i,j}$ is in $\overline {a_i Aa_i}$, and is thus orthogonal with $e_{i-1}$. To see that the sums $R_j := \sum _{i} d_{i,j}$ are strictly convergent, we proceed as in [Reference Kaftal, Ng and Zhang22, Proposition 4.4]. Using that $c_{i,j}\leq a_i\otimes 1_m$ at the third step, we have

\begin{align*} d_{i,j}^2& = ((a_i - 2\varepsilon /3 )_+{\otimes} 1_m) \left(c_{i,j} ((a_i - 2\varepsilon /3)_+{\otimes} 1_m)^2c_{i,j}\right)((a_i - 2\varepsilon /3 )_+{\otimes} 1_m)\\ & \leq \Vert c_{i,j} \Vert^2 \Vert (a_i - 2\varepsilon /3 )_+{\otimes} 1_m \Vert^3 ((a_i - 2\varepsilon /3 )_+{\otimes} 1_m)\\ & \leq \left(\sup_i \Vert a_i \otimes 1_m \Vert^5\right)((a_i - 2\varepsilon /3 )_+{\otimes} 1_m), \end{align*}

where note that the supremum over $i$ is finite because the sequence $(a_i)_i$ is bounded.

Now, given any $t\in M_m (A)$ and any pair $n< k$, and using that the $d_{i,j}$'s are pairwise orthogonal on $i$ (and that so are the $a_i$'s), one has

\begin{align*} \left\Vert t \sum_{i=n}^{k} d_{i,j}\right\Vert^{2} & = \left\Vert t \left(\sum_{i=n}^{k} d_{i,j}^{2}\right) t^*\right\Vert \\ & \leq \left(\sup_i \Vert a_i \otimes 1_m \Vert^5\right) \left\Vert t\left(\sum_{i=n}^{k} (a_i - 2\varepsilon /3 )_+{\otimes} 1_m\right) t^*\right\Vert\\ & \leq \left(\sup_i \Vert a_i \otimes 1_m \Vert^5\right)\Vert t\Vert \left\Vert t\left(\sum_{i=n}^{k} (a_i - 2\varepsilon /3 )_+{\otimes} 1_m\right)\right\Vert \end{align*}

and, similarly,

\[ \left\Vert \left( \sum_{i=n}^{k} d_{i,j}\right) t\right\Vert^{2}\leq \left(\sup_i \Vert a_i \otimes 1_m \Vert^5\right)\Vert t\Vert \left\Vert \left( \sum_{i=n}^{k} (a_i - 2\varepsilon /3 )_+{\otimes} 1_m \right) t\right\Vert \]

Since $\sum _i a_i$ is strictly convergent, so is $\sum _{i} (a_i - 2\varepsilon /3 )_+\otimes 1_m$. Thus, it follows that $R_j=\sum _i d_{i,j}$ is a strictly convergent sum. Further, note that we have

\[ (R-2\varepsilon/3 )_+^3\otimes 1_m = \sum_{j=1}^n R_j \]

and, therefore, $m[(R-\varepsilon )_+]\ll \sum _{j=1}^n [R_j]$.

Let $\delta >0$ be such that $m[(R-\varepsilon )_+]\leq \sum _{j=1}^n [(R_j -\delta )_+]$. Applying [Reference Kaftal, Ng and Zhang22, Lemma 2.2] at

\[ d_{i,j}\otimes 1_{2m}\precsim c_{i,j}\otimes 1_{2m}\precsim b_{i,j}\otimes 1_{2m}\precsim (a_i-\varepsilon/4 )_+{\otimes} 1_m \]

we find elements $r_{i,j}$ such that

\[ (d_{i,j}-\delta)_+{\otimes} 1_{2m} = r_{i,j}(a_i\otimes 1_m)r_{i,j}^*,\quad \text{ and }\quad \sup_i \Vert r_{i,j}\Vert^2 \leq \frac{4}{\varepsilon} \sup_i \Vert d_{i,j}\otimes 1_{2m}\Vert. \]

Set $x_{i,j}:= (a_i\otimes 1_m)^{1/2}r_{i,j}^* (d_{i,j}\otimes 1_{2m}-\delta )_+$. We get

\[ x_{i,j}^*x_{i,j}= (d_{i,j}-\delta )_+^3\otimes 1_{2m},\quad \text{ and } \quad x_{i,j}x_{i,j}^*\in \overline{(a_i\otimes 1_m) M_m(A) (a_i\otimes 1_m)}, \]

with both $((a_i\otimes 1_m)^{1/2}r_{i,j}^*)_i$ and $( r_{i,j}^* (d_{i,j}\otimes 1_{2m}-\delta )_+)_i$ being bounded sequences for each fixed $j$.

Let $j$ be fixed. To see that $\sum _i x_{i,j}$ is strictly convergent, we argue as before. First, note that one has

\begin{align*} x_{i,j}^*x_{i,j}& \leq \sup_i \Vert d_{i,j}\otimes 1_{2m}\Vert^2 ((d_{i,j}-\delta )_+{\otimes} 1_{2m}),\quad \text{ and } \\ x_{i,j}x_{i,j}^*& \leq \frac{4}{\varepsilon} \sup_i \Vert d_{i,j}\otimes 1_{2m}\Vert^3 (a_i\otimes 1_m), \end{align*}

where each suprema is bounded because $(d_{i,j})_i$ is a bounded sequence.

For any given pair $n< k$ and any $t$, one has

\begin{align*} \left\Vert \left(\sum_{i=n}^{k}x_{i,j}\right)t \right\Vert^2 & \leq \left(\sup_i \Vert d_{i,j}\otimes 1_{2m}\Vert^2\right) \Vert t\Vert \left\Vert \left(\sum_{i=n}^{k}(d_{i,j}-\delta )_+{\otimes} 1_{2m}\right)t \right\Vert,\text{ and } \\ \left\Vert t\left(\sum_{i=n}^{k}x_{i,j}\right) \right\Vert^2& \leq \frac{4}{\varepsilon} \sup_i \Vert d_{i,j}\otimes 1_{2m}\Vert^3 \left\Vert t\left(\sum_{i=n}^{k}a_i\otimes 1_m\right) \right\Vert . \end{align*}

Let $x_{i,j}(s,\,r)$ denote the $(s,\,r)$-th entry of $x_{i,j}$. Then, the sums $x_{j}(s,\,r):=\sum _i x_{i,j}(s,\,r)$ are all strictly convergent. Set $x_{j}:=(x_{j}(s,\,r))_{s,r}\in M_{m,2m^2}(\mathcal {M}(A))$. One gets

\[ x_j^* x_j = (R_j -\delta)_+^3\otimes 1_{2m}\sim (R_j -\delta)_+{\otimes} 1_{2m}, \]

and

\[ x_j x_j^*\in \overline{(R\otimes 1_m)M_m (\mathcal{M}(A) )(R\otimes 1_m)}. \]

It follows that, in $\operatorname {Cu} (\mathcal {M}(A))$, we have $2m[(R_j-\delta )_+]\leq m[R]$ for each $j\in {\mathbb {N}}$. Since $[R]=\sup _{0\leq \varepsilon \leq \varepsilon _0}[(R-\varepsilon )_+]$, we have that $m[R]$ is weakly $(2m,\,n)$-divisible, as desired.

Proof. Let $\pi \colon \mathcal {M}(A)\to \mathcal {M}(A)/A$ denote the quotient map. Using lemmas 5.6 and 5.8, we deduce that each positive element in the corona algebra can be written as $\pi (R_1)+\pi (R_2)$ with $[R_1],\, [R_2]$ having multiples of finite weak divisibility.

By lemma 5.7, every element of the form $[\pi (T)]$ in $\operatorname {Cu} (\mathcal {M}(A)/A)$ with $T\in (\mathcal {M}(A)/A)_+$ has a multiple of finite weak divisibility. Lemma 5.4 shows that the corona algebra is nowhere scattered, as desired.

Proof. Using that nowhere scatteredness works well with extensions (see [Reference Thiel and Vilalta49, Proposition 4.2]), the result is a direct consequence of proposition 5.9.

Proof. Assume that $\mathcal {M}(A)$ is nowhere scattered. [Reference Thiel and Vilalta49, Proposition 4.2] implies that $A$ is nowhere scattered as well.

Conversely, assume that $A$ is nowhere scattered. It follows from example 3.11, proposition 4.1 and proposition 4.3 respectively that (i), (ii) and (iii) imply that every sequence of positive elements (orthogonal or not) in $A_+$ has uniformly bounded multiple divisibility. Thus, we can use theorem 5.11 to get the desired result.

Proof. Using lemma 5.8, we see that the unit $1=\sum _i (e_i - e_{i-1})$ in $\mathcal {M}(A)$ satisfies $\operatorname {div}_{2m} (m[1])<\infty$.

The result now follows from [Reference Robert and Rørdam40, Corollary 5.6].