Examples 2.4

1. (1) By [Reference Herstedt11, Proposition 1.18], zero-dimensional systems in which all points are positively and negatively recurrent are fiberwise essentially minimal. In particular, this includes systems in which orbit closures are minimal, called ‘semisimple’ by Furstenberg in [Reference Furstenberg8]. Thus, examples of fiberwise essentially minimal zero-dimensional systems can be found by looking at the local structure of orbit closures.

2. (2) The following is [Reference Herstedt11, Example 1.21(c)]. Let $ Z = \mathbb {Z} \cup \{ \infty \} $ be the one-point compactification of the integers and let $ ( Y , h' ) $ be an essentially minimal zero-dimensional system. Let $ X = ( Y \times Z ) / ( Y \times \{ \infty \} ) $ and let $ \pi : Y \times Z \to X $ be the quotient map. Let $ \widetilde { h } = \pi ( h' \times \text {id} ) : Y \times Z \to X $ and let $ h : X \to X $ be the continuous map satisfying $ h \circ \pi = \widetilde { h } $ , which is obtained from the universal property of the quotient map. One checks that $ h $ is a homeomorphism. Define $ \widetilde { \psi } : Y \times Z \to Z $ by $ \widetilde { \psi } ( ( y , z ) ) = z $ and then let $ \psi : X \to Z $ be the continuous map satisfying $ \psi \circ \pi = \widetilde { \psi } $ , which is obtained from the universal property of the quotient map. One checks that $ \psi $ itself is a quotient map, and then one checks that $ ( X , h ) $ is a fiberwise essentially minimal zero-dimensional system (using $ Z $ and $ \psi $ as defined above). See this in Figure 2.

   Figure 2

   An illustration of Example 2.4(2). Each fiber over $ z \in Z $ is a copy of $ Y $ , aside from the fiber at $ \infty \in Z $ , which is a singleton (this is pictured as the middle fiber). The picture is meant to depict how the singleton is connected to nearby fibers topologically.

   

Proof. Since $ K_1 ( C ( X ) ) = 0 $ , this follows immediately from [Reference Pimsner and Voiculescu14, Theorem 2.4].

Proof. We check the conditions of Definition 3.1. Conditions (1) and (2) follow from surjectivity of $ \pi $ . For condition (3), let $ g \in K^0 ( X , h )^+ \cap -K^0 ( X , h )^+ $ . This means that there is $ f_1 \in C ( X , \mathbb {Z} )^+ $ such that $ \pi ( f_1 ) = g $ and $ f_2 \in C ( X , \mathbb {Z} )^+ $ such that $ \pi ( - f_2 ) = g $ . However, then $ \pi ( f_1 + f_2 ) = 0 $ and so $ f_1 + f_2 \in \text {im} ( \text {id} - \alpha _* ) $ . Let $ f \in C ( X , \mathbb {Z} ) $ satisfy $ f - \alpha _* ( f ) = f_1 + f_2 $ . Let $ E = f^{ -1 } ( \max _{ x \in X } f ( x ) ) $ . Since $ f - \alpha _* ( f ) \geq 0 $ , we must have $ h ( E ) \subset E $ . Let $ \psi $ be as in Definition 2.2 and let $ z \in \psi ( E ) $ and define $ E_z = E \cap \psi ^{ -1 } ( z ) $ . Since $ h ( E_z ) \subset E_z $ , $ E_z $ is invariant so must intersect the minimal set. However, then by [Reference Herman, Putnam and Skau10, Theorem 1.1], since $E_z\neq \varnothing $ , we have $ \bigcup _{ n \in \mathbb {Z}_{ \geq 0 } } h^n ( E_z ) = \psi ^{ -1 } ( z ) $ , and so $ E_z = \psi ^{ -1 } ( z ) $ . Since this holds for all $ z \in \psi ( E ) $ , we see that $ f $ is constant on $ \psi ^{ -1 } ( z ) $ for all $ z \in Z $ . Since $ \psi ^{ - 1 } ( z ) $ is invariant for all $z\in Z$ , we must have $ f = \alpha _* ( f ) $ , and so $ f_1 + f_2 = 0 $ , and since $ f_1 , f_2 \geq 0 $ , we see $ f_1 = f_2 = 0 $ , and finally we see $ g = 0 $ . This proves condition (3).

Finally, the fact that $ \pi ( 1 ) $ is an order unit is also clear from the surjectivity of $ \pi $ .

Proof. Since $ \ker ( i_* ) = \text {im} ( \text {id} - \alpha _* ) $ , and since $ i : C ( X ) \to C^* ( \mathbb {Z} , X , h ) $ is the natural inclusion, the map $ i_* $ induces a map $ \varphi : K^0 ( X , h ) \to K_0 ( C^* ( \mathbb {Z} , X , h ) ) $ which is an isomorphism of groups and satisfies $ K^0 ( X , h )^+ \subset K_0 ( C^* ( \mathbb {Z} , X , h ) )^+ $ .

Let $ p \in C^* ( \mathbb {Z} , X , h ) $ be a projection. By applying Corollary 2.6 with $ a_1 = p $ and $ \varepsilon = 1/2 $ , $ p $ is unitarily equivalent to $ \chi _U $ for some compact open $ U \subset X $ . Let $ q $ be the image of $ \chi _U $ under the quotient map $ C ( X ) \to C ( X ) / \text {im} ( \text {id} - \alpha ) $ . Then $ [ \varphi ( q ) ] = [ \chi _U ] = [ p ] $ . Repeating this argument for $ M_n ( C^* ( \mathbb {Z} , X , h ) ) $ , we see that $ K_0 ( C^* ( \mathbb {Z} , X , h ) )^+ \subset K^0 ( X , h )^+ $ .

Finally, that $ \varphi ( 1 ) = 1 $ is clear, proving the theorem.

Proof. Let $ ( \mathcal {P}^{ ( n ) } ) $ be a generating sequence of partitions of $ X $ . For each $ n \in \mathbb {Z}_{> 0 } $ , we inductively define systems $ \mathcal {S}^{ ( n ) } = ( T^{ ( n ) } , ( X_t^{ ( n ) } ) , ( K_t^{ ( n ) } ) , ( Y_{ t , k }^{ ( n ) } ) , ( J_{ t , k }^{ ( n ) } ) ) $ of finite first return time maps. First, let $ \mathcal {S}^{ ( 1 ) } = ( T^{ ( 1 ) } , ( X_t^{ ( 1 ) } ) , ( K_t^{ ( 1 ) } ) , ( Y_{ t , k }^{ ( 1 ) } ) , ( J_{ t , k }^{ ( 1 ) } ) ) $ be any system of finite first return time maps subordinate to $ \mathcal {P}^{ ( n ) } $ such that $ \mathcal {P}_1 ( \mathcal {S}^{ ( 1 ) } ) $ is finer than $ \mathcal {P}^{ ( 1 ) } $ and such that $ \bigsqcup _{ t = 1 }^{ T^{ ( 1 ) } } X_t^{ ( 1 ) } \supset Z $ (the former is possible by [Reference Herstedt11, Proposition 1.13] and the latter is possible by [Reference Herstedt11, Lemma 4.12]). Now, let $ n \in \mathbb {Z}_{> 0 } $ and suppose we have chosen $ \mathcal {S}^{ ( n ) } = ( T^{ ( n ) } , ( X_t^{ ( n ) } ) , ( K_t^{ ( n ) } ) , ( Y_{ t , k }^{ ( n ) } ) , ( J_{ t , k }^{ ( n ) } ) ) $ . Let $ \mathcal {S}^{ ( n + 1 ) } = ( T^{ ( n + 1 ) } , ( X_t^{ ( n + 1 ) } ) , ( K_t^{ ( n + 1 ) } ) , ( Y_{ t , k }^{ ( n + 1 ) } ) , ( J_{ t , k }^{ ( n + 1 ) } ) ) $ be any system of finite first return time maps subordinate to $ \mathcal {P}^{ ( n + 1 ) } $ such that $ \mathcal {P}_1 ( \mathcal {S}^{ ( n + 1 ) } ) $ is finer than $ \mathcal {P}^{ ( n + 1 ) } $ and finer than $ \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) $ and such that $ \bigsqcup _{ t = 1 }^{ T^{ ( n + 1 ) } } X_t^{ ( n + 1 ) } \supset Z $ .

Let $ n \in \mathbb {Z}_{> 0 } $ . Let $ A^{ ( n ) } $ be the finite dimensional $ C^* $ -subalgebra of $ C^* ( \mathbb {Z} , X , h ) $ spanned by the matrix units $ u^{ i - j } \chi _{ h^j ( Y_{ t , k }^{ ( n ) } ) } $ for $ t \in \{ 1 , \ldots , T^{ ( n ) } \} $ , $ k \in \{ 1 , \ldots , K_t^{ ( n ) } \} $ , and $ i , j \in \{ 0 , \ldots , J_{ t , k }^{ ( n ) } - 1 \} $ . We see $ A^{ ( n ) } \cong \bigoplus _{ t = 1 }^{ T^{ ( n ) } } \bigoplus _{ k = 1 }^{ K_t^{ ( n ) } } M_{ J_{ t , k }^{ ( n ) } } $ . Notice that $ C ( \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) ) \subset A^{ ( n ) } $ as the diagonal matrices. Set $ Z^{ ( n ) } = \bigsqcup _{ t = 1 }^{ T^{ ( n + 1 ) } } X_t^{ ( n ) } $ and then notice that $ u C ( X \setminus Z^{ ( n ) } ) \subset A^{ ( n ) } $ as the superdiagonal matrices, so $ A^{ ( n ) } $ is generated by $ C ( \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) ) $ and $ u C ( X \setminus Z^{ ( n ) } ) $ .

Notice that $ A^{ ( n ) } \subset A^{ ( n + 1 ) } $ , so we get a directed system of finite dimensional $ C^* $ -algebras, whose limit $ A^{ ( \infty ) } $ contains $ C ( X ) $ since $ ( \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) ) $ is a generating sequence of partitions, and since $ \bigcap _{ n = 1 }^\infty Z^{ ( n ) } = Z $ , we have $ u C ( X \setminus Z^{ ( n ) } ) \to u C_0 ( X \setminus Z ) \subset A_Z $ . It now clear that $ A^{ ( \infty ) } $ is generated by $ C ( X ) $ and $ u C_0 ( X \setminus Z ) $ , and is therefore equal to $ A_Z $ .

Proof. Let $ i : A_Z \to C^* ( \mathbb {Z} , X , h ) $ denote the inclusion map, and let $ i_* : K_0 ( A_Z ) \to K_0 ( C^* ( \mathbb {Z} , X , h ) ) $ denote the map induced by $ i $ on $ K_0 $ . Let $ i_1 : C ( X ) \to A_Z $ denote the canonical inclusion, let $ i_2 : C ( X ) \to C^* ( \mathbb {Z} , X , h ) $ denote the canonical inclusion, and let $ ( i_1 )_* $ and $ ( i_2 )_* $ denote the induced maps on $ K_0 $ . We then clearly have $ i \circ i_1 = i_2 $ . By Corollary 3.5, $ ( i_2 )_* : K_0 ( C ( X ) ) \to K_0 ( C^* ( \mathbb {Z} , X , h ) ) $ is a surjective map between ordered groups, and therefore so is $ i^* $ .

By Proposition 3.2, $ \ker ( ( i_2 )_* ) = \text {ran} ( \text {id} - \alpha _* ) $ . Thus, since $ ( i_2 )_* = ( i_1 )_* \circ i_* $ , we have $ ( i_1 )_* ( \text {ran} ( \text {id} - \alpha _* ) ) \subset \ker ( i_* ) $ . Now suppose that $ a \in \ker ( i_* ) $ . Because $ ( i_1 )_* $ is surjective, we can find $ g \in C ( X , \mathbb {Z} ) $ such that $ ( i_1 )_* ( g ) = a $ . Then $ ( i_2 )_* ( g ) = i_* ( a ) = 0 $ , so $ g \in \ker ( ( i_2 )_* ) = \text {ran} ( \text {id} - \alpha _* ) $ , so $ a \in ( i_1 )_* ( \text {ran} ( \text {id} - \alpha _* ) ) $ . Altogether, we have

(3.1) $$ \begin{align} ( i_1 )_* ( \text{ran} ( \text{id} - \alpha_* ) ) = \ker ( i_* ). \end{align} $$

Let $ ( \mathcal {P}^{ ( n ) } ) $ be a sequence of partitions, let $ ( \mathcal {S}^{ ( n ) } ) $ be a sequence of systems of finite first return time maps, and let $ ( A^{ ( n ) } ) $ be a sequence of subalgebras of $ C^* ( \mathbb {Z} , X , h ) $ as in the proof of Theorem 3.7. We now claim that $ ( i_1 )_* ( \text {ran} ( \text {id} - \alpha _* ) ) = 0 $ . Suppose $ g_1 , g_2 \in C ( X , \mathbb {Z} ) $ satisfy $ g_1|_Z = g_2|_Z $ . Since $ ( \mathcal {P}^{ ( n ) } ) $ is a generating sequence of partitions and for each $ n \in \mathbb {Z}_{> 0 } $ , we have $ \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) $ is finer than $ \mathcal {P}^{ ( n ) } $ , there is some $ n \in \mathbb {Z}_{> 0 } $ such that $ g_1 , g_2 \in A^{ ( n ) } $ and such that $ g_1|_{ Z^{ ( n ) } } = g_2|_{ Z^{ ( n ) } } $ (where $ Z^{ ( n ) } $ is defined in the proof of Theorem 3.7). So $ g_1 - g_2 $ is $ 0 $ on $ Z^{ ( n ) } $ , so we can write $ g_1 - g_2 $ as a linear combination of projections in $ C ( X ) \cap A^{ ( n ) } $ each of which is zero on $ Z^{ ( n ) } $ . So by Lemma 3.8, we have $ [ \alpha ( g_1 - g_2 ) ] = [ g_1 - g_2 ] $ in $ K_0 ( A^{ ( n ) } ) $ , and so $ ( i_1 )_* ( g_1 - \alpha ( g_1 ) ) = ( i_1 )_* ( g_2 - \alpha ( g_2 ) ) $ in $ K_0 ( A_Z ) $ .

So let $ g \in C ( X , \mathbb {Z} ) $ . Define $ f \in C ( X , \mathbb {Z} ) $ by $ f = g \circ \psi $ . Then $ f|_Z = g|_Z $ , and so by the above paragraph, we have $ ( i_1 )_* ( g - \alpha ( g ) ) = ( i_1 )_* ( f - \alpha ( f ) ) $ . However, then notice that $ \alpha ( f ) = g \circ \psi \circ h^{ -1 } = g \circ \psi = f $ , and so $ ( i_1 )_* ( f - \alpha ( f ) ) = 0 $ . Thus, we have $ ( i_1 )_* ( \text {ran} ( \text {id} - \alpha _* ) )~=~0$ . Combining this with equation (3.1), we see $ \ker ( i_* ) = 0 $ . Altogether, we have shown that $ i_* $ is an isomorphism of ordered groups.

Proof. Adopt the notation of Proposition 3.2. Then $ K_1 ( C^* ( \mathbb {Z} , X , h ) ) \cong \ker ( \text {id} - \alpha _* ) $ . Identifying $ K_0 ( C ( X ) ) $ with $ C ( X , \mathbb {Z} ) $ , we may replace $ \alpha _* $ with $ \alpha $ .

Let $ f \in \ker ( \text {id} - \alpha ) $ and let $ z \in Z $ . Suppose $ f|_{ \psi ^{ -1 } ( z ) } $ is not constant. Then there is some $ x \in \psi ^{ -1 } ( z ) $ such that $ f ( z ) \neq f ( x ) $ . Let $ U $ be a compact open subset of $ \psi ^{ -1 } ( z ) $ such that $ f ( U ) = f ( z ) $ . Since $ ( \psi ^{ -1 } ( z ) , h|_{ \psi ^{ -1 } ( z ) } ) $ is an essentially minimal zero-dimensional system, by [Reference Herman, Putnam and Skau10, Theorem 1.1], there is an $ n \in \mathbb {Z}_{> 0 } $ such that $ x \in h^{ -n } ( U ) $ . Let $ x' = h^{ n } ( x ) \in U $ . Then $ f ( x' ) \neq f ( x ) = f ( h^{ -n } ( x' ) ) = \alpha ^n ( f ) ( x' ) $ , and so $ f \neq \alpha ^n ( f ) $ , and so $ f \neq \alpha ( f ) $ , which is a contradiction to $ x \in \ker ( \text {id} - \alpha ) $ . Therefore, $ f|_{ \psi ^{ -1 } ( z ) } $ is constant.

Now suppose $ f \in C ( X , \mathbb {Z} ) $ and suppose $ f|_{ \psi ^{ -1 } ( z ) } $ is constant for each $ z \in Z $ . Then for each $ x \in X $ , we have $ \alpha ( f ) ( x ) = f ( h^{ -1 } ( x ) ) = f ( \psi ( x ) ) = f ( x ) $ , and so $ f = \alpha ( f ) $ , and so $ f \in \ker ( \text {id} - \alpha ) $ .

Thus, we have

$$ \begin{align*} \ker ( \text{id} - \alpha ) = \{ f \in C ( X , \mathbb{Z} ) \,|\, f|_{ \psi^{ -1 } ( z ) } \ \mbox{is constant for each}\ z \in Z \} \cong C ( Z , \mathbb{Z} ) \end{align*} $$

as desired.

Notation 4.2. Let $ B = ( V , E ) $ be a Bratteli diagram. For each $ v \in V $ , we denote $ R ( v ) = r ( s^{ -1 } ( v ) ) $ , and for each $ v \in V \setminus V_0 $ , we denote $ S ( v ) = s ( r^{ -1 } ( v ) ) $ . If $ v \in V_n $ , then $ R ( v ) $ is the set of all vertices in $ V_{ n + 1 } $ that are connected to $ v $ by an edge, and $ S ( v ) $ is the set of all vertices in $ V_{ n - 1 } $ connected to $ v $ by an edge. In a reasonable sense, this gives us range and source maps for vertices.

For each $ k , k' \in \mathbb {Z}_{> 0 } $ with $ k < k' $ , we denote by $ P_{ k , k' } $ the set of all paths from $ V_k $ to $ V_{ k' } $ . Formally, $ P_{ k , k' } $ is the set of $ ( e_{ k + 1 } , \ldots , e_{ k' } ) $ such that for all $ j \in \{ k + 1 , \ldots , k' \} $ , $ e_j \in E_j $ and for all $ j \in \{ k + 1 , \ldots , k' - 1 \} $ , we have $ r ( e_j ) = s ( e_{ j + 1 } ) $ .

Examples 4.9. We provide a couple of examples illustrating the definitions above.

1. (1) Consider the ordered Bratteli diagram $ B $ in Figure 3. Assume the pattern shown in the diagram continues forever. As we can see, there is a linear order on the set of edges that share a range vertex. For example, there are three edges going into v, and those three edges are ordered 1 to 3. There is exactly one minimal infinite path and one maximal infinite path; these are the same path, shown in red. It is not too hard to see that the ordering on $ B $ is perfect. The dynamical system $ ( X_B , h_B ) $ is conjugate to the shift on the one-point compactification of the integers, where the path in red on the right corresponds to the point at $ \infty $ and the path on the left (consisting of all edges labeled ‘2’) corresponds to $ 0 \in \mathbb {Z} $ . This is not the only possible diagram that yields such a Vershik system, but one that reflects a nice choice of a sequence of systems of finite first return time maps.

2. (2) We now show an perfectly ordered Bratteli diagram $ B $ such that $ ( X_B , h_B ) $ is conjugate to the dynamical system in Example 2.4(2), where we take $ Y $ to be $ \mathbb {Z} \cup \{ \infty \} $ and $ h' $ to be the shift. See Figure 4. The straight line down the middle is the infinite path corresponding to the crushed point in Figure 2. We can see a bunch of subtrees branching off that look like Figure 3; these correspond to the fibers, which are conjugate to the shift on the one-point compactification of the integers.

   Figure 4

   An illustration of Example 4.9(2). This is a Bratteli diagram associated to the fiberwise essentially minimal zero-dimensional system in Example 4.9(2).

   

   There are many minimal and maximal paths, and we color these with red. However, since the orbit closures are the fibers, each orbit closure has exactly one minimal and maximal path; these paths are the same, like in Figure 3. Therefore, the order is perfect, and the Vershik system does in fact turn out to be conjugate to the system in Example 2.4(2).

Proof. We first construct $ \mathcal {S}' $ and then use it to modify $ \mathcal {S} $ to obtain $ \mathcal {S}^0 $ . By applying [Reference Herstedt11, Lemma 4.13], we may assume that $ \mathcal {S} $ satisfies its conclusions; in particular, for all $ t \in \{ 1 , \ldots , T \} $ , we have $ \psi ( X_t ) \subset Y_{ t , 1 } $ , and the partitions $ \mathcal {P}_1 ( \mathcal {S} ) $ and $ \mathcal {P}_2 ( \mathcal {S} ) $ are finer than $ \mathcal {P} $ . Let $ \mathcal {P}^{\prime \prime } $ be a partition finer than $ \mathcal {P}' $ , $ \mathcal {P}_1 ( \mathcal {S} ) $ , and $ \mathcal {P}_2 ( \mathcal {S} ) $ . Then apply [Reference Herstedt11, Lemma 3.2] to obtain a system $ \mathcal {S}^{ \prime } = ( T^{ \prime } , ( X_t^{ \prime } ) , ( K_t^{ \prime } ) , ( Y_{ t , k }^{ \prime } ) , ( J_{ t , k }^{ \prime } ) ) $ of finite first return time maps subordinate to $ \mathcal {P}^{\prime \prime } $ such that for all $ t' \in \{ 1 , \ldots , T' \} $ , there is a $ t \in \{ 1 , \ldots , T \} $ such that $ X_{ t^{\prime } }' \subset X_t $ . Since $ \mathcal {P}^{\prime \prime } $ is finer than $ \mathcal {P}' $ , $ \mathcal {S}' $ is also subordinate to $ \mathcal {P}' $ . By applying [Reference Herstedt11, Proposition 1.13], we may assume that $ \mathcal {P}_1 ( \mathcal {S}' ) $ is finer than $ \mathcal {P}^{\prime \prime } $ .

Let $ t' \in \{ 1 , \ldots , T' \} $ and let $ t \in \{ 1 , \ldots , T \} $ be such that $ X_{ t^{\prime } }' \subset X_t $ . Since $ \mathcal {P}^{\prime \prime } $ is finer than $ \mathcal {P}_1 ( \mathcal {S} ) $ , there is some $ k \in \{ 1 , \ldots , K_t \} $ such that

(4.1) $$ \begin{align} X_{ t^{\prime} }' \subset Y_{ t , k }. \end{align} $$

Since $ \mathcal {P}^{\prime \prime } $ is finer than $ \mathcal {P}_2 ( \mathcal {S} ) $ , there is some $ l \in \{ 1 , \ldots , K_t \} $ such that

(4.2) $$ \begin{align} X_{ t^{\prime} }' \subset h^{ J_{ t , l } } ( Y_{ t , l } ). \end{align} $$

Define $ T^0 = T $ and for each $ t \in \{ 1 , \ldots , T \} $ , define $ X_t^0 = X_t $ (note that after finishing this construction, this verifies conclusion (c) of the lemma). Let $ t \in \{ 1 , \ldots , T \} $ and let $ \{ s ( 1 ) , \ldots , s ( N_t ) \} $ be the set of all $ t' \in \{ 1 , \ldots , T' \} $ such that $ X_{ t^{\prime } } \subset X_t $ . For each $ n \in \{ 1 , \ldots , N_t \} $ , let $ \{ a ( n , 1 ) , \ldots , a ( n , C_n ) \} $ be the set of all $ k \in \{ 1 , \ldots , K_t \} $ such that $ {Y_{ t , k } \cap \bigcup _{ j \in \mathbb {Z} } h^j ( X_{ s ( n ) }' ) \neq \varnothing }$ . Define $ K_t^0 = \sum _{ n = 1 }^{ N_t } C_n $ and define $ C_0 = 0 $ . Let $ {k \in \{ 1 , \ldots , K_t^0 \} }$ and let $ n \in \{ 1 , \ldots , N_t \} $ , and $ c \in \{ 1 , \ldots , C_n \} $ be such that $ k = C_{ n - 1 } + c $ . Then define

(4.3) $$ \begin{align} Y_{ t , k }^0 = Y_{ t , a ( n , c ) } \cap \bigcup_{ j \in \mathbb{Z} } h^j ( X_{ s ( n ) }' ) \end{align} $$

and define $ J_{ t , k }^0 = J_{ t , a ( n , c ) } $ . It is routine to verify that $ \mathcal {S}^{ 0 } = ( T^{ 0 } , ( X_t^{ 0 } ) , ( K_t^{ 0 } ) , ( Y_{ t , k }^{ 0 } ) , ( J_{ t , k }^{ 0 } ) ) $ is a system of finite first return time maps subordinate to $ \mathcal {P} $ . By applying [Reference Herstedt11, Proposition 1.9], we may assume that $ \mathcal {P}_1 ( \mathcal {S}^0 ) $ is finer than $ \mathcal {P} $ (note that this proves conclusion (2) of the lemma).

We now verify the conclusions of the lemma. Conclusions (1), (2), and (3) have already been verified. Conclusion (4) follows from equation (4.1) and from the fact that $ \mathcal {P}_1 ( \mathcal {S}^0 ) $ is finer than $ \mathcal {P}_1 ( \mathcal {S} ) $ . Conclusion (5) follows from equation (4.1) and the fact that that $ \mathcal {P}_2 ( \mathcal {S}^0 ) $ is finer than $ \mathcal {P}_2 ( \mathcal {S} ) $ . Conclusion (6) is shown by equation (4.3). This proves the lemma.

Proof. Let $ Z $ and $ \psi $ correspond to $ ( X , h ) $ as in Definition 2.2, and let $ ( \mathcal {P}^{ ( n ) } ) $ be a generating sequence of partitions of $ X $ . We will construct an ordered Bratteli diagram $ B = ( V , E , \leq ) $ such that $ ( X_B , h_B ) $ is conjugate to $ ( X , h ) $ via a map $ F : X \to X_B $ that satisfies $ F ( Z ) = X_{ B , \min } $ .

First, we construct a sequence $ ( \mathcal {S}^{ ( n ) } ) $ of finite first return time maps subordinate to $ ( \mathcal {P}^{ ( n ) } ) $ . First, let $ \mathcal {S}^{ ( 1 ) \prime } $ be any system of finite first return time maps subordinate to $ \mathcal {P}^{ ( 1 ) } $ such that $ \mathcal {P}_1 ( \mathcal {S}^{ ( 1 ) \prime } ) $ is finer than $ \mathcal {P}^{ ( 1 ) } $ (such a system exists by [Reference Herstedt11, Proposition 1.13]). We construct the other systems inductively. For each $ n \in \mathbb {Z}_{> 0 } $ , we apply Lemma 4.10 with $\mathcal {S}^{(n)\prime }$ in place of $\mathcal {S}$ , $\mathcal {P}^{(n)}$ in place of $\mathcal {P}$ , $\mathcal {P}^{( n + 1 )}$ in place of $ \mathcal {P}'$ , and get $ \mathcal {S}^{ ( n ) } $ (that is, $\mathcal {S}^0 $ in the lemma) and $ \mathcal {S}^{ ( n + 1 )\prime } $ (that is, $\mathcal {S}'$ in the lemma) satisfying the conclusions of the lemma. Thus, to construct the sequence of systems, we only need to define $\mathcal {S}^{(0)}$ by $T^{(0)}=1$ , $X_1^{(0)}= X$ , $K_1^{(0)}= 1$ , $Y_{1,1}^{(0)} = 1$ , and $J_{1,1 }^{(0)} = 1$ .

Now we begin to define $ B $ . For each $ n \in \mathbb {Z}_{ \geq 0 } $ , define

$$ \begin{align*} V_n = \{ ( n , t , k ) \,|\, t \in \{ 1 , \ldots , T^{ ( n ) } \} \ \mbox{and}\ k \in \{ 1 , \ldots , K_t^{ ( n ) } \} \}. \end{align*} $$

The set of edges from $ ( n , t , k ) \in V_n $ to $ ( n + 1 , t' , k' ) \in V_{ n + 1 } $ is the set of all $ ( n + 1 , t' , k' , j ) $ such that $ h^j ( Y_{ t^{\prime } , k' }^{ ( n + 1 ) } ) \subset Y_{ t , k }^{ ( n ) } $ . Note that this is well defined; by assumption, $ h^j ( Y_{ t^{\prime } , k' }^{ ( n + 1 ) } ) $ is a subset of an element of $ \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) $ , so we do not need to include $ t $ and $ k $ in the tuple defining this edge. We define an order on the edges $ r^{ -1 } ( ( n , t , k ) ) $ by $ ( n , t , k , j_1 ) \leq ( n , t , k , j_2 ) $ if $ j_1 < j_2 $ .

We now construct the orbit map $ F : X \to X_B $ . Let $ x \in X $ . Then for each $ n \in \mathbb {Z}_{> 0 } $ , there is precisely one $ t \in \{ 1 , \ldots , T^{ ( n ) } \} $ , one $ k \in \{ 1 , \ldots , K_t^{ ( n ) } \} $ , and one $ j \in \{ 0 , \ldots , J_{ t , k }^{ ( n ) } \} $ such that $ x \in h^j ( Y_{ t , k }^{ ( n ) } ) $ . If $ x \in h^j ( Y_{ t , k }^{ ( n ) } ) \cap h^{ j' } ( Y_{ t^{\prime } , k' }^{ ( n + 1 ) } ) $ , then $ j' \geq j $ , since otherwise we would have $ h^{ j - j' } ( Y_{ t , k }^{ ( n ) } ) \subset X_t^{ ( n + 1 ) } $ (this follows from Lemma 4.10(3) and (4)), which is not possible since by definition, $ h^i ( Y_{ t , k }^{ ( n ) } ) \cap X_t^{ ( n + 1 ) } = \varnothing $ for $ i \in \{ 1 , \ldots , J_{ t , k }^{ ( n ) } - 1 \} $ . This, combined with the fact that $ \mathcal {P}_1 ( \mathcal {S}^{ ( n + 1 ) } ) $ is finer than $ \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) $ , tells us that $ h^{ j' - j } ( Y_{ t^{\prime } , k' }^{ ( n + 1 ) } ) \subset Y_{ t , k }^{ ( n ) } $ , and therefore there is an edge from $ ( n , t , k ) $ to $ ( n + 1 , t' , k' ) $ ; namely, $ ( n + 1 , t' , k' , j' - j ) $ . Thus, this gives us an infinite path in $ X_B $ associated to $ x $ . We define $ F $ by sending $ x $ to this infinite path.

We now show that $ F $ is injective. Suppose $ x , x' \in X $ and $ F ( x ) = F ( x' ) = ( e_1 , e_2 , \ldots ) $ where we write $ e_n = ( n , t_n , k_n , j_n ) $ for $ n \in \mathbb {Z}_{> 0 } $ . By definition, it is clear that for each $ n \in \mathbb {Z}_{> 0 } $ , there are $ i_n , i_n' \in \{ 0 , \ldots , J_{ t_n , k_n }^{ ( n ) } - 1 \} $ such that $ x \in h^{ i_n } ( Y_{ t_n , k_n }^{ ( n ) } ) $ and $ x' \in h^{ i_n' } ( Y_{ t_n , k_n }^{ ( n ) } ) $ . First, notice that $ i_1 = j_1 $ and $ i_1' = j_1 $ by definition of $ F $ . Then, by definition of $ F $ , we have $ j_2 = i_2 - j_1 $ and $ j_2 = i_2' - j_2 $ ; in particular, $ i_2 = i_2' $ . Proceeding like this, we see that $ i_n = i_n' $ for all $ n \in \mathbb {Z}_{> 0 } $ . Since $ ( \mathcal {P}^{ ( n ) } ) $ is a generating sequence of partitions, so is $ ( \mathcal {P}_1 ( \mathcal {S}^{ ( n ) } ) ) $ , and therefore $ \bigcap _{ n = 0 }^\infty h^{ i_n } ( Y_{ t_n , k_n }^{ ( n ) } ) $ contains at most one element of $ X $ . Thus, $ x = x' $ , and therefore $ F $ is injective.

Next, we show that $ F $ is surjective. Let $ e = ( e_1 , e_2 , \ldots ) \in X_B $ and write $ e_n = ( n , t_n , k_n , j_n ) $ for $ n \in \mathbb {Z}_{> 0 } $ . We construct a sequence $ ( i_n ) $ with $ i_n \in \{ 0 , \ldots , J_{ t_n , k_n }^{ ( n ) } - 1 \} $ for all $ n \in \mathbb {Z}_{> 0 } $ such that $ \bigcap _{ n = 0 }^\infty h^{ i_n } ( Y_{ t_n , k_n }^{ ( n ) } ) $ is non-empty and contains the element of $ X $ that $ F $ maps to $ e $ . First, let $ i_1 = j_1 $ . Then, for all $ n \in \mathbb {Z}_{> 1 } $ , let $ i_n = j_n + i_{ n - 1 } $ (note that this can be rewritten as $ i_n = \sum _{ k = 1 }^n j_k $ ). The claim now follows from the definition of $ B $ , and $ F $ is therefore surjective.

So far, we have shown that $ F $ is bijective. We now show that $ F $ is a homeomorphism. Let $ U ( e_1 , \ldots , e_N ) $ be an element of the basis of the topology of $ X_B $ . For each $ n \in \{ 1 , \ldots , N \} $ , write $ e_n = ( n , t_n , k_n , j_n ) $ . For each $ n \in \{ 1 , \ldots , N \} $ , write $ j_n' = \sum _{ k = 1 }^n j_k $ . We claim that if $ x \in h^{ j_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) $ , then $ F ( x ) \in U ( e_1 , \ldots , e_N ) $ . So let $ x \in h^{ j_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) $ . First notice that by definition of $ e_N $ ,

$$ \begin{align*} h^{ j_N } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset Y_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) }. \end{align*} $$

Thus,

$$ \begin{align*} h^{ j_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset h^{ j_{ N - 1 }' } ( Y_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } ) , \end{align*} $$

since $ j_N + j_{ N - 1 }' = j_N' $ . Similarly, we have

$$ \begin{align*} h^{ j_{ N - 1 } } ( Y_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } ) \subset Y_{ t_{ N - 2 } , k_{ N - 2 } }^{ ( N - 2 ) }. \end{align*} $$

Thus, for every $ n \in \{ 1 , \ldots , N \} $ , we have

$$ \begin{align*} x \in h^{ j_n' } ( Y_{ t_n , k_n }^{ ( N ) } ) , \end{align*} $$

and so since $ j_n' - j_{ n - 1 }' = j_n $ , the $ n $ th edge of $ F ( x ) $ is indeed $ e_n $ , and $ F ( x ) \in U ( e_1 , \ldots , e_N ) $ as desired. Next, we claim that if $ x \in X $ satisfies $ F ( x ) \in U ( e_1 , \ldots , e_N ) $ , then $ x \in h^{ j_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) $ . So let $ x \in X $ satisfy $ F ( x ) \in U ( e_1 , \ldots , e_N ) $ . Then for each $ n \in \{ 1 , \ldots , N \} $ , $ x \in h^{ i_n } ( J_{ t_n , k_n }^{ ( n ) } ) $ for some $ i_n \in \{ 0 , \ldots , J_{ t_n , k_n }^{ ( n ) } - 1 \} $ . It is clear that $ i_1 = j_1 $ . Then, notice that $ i_2 $ is such that $ j_2 = i_2 - j_1 $ , and so $ i_2 = j_1 + j_2 = j_2' $ . Repeating this process inductively, we see that $ i_N = j_N' $ , and so $ x \in h^{ j_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) $ as desired. Altogether, this shows that $ F $ is a homeomorphism.

If $ x \in Z $ , then there are sequences of integers $ ( t_n ) $ and $ ( k_n ) $ such that $ x \in \bigcap _{ n = 0 }^\infty Y_{ t_n , k_n }^{ ( n ) } $ . By definition of the order on $ B $ , this means that $ F ( x ) \in X_{ B , \min } $ . Conversely, suppose $ x \in X $ satisfies $ F ( x ) \in X_{ B , \min } $ . Write $ F ( x ) = ( e_1 , e_2 , \ldots ) $ and for $ n \in \mathbb {Z}_{> 0 } $ , write $ e_n = ( n , t_n , k_n , i_n ) $ . Since $ e_n $ is minimal, $ i_n $ is the minimal element of $ \{ 0 , \ldots , J_{ t_n , k_n }^{ ( n ) } - 1 \} $ such that $ h^{ i_n } ( Y_{ t_n , k_n }^{ ( n ) } ) \subset X_{ t_{ n - 1 } }^{ ( n - 1 ) } $ . However, since $ X_{ t_n }^{ ( n ) } \subset X_{ t_{ n - 1 } }^{ ( n - 1 ) } $ , we have $ i_n = 0 $ . Hence, $ x \in \bigcap _{ n = 0 }^\infty Y_{ t_n , k_n }^{ ( n ) } $ . Thus, $ F ( Z ) = X_{ B , \min } $ . Also notice that if $ x \in h^{ -1 } ( Z ) $ , there are sequences of integers $ ( t_n ) $ and $ ( k_n ) $ such that $ x \in \bigcap _{ n = 0 }^\infty h^{ J_{ t_n , k_n }^{ ( n ) } - 1 } ( Y_{ t_n , k_n }^{ ( n ) } ) $ . By definition of the order on $ B $ , this means that $ F ( x ) \in X_{ B , \max } $ . Similarly, the converse holds, and so $ F ( h^{ - 1 } ( Z ) ) = X_{ B , \max } $ .

We now show that $ ( F \circ h )|_{ X \setminus h^{ -1 } ( Z ) } = ( \widetilde { h }_B \circ F )|_{ X \setminus h^{ -1 } ( Z ) } $ . Let $ x \in X \setminus h^{ -1 } ( Z ) $ . For each $ n \in \mathbb {Z}_{> 0 } $ , let $ t_n \in \{ 1 , \ldots , T^{ ( n ) } \} $ , let $ k_n \in \{ 1 , \ldots , K_{ t_n }^{ ( n ) } \} $ , and let $ j_n \in \{ 0 , \ldots , K_{ t_n }^{ ( n ) } - 1 \} $ satisfy $ x \in h^{ j_n } ( Y_{ t_n , k_n }^{ ( n ) } ) $ . Since $ x \notin h^{ -1 } ( Z ) $ , there is some smallest $ N \in \mathbb {Z}_{> 0 } $ such that $ j_N \neq J_{ t_N , k_N }^{ ( N ) } - 1 $ . We have

$$ \begin{align*} h^{ j_N + 1 } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset J_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } ( Y_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } ) \end{align*} $$

and so by Lemma 4.10(4), there is a $ k_{ N - 1 }' \in \{ 1 , \ldots , K_{ t_{ N - 1 } }^{ ( N - 1 ) } \} $ such that

(4.4) $$ \begin{align} h^{ j_N + 1 } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset Y_{ t_{ N - 1 } , k_{ N - 1 }' }^{ ( N - 1 ) }. \end{align} $$

Inductively, we can find for each $ n \in \{ 1 , \ldots , N - 2 \} $ a $ k_n' \in \{ 1 , \ldots , K_{ t_n }^{ ( n ) } \} $ such that $ Y_{ t_{ n + 1 }' , k_{ n + 1 }' }^{ ( n + 1 ) } \subset Y_{ t_{ n }' , k_n' }^{ ( n ) } $ . Since $ j_N + 1 < J_{ t_N , k_N }^{ ( N ) } $ and since for each $ n \in \mathbb {Z}_{> N } $ we have $ X_{ t_n }^{ ( n ) } \subset X_{ t_N }^{ ( N ) } $ , it follows that $ j_n + 1 < J_{ t_N , k_N }^{ ( N ) } $ as well. So let $ k_n' = k_n $ and let $ j_n' = 0 $ for $ n \in \{ 1 , \ldots , N - 1 \} $ , and let $ j_n' = j_n + 1 $ for $ n \in \mathbb {Z}_{ \geq N } $ . Then for each $ n \in \mathbb {Z}_{> 0 } $ , $ h ( x ) \in h^{ j_n' } ( Y_{ t_n , k_n' }^{ ( n ) } ) $ .

Now write $ F ( x ) = ( e_1 , e_2 , \ldots ) $ and $ \widetilde { h }_B ( F ( x ) ) = ( e_1' , e_2' , \ldots ) $ and for each $ n \in \mathbb {Z}_{> 0 } $ , write $ e_n = ( n , s_n , l_n , i_n ) $ and $ e_n' = ( n , s_n' , l_n' , i_n' ) $ . By definition of $ F $ , for all $ n \in \mathbb {Z}_{> 0 } $ , we have $ s_n = t_n $ , $ l_n = k_n $ , and $ i_n = j_n - j_{ n - 1 } $ where $ j_0 = 0 $ . We also see that $ N $ is the smallest element of $ \mathbb {Z}_{> 0 } $ such that $ e_N \notin E_{ \max } $ , so $ e_N' $ is the successor of $ e_N $ , $ ( e_1', \ldots , e_{ N - 1 }' ) $ is the minimal path such that $ r ( e_{ N - 1 }' ) = s ( e_N' ) $ , and $ e_n' = e_n $ for all $ n \in \mathbb {Z}_{> N } $ . In particular, we see that $ s_n' = t_n $ for all $ n \in \mathbb {Z}_{> 0 } $ , $ l_n' = k_n $ for all $ n \in \mathbb {Z}_{ \geq N } $ , $ i_n' = 0 $ for $ n \in \{ 1 , \ldots , N - 1 \} $ , and $ i_n' = i_n = j_n - j_{ n - 1 } $ for all $ n \in \mathbb {Z}_{> N } $ . Observe that $ i_N' $ is the smallest integer greater than $ i_N $ such that $ h^{ i_N' } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset Y_{ t_{ N - 1 } , l_{ N - 1 }' }^{ ( N - 1 ) } $ . Thus, this combined with

(4.5) $$ \begin{align} h^{ i_N } ( Y_{ t_N , k_N }^{ ( N ) } ) \subset Y_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } \end{align} $$

tells us

$$ \begin{align*} i_N' &= i_N + J_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } \\ &= j_N - j_{ N - 1 } + J_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } \\ &= j_N - ( J_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } - 1 ) + J_{ t_{ N - 1 } , k_{ N - 1 } }^{ ( N - 1 ) } \\ &= j_N + 1 \\ &= j_N'. \end{align*} $$

From equations (4.4) and (4.5), we see $ l_{ N - 1 }' = k_{ N - 1 }' $ . Similarly, for $ n \in \{ 2 , \ldots , N - 1 \} $ , since $ i_n' = j_n' = 0 $ , we have $ l_{ n - 1 }' = k_{ n - 1 }' $ . Write $ F ( h ( x ) ) = ( e_1^{\prime \prime } , e_2^{\prime \prime } , \ldots ) $ and for each $ n \in \mathbb {Z}_{> 0 } $ , write $ e_n^{\prime \prime } = ( n , s_n^{\prime \prime } , l_n^{\prime \prime } , i_n^{\prime \prime } ) $ . For each $ n \in \mathbb {Z}_{> 0 } $ , we have $ s_n^{\prime \prime } = t_n' = t_n $ and $ l_n^{\prime \prime } = k_n' $ . For $ n \in \{ 1 , \ldots , N - 1 \} $ , we have $ i_n^{\prime \prime } = i_n' = 0 $ . We also have $ i_N^{\prime \prime } = j_N' - j_{ N - 1 }' = j_N' - 0 = i_N' $ . For $ n \in \mathbb {Z}_{> N } $ , we have

$$ \begin{align*} i_n^{\prime\prime} = j_n' - j_{ n - 1 }' &= j_n + 1 - ( j_{ n - 1 } - 1 ) \\ &= j_n - j_{ n - 1 } \\ &= i_n \\ &= i_n'. \end{align*} $$

Altogether, we see $ F ( h ( x ) ) = \widetilde { h }_B ( F ( x ) ) $ , and so

(4.6) $$ \begin{align} ( F \circ h )|_{ X \setminus h^{ -1 } ( Z ) } = ( \widetilde{ h }_B \circ F )|_{ X \setminus h^{ -1 } ( Z ) }. \end{align} $$

We now show that the order on $ B $ is perfect. For each $ x \in X $ , $ z $ is in the minimal set of the essentially minimal zero-dimensional system $ ( \psi ^{ -1 } ( Z ) , h|_{ \psi ^{ -1 } ( Z ) } ) $ , $ \overline { \text {orb} } ( x ) $ contains exactly one element of $ Z $ and exactly one element of $ h^{ -1 } ( Z ) $ . Now, equation (4.6) combined with $ F ( Z ) = X_{ B , \min } $ and $ F ( h^{ -1 } ( Z ) ) = X_{ B , \max } $ tells us that the ordering on $ B $ is perfect. It is now clear that $ F \circ h = h_B \circ F $ . This proves conclusion (1) of the proposition.

Before we prove the rest, we first prove two claims that will be used a few times in the proof.

Claim ( $\ast $ ): let $ n \in \mathbb {Z}_{> 0 } $ , $ t_n \in \{ 1 , \ldots , T^{ ( n ) } \} $ , $ k_n \in \{ 1 , \ldots , K_{ t_n }^{ ( n ) } \} $ , and $ t_{ n + 1 } \in \{ 1 , \ldots , T^{ ( n + 1 ) } \} $ . If $ X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ) $ , then for any $ k_{ n + 1 } \in \{ 1 , \ldots , K_{ t_{ n + 1 } }^{ ( n + 1 ) } \} $ , there is an $ i_{ n + 1 } \in \{ 1 , \ldots , J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } \} $ such that $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ is a maximal edge with $ s ( e ) = ( n , t_n , k_n ) $ .

We now prove claim ( $\ast $ ). Let $ k_{ n + 1 } \in \{ 1 , \ldots , K_{ t_{ n + 1 } }^{ ( n + 1 ) } \} $ . Then

$$ \begin{align*} h^{ J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } } ( Y_{ t_{ n + 1 } , k_{ n + 1 } } ) \subset X_{ t_{ n + 1 }^{ ( n + 1 ) } } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ), \end{align*} $$

so we have

$$ \begin{align*} h^{ J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } - J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_{ n + 1 } , k_{ n + 1 } } ) \subset Y_{ t_n , k_n }^{ ( n ) }. \end{align*} $$

Set $ i_{ n + 1 } = J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } - J_{ t_n , k_n }^{ ( n ) } $ . Let $ j \in \{ 1 , \ldots , J_{ t_n , k_n }^{ ( n ) } - 1 \} $ . Then

$$ \begin{align*} h^{ i_{ n + 1 } + j } ( Y_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } ) \subset h^j ( Y_{ t_n , k_n }^{ ( n ) } ), \end{align*} $$

and since $ h^j ( Y_{ t_n , k_n }^{ ( n ) } ) \cap ( \bigsqcup _{ t = 1 }^{ T^{ ( n ) } } X_t^{ ( n ) } ) = \varnothing $ , we indeed see that $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) \in ~E_{ \max } $ . This proves claim ( $\ast $ ).

Claim ( $\ast \ast $ ): if there is a $ k_{ n + 1 } \in \{ 1 , \ldots , K_{ t_{ n + 1 } }^{ ( n + 1 ) } \} $ and a $ i \in \{ 1 , \ldots , J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } \} $ such that $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ is a maximal edge with $ s ( e ) = ( n , t_n , k_n ) $ , then $ X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ) $ .

We now prove claim ( $\ast \ast $ ). Since $ s ( e ) = ( n , t_n , k_n ) $ , we have

$$ \begin{align*} h^{ i_{ n + 1 } } ( Y_{ t_{ n + 1 }, k_{ n + 1 } }^{ ( n + 1 ) } ) \subset Y_{ t_n , k_n }^{ ( n ) }. \end{align*} $$

Since $ e $ is maximal, there is no $ j \in \{ i_{ n + 1 } + 1 , \ldots , J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } - 1 \} $ with $ h^j ( Y_{ t_{ n + 1 }, k_{ n + 1 } }^{ ( n + 1 ) } ) \subset X_{ t_n , k_n }^{ ( n ) } $ . However, notice that

$$ \begin{align*} h^{ i_{ n + 1 } + J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } ) \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ) \subset X_{ t_n , k_n }^{ ( n ) }. \end{align*} $$

Thus, we must have $ i_{ n + 1 } + J_{ t_n , k_n }^{ ( n ) } = J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } $ . Thus, $ X_{ t_{ n + 1 } }^{ ( n + 1 ) } \cap h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ) \neq \varnothing $ , and so by Lemma 4.10(5), we actually have $ X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ) $ . This proves claim ( $\ast \ast $ ).

We now prove conclusion (2) of the proposition. Let $ v \in V_{ \min } $ . Write $ v = ( n , t_n , k_n ) $ . Since $ v \in V_{ \min } $ , there is some $ v^{\prime \prime } \in V_{ n + 1 } $ and some $ e' \in E_{ \min } $ with $ s ( e' ) = v $ and $ r ( e' ) = v^{\prime \prime } $ . Write $ v^{\prime \prime } = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } ) $ and then $ e' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ . By Lemma 4.10(4), since $ e' \in E_{ \min } $ , we have $ i_{ n + 1 } = 0 $ , so $ Y_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) } $ . Again by Lemma 4.10(4), we have

(4.7) $$ \begin{align} X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) }. \end{align} $$

Now, let $ t_{ n + 2 } \in \{ 1 , \ldots , T^{ ( n + 2 ) } \} $ satisfy

$$ \begin{align*} X_{ t_{ n + 2 } }^{ ( n + 2 ) } \cap \bigg( \bigcup_{ j \in \mathbb{Z} } h^j ( X_{ t_{ n + 1 } }^{ ( n + 1 ) } ) \bigg) \neq \varnothing. \end{align*} $$

By Lemma 4.10(4), there is actually a $ k_{ n + 1 }' \in \{ 1 , \ldots , K_{ t_{ n + 1 } }^{ ( n + 1 ) } \} $ such that

$$ \begin{align*} X_{ t_{ n + 2 } }^{ ( n + 2 ) } \subset Y_{ t_{ n + 1 } , k_{ n + 1 }' }^{ ( n + 1 ) }. \end{align*} $$

So for any $ k_{ n + 2 } \in \{ 1 , \ldots , K_{ t_{ n + 2 } }^{ ( n + 2 ) } \} $ , we have a minimal edge $ e^{\prime \prime } = ( n + 2 , t_{ n + 2 } , k_{ n + 2 } , 0 ) $ with $ s ( e^{\prime \prime } ) = v' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 }' ) $ . Thus, $ v' \in V_{ \min } $ . By equation (4.7), $ Y_{ t_{ n + 1 } , k_{ n + 1 }' }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) } $ , and so there is a minimal edge $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 }' , 0 ) $ with $ s ( e ) = v $ and $ r ( e ) = v' $ .

Now let $ v \in V_{ \max } $ and write $ v = ( n , t_n , k_n ) $ . Since $ v \in V_{ \max } $ , there is some $ v^{\prime \prime } \in V_{ n + 1 } $ and some $ e' \in E_{ \max } $ with $ s ( e' ) = V $ and $ r ( e' ) = v^{\prime \prime } $ . Write $ v^{\prime \prime } = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } ) $ and then $ e' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ . By claim $(\ast \ast )$ , we have

$$ \begin{align*} X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ). \end{align*} $$

Now, let $ t_{ n + 2 } \in \{ 1 , \ldots , T^{ ( n + 2 ) } \} $ satisfy

$$ \begin{align*} X_{ t_{ n + 2 } }^{ ( n + 2 ) } \cap \bigg( \bigcup_{ j \in \mathbb{Z} } h^j ( X_{ t_{ n + 1 } }^{ ( n + 1 ) } ) \bigg) \neq \varnothing. \end{align*} $$

By Lemma 4.10(5), there is actually a $ k_{ n + 1 }' \in \{ 1 , \ldots , K_{ t_{ n + 1 } }^{ ( n + 1 ) } \} $ such that

$$ \begin{align*} X_{ t_{ n + 2 } }^{ ( n + 2 ) } \subset h^{ J_{ t_{ n + 1 } , k_{ n + 1 }' }^{ ( n + 1 ) } } ( Y_{ t_{ n + 1 } , k_{ n + 1 }' }^{ ( n + 1 ) } ). \end{align*} $$

By claim $(\ast )$ , $ v' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 }' ) \in V_{ \max } $ . By claim $(\ast )$ , there is $ e \in E_{ \max } $ with $ s ( e ) = v $ and $ r ( e ) = v' $ . This completes the proof of conclusion (2).

We now prove conclusion (3) of the proposition. Let $ v \in V_{ \min } $ and $ e \in E_{ \min } $ satisfy $ r ( e ) \in R ( v ) $ . Write $ v = ( n , t_n , k_n ) $ and $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ . Since $ r ( e ) \in R ( v ) $ , there is an edge $ e' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 }' ) $ with $ s ( e' ) = v $ , which tells us there is $ j_{ n + 1 } \in \{ 0 , \ldots , J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } - 1 \} $ such that $ h^{ j_{ n + 1 } } ( Y_{ t_{ n + 1 } , k_{ n + 1 } } ) \subset Y_{ t_n , k_n }^{ ( n ) } $ . By Lemma 4.10(6), $ Y_{ t_n , k_n }^{ ( n ) } \subset \bigcup _{ j \in \mathbb {Z} } h^j ( X_{ t_{ n + 1 } }^{ ( n + 1 ) } ) $ . Since $ v \in V_{ \min } $ , there is some edge $ e^{\prime \prime } = ( n + 1 , t_{ n + 1 } , k_{ n + 1 }^{\prime \prime } , i_{ n + 1 }^{\prime \prime } ) \in E_{ \min } $ with $ s ( e ) = v $ . By Lemma 4.10(4), since $ e^{\prime \prime } \in E_{ \min } $ , we have $ i_{ n + 1 }^{\prime \prime } = 0 $ , so $ Y_{ t_{ n + 1 } , k_{ n + 1 }^{\prime \prime } }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) } $ . Again by Lemma 4.10(4), we actually have

(4.8) $$ \begin{align} X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) }. \end{align} $$

Now, since $ e \in E_{ \min } $ , by Lemma 4.10(4), we have $ i_{ n + 1 } = 0 $ . But by (4.8), we must have $ Y_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } \subset Y_{ t_n , k_n }^{ ( n ) } $ , which means that $ s ( e ) = v $ .

Let $ v \in V_{ \max } $ and $ e \in E_{ \max } $ satisfy $ r ( e ) \in R ( v ) $ . Write $ v = ( n , t_n , k_n ) $ and $ e = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 } ) $ . Since $ r ( e ) \in R ( v ) $ , there is an edge $ e' = ( n + 1 , t_{ n + 1 } , k_{ n + 1 } , i_{ n + 1 }' ) $ with $ s ( e' ) = v $ , which tells us there is $ j_{ n + 1 } \in \{ 0 , \ldots , J_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } - 1 \} $ such that $ h^{ j_{ n + 1 } } ( Y_{ t_{ n + 1 } , k_{ n + 1 } }^{ ( n + 1 ) } ) \subset Y_{ t_n , k_n }^{ ( n ) } $ . By Lemma 4.10(6), $ Y_{ t_n , k_n }^{ ( n ) } \subset \bigcup _{ j \in \mathbb {Z} } h^j ( X_{ t_{ n + 1 } }^{ ( n + 1 ) } ) $ . Since $ v \in V_{ \max } $ , there is some edge $ e^{\prime \prime } = ( n + 1 , t_{ n + 1 } , k_{ n + 1 }^{\prime \prime } , i_{ n + 1 }^{\prime \prime } ) \in E_{ \max } $ with $ s ( e ) = v $ . By claim $(\ast \ast )$ , we have

$$ \begin{align*} X_{ t_{ n + 1 } }^{ ( n + 1 ) } \subset h^{ J_{ t_n , k_n }^{ ( n ) } } ( Y_{ t_n , k_n }^{ ( n ) } ). \end{align*} $$

By claim ( $\ast $ ), $ s ( e ) = v $ . This completes the proof of conclusion (3).

Now we prove conclusion (4) of this lemma. Let $ v \in V_{ \min } $ , let $ m \in \mathbb {Z}_{> 0 } $ , and write $ v = ( n , t_n , k_n ) $ . It is clear that $ v \in ( S^m \circ R^m ) ( v ) $ , and so

$$ \begin{align*} R^m ( v ) \subset ( R^m \circ S^m \circ R^m ) ( v ). \end{align*} $$

Now, let $ w = ( n + m , t_{ n + m } , k_{ n + m } ) \in R^m ( v ) $ . By Lemma 4.10(4) (applied $ m $ times), we have

$$ \begin{align*} \bigcup_{ j \in \mathbb{Z} } h^j ( X_{ t_{ n + m } }^{ ( n + m ) } ) \subset \bigcup_{ j \in \mathbb{Z} } h^j ( X_{ t_n }^{ ( n ) } ). \end{align*} $$

So every element of $ ( S^m \circ R^m ) ( v ) $ has the form $ ( n , t_n , k_n' ) $ for some $ k_n' \in \{ 1 , \ldots , K_{ t_n }^{ ( n ) } \} $ . Let $ w' = ( n + m , t_{ n + m }' , k_{ n + m }' ) \in ( R^m \circ S^m \circ R^m ) ( v ) $ , meaning there is an path $ ( e_1 , \ldots , e_m ) $ with $ s ( e_1 ) = ( n , t_n , k_n' ) $ for some $ k_n' \in \{ 1 , \ldots , K_{ t_n }^{ ( n ) } \} $ and $ r ( e_m ) = w' $ . Write $ e_m = ( n + m , t_{ n + m }' , k_{ n + m }' , i_{ n + m }' ) $ . This means that $ h^{ i_{ n + m }' } ( Y_{ t_{ n + m }' , k_{ n + m }' }^{ ( n + m ) } ) \subset Y_{ t_n , k_n' }^{ ( n ) } $ . However, then by Lemma 4.10(4) (applied $ m $ times), we must have $ Y_{ t_{ n + m }' , k_{ n + m }' }^{ ( n + m ) } \subset Y_{ t_n , k_n }^{ ( n ) } $ . Therefore, $ w' \in R^m ( v ) $ (by a minimal path), as desired. An identical argument using Lemma 4.10(5) in place of conclusion (4) shows that this equation also holds when $ v \in V_{ \max } $ . This proves conclusion (4) of this proposition and therefore finishes the proof of the proposition.

Proof. Let $ ( k_n ) $ be the telescoping sequence corresponding to $ B' $ , so that $ V_n' = V_{ k_n } $ and $ E_n' = P_{ k_n + 1 , k_{ n + 1 } } $ for all $ n \in \mathbb {Z}_{> 0 } $ (where $ k_0 = 0 $ ). This identification induces a map $ \varphi : X_B \to X_{ B' } $ by sending $ e = ( e_1 , e_2 , \ldots ) $ to $ \varphi ( e ) = ( ( e_1 , \ldots , e_{ k_1 } ) , ( e_{ k_1 + 1 } , \ldots , e_{ k_2 } ) , \ldots ) $ . By definition of the induced order on a telescoped Bratteli diagram, $ \varphi $ is a conjugation, and therefore $ B' $ satisfies conclusion (1) of Proposition 4.11.

We now show that conclusion (2) holds. Let $ v \in V_{ n , \min }' $ . What we are looking for is a $ v' \in V_{ n + 1 , \min }' $ and $ e \in E_{ n + 1 , \min }' $ such that $ s ( e ) = v $ and $ r ( e ) = v' $ . We can regard all of this as happening in $ B $ instead of $ B' $ , so that we have $ v \in V_{ k_n , \min } $ , and we are looking for $ v' \in V_{ k_{ n + 1 } , \min } $ and a minimal path $ ( e_{ k_n + 1 } , \ldots , e_{ k_{ n + 1 } } ) \in P_{ k_n + 1 , k_{ n + 1 } } $ . By Proposition 4.11(2), there is a $ e_{ k_n + 1 } \in E_{ k_n + 1 , \min } $ and a $ v_{ k_n + 1 } \in V_{ k_n + 1 , \min } $ with $ s ( e ) = v $ and $ r ( e ) = v_{ k_n + 1 } $ . Proceeding inductively, we construct the desired result, with $ v' = v_{ k_{ n + 1 } } $ . The same argument works for $ V_{ \max }' $ in place of $ V_{ \min }' $ . This proves that conclusion (2) holds.

Next, we prove that conclusion (3) holds. Let $ v \in V_{ n , \min }' $ and let $ e \in E_{ n + 1 , \min }' $ satisfy $ r ( e ) \in R ( v ) $ . We want to show that $ s ( e ) = v $ . We once again regard all of this as happening in $ B $ instead of $ B' $ . What this means is that we have $ v \in V_{ k_n , \min }' $ and a minimal path $ ( e_{ k_n + 1 } , \ldots , e_{ k_{ n + 1 } } ) \in P_{ k_n + 1 , k_{ n + 1 } } $ such that there is some path $ ( e_{ k_n + 1 }' , \ldots , e_{ k_{ n + 1 } }' ) \in P_{ k_n + 1 , k_{ n + 1 } } $ such that $ s ( e_{ k_n + 1 }' ) = v $ and $ r ( e_{ k_{ n + 1 } }' ) = r ( e_{ k_{ n + 1 } } ) $ , and we want to show that $ s ( e_{ k_n + 1 } ) = v $ . Suppose not. Then by Proposition 4.11(3), we have $ r ( e_{ k_n + 1 } ) \notin R ( v ) $ ; in particular, we have $ r ( e_{ k_n + 1 } ) \neq r ( e_{ k_n + 1 }' ) $ . Proceeding like this, we eventually see $r(e_{k_{n + 1}})~\neq ~r(e_{k_{n+1}}')$ , which is a contradiction. Thus, $ s ( e_{ k_n + 1 } ) = v $ . The proof for $ V_{ \max }' $ in place of $ V_{ \min }' $ is analogous. This proves that conclusion (3) holds.

That conclusion (4) holds is immediate; replace $ m $ with $ k_{ n + m } - k_n $ . This completes the proof of the proposition.

Proof. Write $ e = ( e_1 , e_2 , \ldots ) $ and write $ f = ( f_1 , f_2 , \ldots ) $ . Let $ n_1 $ be the largest element of $ \{ 1 , \ldots , k \} $ such that $ e_{ n_1 } \neq f_{ n_1 } $ .

Suppose $ e_{ n_1 } < f_{ n_1 } $ . Then there is an $ N_1 \in \mathbb {Z}_{> 0 } $ such that

$$ \begin{align*} h_B^{ N_1 } ( e ) = ( e_1' , \ldots , e_{ n_1 - 1 }' , f_{ n_1 } , e_{ n_1 + 1 } , e_{ n_1 + 2 } , \ldots ) , \end{align*} $$

where $ ( e_1' , \ldots , e_{ n_1 - 1 }' ) $ is the minimal path from $ v_0 \in V_0 $ to $ s ( f_{ n_1 } ) $ . Now, $ h_B^{ M } ( e ) $ and $ f $ pass through the same vertex at level $ n_1 - 1 $ . So let $ n_2 $ be the largest element of $ \{ 1 , \ldots , n_1 - 1 \} $ such that $ e_{ n_2 }' \neq f_{ n_2 } $ . Clearly, $ e_{ n_2 }' < f_{ n_2 } $ . So repeating the above process, we find an integer $ N_2 \in \mathbb {Z}_{> 0 } $ such that $ h_B^{ N_1 + N_2 } ( e ) = ( e_1^{\prime \prime } , \ldots , e_{ n_2 - 1 }^{\prime \prime } , f_{ n_2 } , \ldots , f_{ n_1 } , e_{ n_1 + 1 } , \ldots ) $ . Repeating this process inductively, we arrive at an integer $ N $ such that $ h_B^N ( e ) = f $ .

Proof. ( $ \Leftarrow $ ). Let $ B_1 $ and $ B_2 $ be the Bratteli diagrams satisfying the conclusions of Proposition 4.11 for $ ( X_1 , h_1 , Z_1 ) $ and $ ( X_2 , h_2 , Z_2 ) $ , respectively. By Theorem 3.4, we have $ K^0 ( X_1 , h_1 ) \cong K^0 ( X_2, h_2 ) $ . By Proposition 4.11(1), we have $ K^0 ( X_{ B_1 } , h_{ B_1 } ) \cong K^0 ( X_{ B_2 } , h_{ B_2 } ) $ . By a slight but trivial extension of [Reference Herman, Putnam and Skau10, Theorem 5.4], we have $ K_0 ( B_1 ) \cong K_0 ( B_2 ) $ . By [Reference Elliott6], we have $ B_1 \sim B_2 $ (in the equivalence class of Bratteli diagrams generated by telescoping and isomorphism), which tells us that there is a (non-ordered) Bratteli diagram $ B $ such that telescoping $ B $ to odd levels yields a telescoping of $ B_1 $ and telescoping $ B $ to even levels yields a telescoping of $ B_2 $ . By replacing $ B_1 $ and $ B_2 $ with their telescopings (which can be done without changing the above due to Proposition 4.13), we may assume that telescoping $ B $ to odd levels yields $ B_1 $ and telescoping $ B $ to even levels yields $ B_2 $ . Let $ B' $ be the telescoping of $ B $ by the sequence $ ( 3n - 2 ) $ , so that telescoping $ B $ to odd levels yields a telescoping of $ B_1 $ by $ ( 3n - 2 ) $ and telescoping $ B $ to even levels yields a telescoping of $ B_2 $ by $ ( 3n - 1 ) $ . Note that by Proposition 4.13, these telescopings of $ B_1 $ and $ B_2 $ also satisfy the conclusions of Proposition 4.11.

We denote by $ V_{ \min }' $ ( $ V_{ \max }' $ ) the minimal (respectively maximal) vertices as inherited by $ B_1 $ and $ B_2 $ . We claim that $ B' $ has the following property:

• (*) let $ v \in V_{ \min }' $ . There is precisely one $ v' \in V_{ \min }' $ with $ v' \in S ( v ) $ .

We now prove property ( $\ast $ ). Let $ v \in V_{ \min , n }' $ and, without loss of generality, suppose that $ n $ is odd. The case $ n = 1 $ is trivial so suppose $ n> 1 $ . View $ v $ as a vertex in $ V_{ \min , 3n - 2 } $ , so that the statement we are trying to prove is that $ S^3 ( v ) $ contains precisely one element. Let $ w \in S ( v ) $ and let $ ( e_1 , e_2 ) $ be a minimal path with $ r ( e_2 ) = w $ . Since $ s ( e_1 ) \in V_{ \min } $ , we have $ S^3 ( v ) \cap V_{ \min } \neq \varnothing $ . Now, suppose $ S^3 ( v ) \cap V_{ \min } $ contains two elements, $ v_1 $ and $ v_2 $ . By Proposition 4.11(3), we have $ R^2 ( v_1 ) \cap R^2 ( v_2 ) = \varnothing $ . By Proposition 4.11(2), $ R^2 ( v_1 ) \cap V_{ \min } \neq \varnothing $ and $ R^2 ( v_2 ) \cap V_{ \min } \neq \varnothing $ . By Proposition 4.11(3) again,

(5.1) $$ \begin{align} R^2 ( R^2 ( v_1 ) \cap V_{ \min } ) \cap R^2 ( R^2 ( v_2 ) \cap V_{ \min } ) = \varnothing. \end{align} $$

Since $ v \in R^3 ( v_1 ) $ and $ v \in R^3 ( v_2 ) $ , we have

(5.2) $$ \begin{align} R^4 ( v_1 ) \cap R^4 ( v_2 ) \neq \varnothing. \end{align} $$

However, now notice that by Proposition 4.11(4), we have $ R^4 ( v_1 ) = R^2 ( R^2 ( v_1 ) \cap V_{ \min } ) $ and $ R^4 ( v_2 ) = R^2 ( R^2 ( v_2 ) \cap V_{ \min } ) $ . Thus, we see that equations (5.1) and (5.2) yield a contradiction. Therefore, $ S^3 ( v ) \cap V_{ \min } $ contains precisely one element. The proof for $ v \in V_{ \max }' $ is analogous. This proves property ( $\ast $ ).

For convenience, replace $ B^{ ( 1 ) } $ with its telescoping by $ ( 3 n - 2 ) $ and replace $ B^{ ( 2 ) } $ with its telescoping by $ ( 3 n - 1 ) $ . Now, let $ e = ( e_1 , e_2 , \ldots ) \in X_{ B^{ ( 1 ) } , \min } $ and let $ ( v_1 , v_2 , \ldots ) $ be its associated vertices. Let $ n \in \mathbb {Z}_{> 1 } $ . We view $ v_n $ as a vertex in $ V_{ 2n - 1 }' $ . By property ( $\ast $ ), there is a unique vertex $ v_{ n - 1 }' \in V_{ \min , 2n - 2 }' $ such that $ v_{ n - 1 }' \in S ( v_n ) $ . We claim that $ v_{ n - 1 } \in S ( v_{ n - 1 }' ) $ . If not, then by property ( $\ast $ ), there is some $ w \in S ( v_{ n - 1 }' ) $ such that $ w \neq v_{ n - 1 } $ . However, then $ R^2 ( w ) \cap R^2 ( v_{ n - 1 } ) \neq \varnothing $ , which is a contradiction again by Proposition 4.11(3). Thus, for each $ n \in \mathbb {Z}_{> 1 } $ , $ v_{ n - 1 }' $ is connected by a path to $ v_n' $ , and there is an $ e_n' \in E_{ \min , n }^{ ( 2 ) } $ with $ s ( e_n' ) = v_{ n - 1 }' $ and $ r ( e_n' ) = v_n' $ by Proposition 4.11(3). Thus, this gives us $ e' = ( e_1' , e_2' , \ldots ) \in X_{ B^{ ( 2 ) } , \min } $ such that for each $ n \in \mathbb {Z}_{> 0 } $ , we have $ s ( e_n' ) = v_{ n - 1 }' $ and $ r ( e_n' ) = v_n' $ .