Proof. (1). Since $v$ is a vertex on $\mu$, we have $v=\mu (p)$ for some $p\leq d(\mu )$. Set $\iota _v{:=} \mu (0,\,p)$ and $\tau _v{:=} \mu (p,\,d(\mu ))$. Then $\mu =\iota _v\tau _v$ and $s(\iota _v)=v=r(\tau _v)$.

(2). By property (1), $s(\iota _v)=v=r(\tau _v)$, so the path $\mu _v{:=} \tau _v\iota _v\in \Lambda$ satisfies $r(\mu _v)=s(\mu _v)$. Suppose $r(\mu _v) \Lambda ^{e_i}$ is non-empty, say $\alpha \in r(\mu _v) \Lambda ^{e_i}$. Then $(\iota _v\alpha )(0,\,e_i)\in r(\mu )\Lambda ^{e_i}$. Since $\mu$ is an initial cycle, it follows that $d(\mu )_i \neq 0$.

(3). Suppose $f\in \Lambda ^{e_i}$ is an edge with range $v$ on $\mu$. Since $r(f)=v=r(\mu _{v})$, we have $f \in r(\mu _{v})\Lambda ^{e_i}$. Now property (2) ensures that $d(\mu _v)_i \neq 0$. Since $d(\mu _v)_i > 0$, there exists a path $\lambda \in s(f)\Lambda ^{\leq d(\mu _v)-e_i}$. Hence $f\lambda \in \Lambda ^{\leq e_i}\Lambda ^{\leq d(\mu _v)-e_i}=\Lambda ^{\leq d(\mu _v)}$. Now using that $\mu _v\in v\Lambda$ is a cycle and that $\Lambda$ has no cycle with an entrance, it follows that $v\Lambda ^{\leq d(\mu _v)}= \{\mu _v\}$. Hence $\mu _v=f\lambda$. Since $f\lambda =\mu _v=\tau _v\iota _v$, we get $f=\tau _v(0,\,e_i)$ if $d(\tau _v)_i>0$ and $f=\iota _v(0,\,e_i)$ if $d(\tau _v)_i=0$.

(4). Fix $n\leq d(\mu )$ and $\lambda \in v\Lambda ^{\leq n}$. If $d(\lambda ) = 0$ then $\lambda = v$ and the statement is trivial, so we may assume that $\lambda \not \in \Lambda ^{0}$. Write $d(\lambda )=e_{i_1}+\ldots +e_{i_m}$ where $i_1\ldots ,\,i_m\in \{1,\,\ldots ,\,k\}$. By the factorization property $\lambda =\lambda _1\ldots \lambda _m$ for some $\lambda _j \in \Lambda ^{e_{i_j}}$. Repeated applications of part (3) give $\lambda _j = \mu _{r(\lambda _j)}(0,\, e_{i_j})$ for $j \le m$. Since $d(\lambda )\leq d(\mu )$, it follows that $\lambda =\mu _{v}(0,\, d(\lambda ))$.

(5). Since $s(\tau _v)=s(\mu )$ we have $s_{\tau _v}^{*}s_{\tau _v}=s_{s(\mu )}$. For $n{:=}d(\tau _v)$ notice that $n\leq d(\mu )$. Fix $\lambda \in v\Lambda ^{\leq n}$. Using part (4) we have $\lambda =\mu _{v}(0,\, d(\lambda ))$. Since $\tau _v\in v\Lambda ^{\leq n}$ and since $d(\lambda )\leq n$, we have $\tau _v=\mu _{v}(0,\, n)=\lambda \mu _{v}(d(\lambda ),\, n)$. But both $\tau _v$ and $\lambda$ belong to $\Lambda ^{\leq n}$, so $\tau _v=\lambda$. Consequently $v\Lambda ^{\leq n}=\{\tau _v\}$ and $s_{\tau _v}s_{\tau _v}^{*}=\sum \nolimits _{\lambda \in v\Lambda ^{\leq n}}s_\lambda s_\lambda ^{*}=s_{v}$.

Proof. For part (1) use $1=\sum \nolimits _{v\in \Lambda ^{0}}s_v=\sum \nolimits _{v\in \Lambda ^{0}}\sum \nolimits _{\lambda \in v\Lambda ^{\leq N}}s_\lambda s_\lambda ^{*}=\sum \nolimits _{\lambda \in \Lambda ^{\leq N}}s_\lambda s_\lambda ^{*}.$ For part (2), we refer to the second paragraph of the proof of [Reference Evans and Sims14, Proposition 5.9].

Proof. Firstly, note that each $\theta _{\lambda ,\nu }$ makes sense because the source of $\lambda \in \Lambda ^{\leq N}(\mu ^{\infty })^{0}$ is a vertex on $\mu$ and $\tau _{s(\lambda )}$ is a path on the initial cycle $\mu$ with range $s(\lambda )$.

Fix $\lambda ,\, \nu ,\,\gamma ,\, \eta \in \Lambda ^{\leq N}(\mu ^{\infty })^{0}$. Clearly, $\theta _{\lambda ,\nu }^{*}=\theta _{\nu ,\lambda }$. We claim that $\theta _{\lambda ,\nu }\theta _{\gamma , \eta }=\delta _{\nu ,\gamma }\theta _{\lambda , \eta }$. To see this let $v{:=}r(\mu )$. Then

(2.1)\begin{equation} s_{\nu\tau_{s(\nu)}}^{*}s_{\gamma\tau_{s(\gamma)}}=s_{\tau_{s(\nu)}}^{*}s_\nu^{*} s_\gamma s_{\tau_{s(\gamma)}} \end{equation}

is non-zero only if $r(\nu )=r(\gamma )$. Since $s_{r(\nu )} = \sum \nolimits _{\alpha \in {r(\nu )}\Lambda ^{\leq N}} s_\alpha s_\alpha ^{*}$, we have $(s_\nu s_\nu ^{*})(s_\gamma s_\gamma ^{*})=\delta _{\nu ,\gamma }s_\nu s_\nu ^{*}$, so if (2.1) is non-zero, then $\nu =\gamma$ and then

\[ s_{\nu\tau_{s(\nu)}}^{*}s_{\gamma\tau_{s(\gamma)}}=\delta_{\nu,\gamma}s_{\nu\tau_{s(\nu)}}^{*}s_{\gamma\tau_{s(\gamma)}}=\delta_{\nu,\gamma} s_{s(\nu\tau_{s(\nu)})}=\delta_{\nu,\gamma}s_{s(\tau_{s(\nu)})}=\delta_{\nu,\gamma}s_{v}. \]

Hence $\theta _{\lambda ,\nu }\theta _{\gamma , \eta }=s_{\lambda \tau _{s(\lambda )}}(\delta _{\nu ,\gamma }s_{v})s_{\eta \tau _{s(\eta )}}^{*}=\delta _{\nu ,\gamma }\theta _{\lambda , \eta }$ as claimed.

Proof. Clearly $A\cong \bigoplus _{k=1}^{r}p^{(k)}Ap^{(k)}$ via $a\mapsto (p^{(1)}ap^{(1)},\,\ldots ,\, p^{(r)}ap^{(r)})$ and inverse $(a^{(1)},\, \ldots ,\, a^{(r)})\mapsto \sum \nolimits _{k=1}^{r} a^{(k)}$. Routine calculations show that for each $k\in \{1,\,\ldots ,\,r\}$, the elements $v_i{:=}e_{i1}^{(k)}$, $i=1,\, \ldots ,\, n_k$, satisfy

\[ v_i^{*}v_j=\delta_{i,j}e_{11}^{(k)} \text{ for } 1\leq i,j\leq n_k, \quad\text{and } p^{(k)}=\sum_{i=1}^{n_k}v_iv_i^{*}. \]

By [Reference Kumjian, Pask and Sims22, Lemma 3.3], $p^{(k)}Ap^{(k)}\cong M_{n_k}(e_{11}^{(k)}Ae_{11}^{(k)})$ completing the proof.

Proof. Evidently, $\operatorname {IC}(\Lambda )/\negthickspace \sim$ is finite since $\Lambda ^{0}$ is finite. Let $I$ be a maximal collection of initial cycles satisfying $(\mu ^{\infty })^{0}\cap (\nu ^{\infty })^{0}=\emptyset$ for any $\mu \neq \nu \in I$. For each initial cycle $\mu \in I$ let $\{\theta _{\lambda ,\nu }^{(\mu )}\}$ be the matrix units of Lemma 2.3. We first prove that $\{\theta _{\lambda ,\nu }^{(\mu )}: \mu \in I \}$ is a system of matrix units, i.e.,

1. (1) $\theta _{\lambda \lambda '}^{(\mu )}\theta _{\lambda '\lambda ''}^{(\mu )}=\theta _{\lambda \lambda ''}^{(\mu )}$;

2. (2) $\theta _{\lambda \lambda '}^{(\mu )}\theta _{\eta \eta '}^{(\nu )}=0$ if $\mu \neq \nu$ or if $\lambda '\neq \eta$;

3. (3) $(\theta _{\lambda \lambda '}^{(\mu )})^{*}=\theta _{\lambda '\lambda }^{(\mu )}$; and

4. (4) $\sum \nolimits _{\mu \in I}\sum \nolimits _{\lambda \in \Lambda ^{\leq N}(\mu ^{\infty })^{0}} \theta _{\lambda \lambda }^{(\mu )}=1$.

We start with property (4). Fix $\mu \in I$ and $\lambda \in \Lambda ^{\leq N}(\mu ^{\infty })^{0}$. Using Lemma 2.1(5), we have $s_{\tau _{s(\lambda )}}s_{\tau _{s(\lambda )}}^{*}=s_{s(\lambda )}$. Hence $\theta _{\lambda ,\lambda }^{(\mu )}= s_{\lambda \tau _{s(\lambda )}}s_{\lambda \tau _{s(\lambda )}}^{*}=s_\lambda s_\lambda ^{*}$. Lemma 2.2(1) now gives

(2.2)\begin{equation} 1=\sum_{\lambda\in \Lambda^{{\leq} N}}s_\lambda s_\lambda^{*}=\sum_{\mu\in I} \sum_{\lambda\in \Lambda^{{\leq} N}(\mu^{\infty})^{0}}s_\lambda s_\lambda^{*}=\sum_{\mu\in I}\sum_{\lambda \in \Lambda^{{\leq} N}(\mu^{\infty})^{0}} \theta_{\lambda\lambda}^{(\mu)}. \end{equation}

Properties (1)–(3) follow from Lemma 2.2 and that $\theta _{\lambda \lambda }^{(\mu )}\theta _{\lambda \lambda }^{(\nu )}=0$ whenever $\mu \neq \nu$ in $I$ (the latter is a consequence of property (4)).

For each $\mu \in I$, define $p^{(\mu )}{:=} \sum \nolimits _{\lambda \in \Lambda ^{\leq N}(\mu ^{\infty })^{0}} \theta _{\lambda \lambda }^{(\mu )}$. Then (2.2) gives $\sum \nolimits _{\mu \in I} p^{(\mu )}=1$. We claim that

\[ A := \left\{a\in C^{*}(\Lambda) : a=\sum_{\mu\in I} p^{(\mu)}ap^{(\mu)}\right\} \]

is all of $C^{*}(\Lambda )$. Clearly $A$ is a closed linear subspace of $C^{*}(\Lambda )$. Fix $\alpha ,\, \beta \in \Lambda ^{\leq N}$ such that $s(\alpha )=s(\beta )$. Using Lemma 2.2(2), it follows that $s(\alpha )\in (\mu ^{\infty })^{0}$ for some $\mu \in I$. Since $s_\alpha s_\alpha ^{*}\leq p^{(\mu )}$, we get

\[ s_\alpha =p^{(\mu)}s_\alpha s_\alpha^{*} s_\alpha =p^{(\mu)}s_\alpha, \quad\text{and}\quad s_\alpha s_\beta^{*}=p^{(\mu)}s_\alpha s_\beta^{*}p^{(\mu)}, \]

so $s_\alpha s_\beta ^{*}\in A$. Using that $\operatorname {span}\{s_\alpha s_\beta ^{*}: \alpha ,\, \beta \in \Lambda ^{\leq N},\, s(\alpha )=s(\beta )\}$ is dense in $C^{*}(\Lambda )$, we get $A=C^{*}(\Lambda )$ as claimed.

For each $\mu \in I$, Lemma 2.1(3)–(4) implies that $r(\mu )\Lambda ^{\leq N}$ contains exactly one path which we denote by $\lambda _\mu$. As in the proof of (4), we have $\theta _{\lambda _\mu ,\lambda _\mu }^{(\mu )}=s_{\lambda _\mu } s_{\lambda _\mu } ^{*}$, so $s_{r(\mu )}= \sum \nolimits _{\lambda \in r(\mu )\Lambda ^{\leq N}} s_\lambda s_\lambda ^{*}=\theta _{\lambda _\mu ,\lambda _\mu }^{(\mu )}$. Identifying $I$ with $\operatorname {IC}(\Lambda )/\negthickspace \sim$ via the map $\mu \mapsto [\mu ]$, Lemma 2.4 provides an isomorphism

\[ C^{*}(\Lambda)\cong \bigoplus_{\mu \in \operatorname{IC}(\Lambda)/{\sim}} M_{\Lambda^{{\leq} N}(\mu^{\infty})^{0}}(s_{r(\mu)}C^{*}(\Lambda)s_{r(\mu)}). \]

To see that each $s_{r(\mu )}C^{*}(\Lambda )s_{r(\mu )}\cong C(\mathbb {T}^{\ell _\mu })$, see the proof of [Reference Evans and Sims14, Proposition 5.9].

Proof. By Theorem 2.5 and property (4) from § 1.1,

\[ sr(C^{*}(\Lambda))= \max_{[\mu] \in \operatorname{IC}(\Lambda)/{\sim}} M_{\Lambda^{{\leq} N}(\mu^{\infty})^{0}}(s_{r(\mu)}C^{*}(\Lambda)s_{r(\mu)}), \]

and each $s_{r(\mu )}C^{*}(\Lambda )s_{r(\mu )}\cong C(\mathbb {T}^{\ell _\mu })$. Now using property (2) from § 1.1, we get

\[ sr(C^{*}(\Lambda))= \max_{[\mu] \in \operatorname{IC}(\Lambda)/{\sim}} \left\lceil {\frac{sr(s_{r(\mu)}C^{*}(\Lambda)s_{r(\mu)})-1}{{\Lambda^{{\leq} N}(\mu^{\infty})^{0}}}}\right\rceil+1. \]

Finally, property (1) from § 1.1, gives $sr(s_{r(\mu )}C^{*}(\Lambda )s_{r(\mu )}) -1=\left \lfloor {\ell _\mu /2}\right \rfloor$ for each $[\mu ] \in \operatorname {IC}(\Lambda )/\negthickspace \sim$, completing the proof.

Proof. To prove (1)$\Rightarrow$(2) let $\mu$ be a cycle with an entrance $\tau$, so $\tau \in r(\mu )\Lambda$ satisfies $d(\tau )\leq d(\mu ^{\infty })$ and $\tau \neq \mu ^{\infty }(0,\,d(\tau ))$. Fix $n\geq 1$ such that $nd(\mu )\geq d(\tau )$. Clearly $\tau \in r(\mu ^{n})\Lambda$. Then $\tau \neq \mu ^{\infty }(0,\,d(\tau ))=\mu ^{n}(0,\,d(\tau ))$, so $\operatorname {MCE}(\mu ^{n},\, \tau )= (\mu ^{n}\Lambda \cap \Lambda ^{d(\mu ^{n})\vee d(\tau )})\cap \tau \Lambda \subseteq \{\mu ^{n}\} \cap \tau \Lambda =\emptyset$. For the proof of (2)$\Rightarrow$(3), see [Reference Evans and Sims14, Corollary 3.8].

The implication (3)$\Rightarrow$(4) follows from [Reference Rørdam, Larsen and Laustsen37, Lemma 5.1.2]. It remains to prove (4)$\Rightarrow$(1). We establish the contrapositive. Suppose that condition (1) does not hold, that is $\Lambda$ has no cycle with an entrance. Theorem 2.5 gives that $C^{*}(\Lambda )$ is isomorphic to a direct sum of matrix algebras over commutative $C^{*}$-algebras, hence stably finite, so condition (4) does not hold.

Proof. Let $\mu$ be an initial cycle in $\Lambda$. By Theorem 2.5 and Theorem 2.6, we have

\[ sr(C^{*}(\Lambda))=\left\lceil \frac{\left\lfloor \frac{\ell_\mu}{2}\right\rfloor}{|\Lambda^{{\leq} (n,\ldots, n)}|}\right\rceil+1, \quad C^{*}(\Lambda)\cong M_{\Lambda^{{\leq} (n,\ldots, n)}}(C(\mathbb{T}^{\ell_\mu})), \]

where $\ell _\mu$ is the rank of the periodicity group associated with $\mu$ (Definition 1.5). By the factorization property $\Lambda$ has no sources, so $\Lambda ^{\leq (n,\ldots , n)}=\Lambda ^{(n,\ldots , n)}$. Lemma 2.1(3) implies that $|\Lambda ^{(n,\ldots , n)}|=n$. By Proposition 3.3, $\ell _\mu =\ell$. Combining these results gives $C^{*}(\Lambda )\cong M_n(C(\mathbb {T}^{\ell }))$, and $sr(C^{*}(\Lambda ))=\lceil \lfloor \ell /2\rfloor /n\rceil +1$.

Proof. Suppose that $sr(C^{*}(\Lambda ))=1$. Then $sr(M_n(C^{*}(\Lambda )))=1$ for each positive integer $n$ [Reference Rieffel34, Theorem 6.1]. Hence each $M_n(C^{*}(\Lambda ))$ is finite [Reference Blackadar2, V.3.1.5], which implies that $C^{*}(\Lambda )$ is stably finite. Since $C^{*}(\Lambda )$ is finite, it does not contain any infinite projections [Reference Rørdam, Larsen and Laustsen37, Lemma 5.1.2]. Hence by [Reference Evans and Sims14, Proposition 5.9], there exist $n \ge 1$ and $l_1,\, \ldots ,\, l_n \in \{0,\, \ldots ,\, k\}$ such that $C^{*}(\Lambda )$ is stably isomorphic to $\bigoplus ^{n}_{i=1} C(\mathbb {T}^{l_i})$. Since $sr(C^{*}(\Lambda ))=1$, we deduce that $sr(\bigoplus _{i=1}^{n} C(\mathbb {T}^{l_i}))=1$, because stable rank 1 for unital $C^{*}$-algebras is preserved my stable isomorphism [Reference Rieffel34, Theorem 3.6]. By property 4 in § 1.1, we have $sr(\bigoplus _{i=1}^{n} C(\mathbb {T}^{l_i}))=\max _{i=1}^{n} C(\mathbb {T}^{l_i})$. For each $i=1,\,\ldots ,\,n$ we use [Reference Rieffel34, Proposition 1.7] to deduce that $sr(C(\mathbb {T}^{l_i}))=\lfloor l_i/2\rfloor +1$, where $\lfloor \cdot \rfloor$ denotes ‘integer part of’. Hence $\max _{i=1}^{n} l_i=1$.

By inspection of the proof of [Reference Evans and Sims14, Proposition 5.9], it is clear that each of the integers $l_i$ is the rank of $\mu$ for some $\mu \in IC(\Lambda )$, so $\max _{\mu \in IC(\Lambda )} \ell _{\mu }\geq \max _{i=1}^{n} l_i=1$. For each $\mu ,\,\nu \in IC(\Lambda )$ define $P_\nu {:=} \sum \nolimits _{v\in (\nu ^{\infty })^{0}}s_v$ and $\mu \sim \nu \Leftrightarrow (\mu ^{\infty })^{0}=(\nu ^{\infty })^{0}$. Since $P_\nu =P_\mu$ whenever $\mu \sim \nu$, the proof of [Reference Evans and Sims14, Proposition 5.9] implies that for each $\mu \in IC(\Lambda )$, we have $\ell _\mu =l_i$ for some $i\in \{1,\,\ldots ,\,n\}$. Consequently, $\max _{\mu \in IC(\Lambda )} \ell _{\mu }= \max _{i=1}^{n} l_i$.

Conversely, suppose $C^{*}(\Lambda )$ is finite and $\max _{\mu \in IC(\Lambda )} \ell _{\mu }=1$. By [Reference Rørdam, Larsen and Laustsen37, Lemma 5.1.2], $C^{*}(\Lambda )$ has no infinite projections. So [Reference Evans and Sims14, Corollary 5.7] implies that $C^{*}(\Lambda )$ is stably isomorphic to $\bigoplus _{i=1}^{n} C(\mathbb {T}^{l_i})$ for some $n\geq 1$ and $l_1,\,\ldots ,\,l_n\in \{0,\,\ldots ,\,k\}$ such that $\max _{\mu \in IC(\Lambda )} \ell _{\mu }= \max _{i=1}^{n} l_i$. By the properties in § 1.1, it follows that

\[ sr(C^{*}(\Lambda))=sr\left(\bigoplus_{i=1}^{n} C(\mathbb{T}^{l_i})\right)=\max_{i=1,\ldots,n} \lfloor l_i/2\rfloor+1=1. \]

Proof. Let $I=\{i\leq k: d(\mu )_i> 0\}$. We must show that $\ell _\mu = | I |$. If $I=\emptyset$ then $\ell _\mu = 0= | I |$, so assume that $I$ is non-empty. By (1.1),

\[ G_\mu=\{m-n: n,m\leq d(\mu^{\infty}), \mu^{\infty}(m)=\mu^{\infty}(n)\}. \]

Since each $n,\,m\leq d(\mu ^{\infty })$ satisfy $n,\,m\in \operatorname {span}_{\mathbb {N}}\{e_i : i\in I\}$, the rank of $G_\mu$ is at most $| I |$. Consequently, it suffices to show $G_\mu$ contains a subgroup of rank $| I |$.

Let $v{:=} \mu ^{\infty }(0)$. We claim that for each colour $i\in I$, there exists a positive integer $m_i$ such that $\mu ^{\infty }(0,\,m_ie_i)=v$. Indeed, since $\Lambda ^{0}$ is finite there exists $m< n$ such that $\mu ^{\infty }(m e_i)=\mu ^{\infty }(n e_i)$. Now using that $\mu ^{\infty }\in \Lambda ^{\leq \infty }$ (see Lemma 2.1) and that for every vertex $w$ on $\mu$ there is a unique path in $w\Lambda ^{\leq \infty }$ (see § 1.7) we get that $\sigma ^{m e_i}(\mu ^{\infty })=\sigma ^{n e_i}(\mu ^{\infty })$. Now for $N{:=}md(\mu )$ it follows that $\sigma ^{N}(\mu ^{\infty })=\mu ^{\infty }$. Since

\[ \mu^{\infty}=\sigma^{N}(\mu^{\infty})=\sigma^{N-me_i+me_i}(\mu^{\infty})=\sigma^{N-me_i+ne_i}(\mu^{\infty})=\sigma^{(n-m)e_i}(\mu^{\infty}), \]

we get $(\mu ^{\infty })((n-m)e_i)=v$. Hence $\mu ^{\infty }$ contains a cycle of degree $(n-m)e_i$ based at $v$. In particular, we can use $m_i{:=}n-m$.

By the preceding paragraph, $\{m_ie_i : i\in I\}\subseteq G_\mu$ is a $\mathbb {Z}$-linearly independent set generating a rank-$|I|$ subgroup of $G_\mu$. So the rank of $G_\mu$ is $| I |$.

Proof. Combine Proposition 2.7, Theorem 3.1, and Proposition 3.3.

Proof. If $\Lambda$ contains no cycles then [Reference Evans and Sims14, Corollary 5.7] gives $C^{*}(\Lambda )\cong M_{\Lambda v}(\mathbb {C})$ for some vertex $v \in \Lambda ^{0}$. Using [Reference Rieffel34, Proposition 1.7 and Theorem 3.6], we obtain that $sr(C^{*}(\Lambda ))=1$.

If $\Lambda$ contains a cycle, then another application of [Reference Evans and Sims14, Corollary 5.7] (see also [Reference Brown, Clark and an Huef4, Remark 5.8]) gives that $C^{*}(\Lambda )$ is purely infinite. Since $C^{*}(\Lambda )$ is unital, simple and purely infinite, it contains two isometries with orthogonal ranges, so [Reference Rieffel34, Proposition 6.5] gives $sr(C^{*}(\Lambda ))=\infty$.

Proof. Fix $v\in \Lambda ^{0}$. Suppose $v\not \in H$. Since $\Lambda$ is saturated, for all $i\leq k$, $\{s(\lambda ):\lambda \in v\Lambda ^{\leq e_i}\}\not \subseteq H$. Clearly $s(v\Lambda ^{\leq m})\not \subseteq H$ for $m=0$. Fix any $m\in \mathbb {N}^{k}\setminus \{0\}$. Set $(n^{(0)},\,v^{(0)},\,\lambda ^{(0)})=(m,\,v,\,v)$. Choose $i$ such that $n^{(0)}_i\neq 0$. Since $v^{(0)}\not \in H$, there exists $\mu ^{(1)}\in v^{(0)}\Lambda ^{\leq e_i}\setminus \Lambda H$. Set $(n^{(1)},\,v^{(1)},\,\lambda ^{(1)})=(n^{(0)}-e_i,\,s(\mu ^{(1)}),\,\lambda ^{(0)}\mu ^{(1)})$. Choose $i$ such that $n^{(1)}_i\neq 0$. Since $v^{(1)}\not \in H$, there exists $\mu ^{(2)}\in v^{(1)}\Lambda ^{\leq e_i}\setminus \Lambda H$. Set $(n^{(2)},\,v^{(2)},\,\lambda ^{(2)})=(n^{(1)}-e_i,\,s(\mu ^{(2)}),\,\lambda ^{(1)}\mu ^{(2)})$.

For each step, $|n^{(i)}|=|m|-i$, so $l=|m|$ satisfies $n^{(l)}=0$. Notice that $\lambda ^{(0)}\in v\Lambda ^{\leq (m-n^{(0)})},\, \lambda ^{(1)}\in v\Lambda ^{\leq (m-n^{(1)})},\, \ldots ,\, \lambda ^{(l)}\in v\Lambda ^{\leq (m-n^{(l)})}$. Hence $\lambda ^{(l)}\in v\Lambda ^{\leq m}$ and $s(\lambda ^{(l)})\not \in H$ so $s(v\Lambda ^{\leq m})\not \subseteq H$.

Proof. Firstly, we show that (1)$\Rightarrow$(3). Suppose (1) and suppose that $H\subseteq \Lambda ^{0}$ is a non-empty hereditary, saturated set. We show that $H=\Lambda ^{0}$. Fix $v\in \Lambda ^{0}$. Since $H$ is non-empty, there exists $w\in H$. By (1) there exists $n\in \mathbb {N}^{k}$ such that $s(v\Lambda ^{\leq n})\subseteq s(w\Lambda )$. Since $H$ is hereditary, $s(v\Lambda ^{\leq n})\subseteq s(H\Lambda ) \subseteq H$. Hence Lemma 5.1 gives $v\in H$.

Now we show that (3)$\Rightarrow$(2). Suppose that (2) fails, that is, there exist $v\in \Lambda ^{0}$, and a sequence $(\lambda _i)$ with $\lambda _i\in \Lambda ^{\leq (1,\ldots , 1)},\, s(\lambda _i)=r(\lambda _{i+1})$ for all $i$ such that for all $i\in \mathbb {N}$ and all $n\leq d(\lambda _i)$, we have $v\Lambda \lambda _i(n) = \emptyset$. Let

\[ H=\{w\in \Lambda^{0}: w\Lambda \lambda_i(n) = \emptyset \text{ for all } i\in \mathbb{N} \text{ and }n\leq d(\lambda_i)\}. \]

Then $H$ is non-trivial as $v\in H$ and hereditary because if $u\Lambda w\neq \emptyset$ then $s(w\Lambda )\subseteq s(u\Lambda )$. To show that $H$ is saturated take $u \in \Lambda ^{0}$ and $j\leq k$ such that $s(u{\Lambda }^{\le e_j})\subseteq H$. We must show that $u\in H$. Assume otherwise for contradiction. We have $u\not \in s(u{\Lambda }^{\le e_j})$ because otherwise $u=s(u)$ belongs to $H$, so $u{\Lambda }^{e_j}\neq \emptyset$. Since $u\not \in H$, there exists $\lambda \in u\Lambda$ such that $s(\lambda )=\lambda _i(n)$ for some $i,\,n$. We claim that $d(\lambda )_j=0$. Indeed, if not, then $\lambda =\mu \mu '$ for some $\mu \in u{\Lambda }^{e_j}$. We then have $s(\mu )\in s(u{\Lambda }^{e_j})\subseteq H$, so $s(\mu )\Lambda \lambda _i(n) = \emptyset$ contradicting $\mu '\in s(\mu )\Lambda \lambda _i(n)$. Since $\Lambda$ is locally convex and $d(\lambda )_j=0$ and since $u\Lambda ^{e_j}\neq \emptyset$, we have $\lambda _i(n)\Lambda ^{e_j}=s(\lambda )\Lambda ^{e_j}\neq \emptyset$. Let $\beta =\lambda _i(n,\, d(\lambda _i))$. Since $\Lambda$ is locally convex, either $d(\beta )_j\neq 0$ or $s(\lambda _i)\Lambda ^{e_j}\neq \emptyset$. Since $\lambda _{i+1}\in \Lambda ^{\leq (1,\ldots ,1)}$ it follows that $d(\beta \lambda _{i+1})\geq e_j$. Now $\lambda ':=\lambda \beta \lambda _{i+1} \in u\Lambda$ satisfies $s(\lambda ')=\lambda _{i'}({n'})$ for some ${i'},\,{n'}$. But then, just as we got $d(\lambda )_j=0$, we deduce $d(\lambda ')_j=0$, a contradiction. So $H$ is saturated, so (3) does not hold.

Finally, we prove (2)$\Rightarrow$(1). Given (2), we suppose that (1) fails, and we derive a contradiction. Since (1) fails, there exist $v,\,w\in \Lambda ^{0}$ such that for all $n\in \mathbb {N}^{k}$, we have $s(w\Lambda ^{\leq n})\not \subseteq s(v\Lambda )$. Set

\[ K=\{u\in \Lambda^{0}: s(u\Lambda^{{\leq} n})\not\subseteq s(v\Lambda) \text{ for all } n\in \mathbb{N}^{k}\}. \]

Fix $u\in K$ and $j\leq k$. We claim that there exists $\mu \in u\Lambda ^{\leq e_j}$ such that $s(\mu )\in K$. Indeed if $s(u\Lambda ^{\leq e_j})\subseteq \Lambda ^{0}\setminus K$, then for each $\mu \in u\Lambda ^{\leq e_j}$ there exists $n_\mu \in \mathbb {N}^{k}$ such that $s(\mu \Lambda ^{n_\mu })\subseteq s(v\Lambda )$. Since $s(v\Lambda )$ is hereditary, it follows that $n=\bigvee _{\mu \in u\Lambda ^{\leq e_j}} n_\mu$ satisfies $s(u\Lambda ^{\leq {n+e_j}})=\bigcup _{\mu \in u\Lambda ^{\leq e_j}}s(\mu \Lambda ^{\leq {n}}) \subseteq s(v\Lambda )$, contradicting $u\in K$.

Since $w\in K$, we can construct a sequence $(\lambda _i)$ such that each $\lambda _i\in \Lambda ^{\leq (1,\ldots , 1)}$, each $s(\lambda _i)=r(\lambda _{i+1})$, and for each $n\leq d(\lambda _i)$, we have $\lambda _i(n)\in K$. By (2), there exist $i$ and $n\leq d(\lambda _i)$ such that $v\Lambda \lambda _i(n) \neq \emptyset$, i.e., such that $s(\lambda _i(n)\Lambda ^{\leq 0})\subseteq s(v\Lambda )$. So $\lambda _i(n)\not \in K$, a contradiction.