Proof. To prove (a) we begin with two preliminary facts with similar proofs. First, we show that for any $z'\in {\tilde {Z}}$ , $\pi _k({\operatorname {\mathrm {Cl}}(o(z',\tilde {g}_k))}) = Z$ . Let $z = \pi _k(z')$ and pick $y\in Z$ . By minimality there exists $g^{n_i}(z)\rightarrow y$ . Lifting and using the compactness of $ S_k$ , there are a subsequence $n_{i'}$ and a $y'\in S_k$ with $\tilde {g}_k^{n_{i'}}(z')\rightarrow y'$ . Thus $g_k^{n_{i'}}(z) = \pi _k(\tilde {g}_k^{n_{i'}}(z'))\rightarrow \pi _k(y')$ and so $y = \pi _k(y')$ .

Second, we show that for any $z',y'\in {\tilde {Z}}$ , there exists a p with $T^p(y')\in {\operatorname {\mathrm {Cl}}(o(z',\tilde {g}_k))}$ . Let $z=\pi _k(z')$ and $y = \pi _k(y')$ . By minimality again, we have $g^{n_i}(z)\rightarrow y$ . Lifting and passing to a subsequence, there are a subsequence $n_{i'}$ and a $y"\in S_k$ with $\tilde {g}_k^{n_{i'}}(z')\rightarrow y"$ . Thus $\pi _k(y") = y$ also, so there is a p with $y" = T_k^p(y')$ and so $T_k^p(y')\in {\operatorname {\mathrm {Cl}}(o(z',\tilde {g}_k))}$ .

Now for the main proof, pick $z'\in {\tilde {Z}}$ and let $Z_1' = {\operatorname {\mathrm {Cl}}(o(z',\tilde {g}_k))}$ , so by the first fact, $\pi _k(Z_1') = Z$ . We now show $Z_1'$ is minimal under $\tilde {g}_k$ . If not, there is a $y'\in Z_1'$ with ${\operatorname {\mathrm {Cl}}(o(y',\tilde {g}_k))} {\subsetneq } {\operatorname {\mathrm {Cl}}(o(z',\tilde {g}_k))}$ . By the second preliminary fact, there is some p with

$$ \begin{align*}\operatorname{\mathrm{Cl}}(o(T_k^p(z'), \tilde{g}_k)) \subset {\operatorname{\mathrm{Cl}}(o(y',\tilde{g}_k))} {\subsetneq} {\operatorname{\mathrm{Cl}}(o(z',\tilde{g}_k))}.\end{align*} $$

Acting by the homeomorphism $T_k^p$ and iterating the strict inclusions, we have

$$ \begin{align*} {\operatorname{\mathrm{Cl}}(o(z',\tilde{g}_k))} &= \operatorname{\mathrm{Cl}}(o(T_k^{pk}(z'), \tilde{g}_k)) {\subsetneq} \operatorname{\mathrm{Cl}}(o(T_k^{p(k-1)}(z'), \tilde{g}_k)) {\subsetneq} \cdots {\subsetneq} \operatorname{\mathrm{Cl}}(o(T_k^p(z'), \tilde{g}_k))\\ &{\subsetneq}\, {\operatorname{\mathrm{Cl}}(o(z',\tilde{g}_k))}, \end{align*} $$

a contradiction, so $\tilde {g}_k$ acting on $Z_1'$ is minimal. Thus since $\tilde {g}_k T_k = T_k \tilde {g}_k$ , $\tilde {g}_k$ acting on each $Z_j' := \tilde {g}_k^j(Z_1')$ is minimal. Now minimal sets either coincide or are disjoint, so there is a least m with $T_k^{m+1}Z_1' = Z_1'$ .

For the proof of (b), assume $\pi _k(Z') = \pi _k(Z")$ . Now $Z:= \pi _k(Z')$ is minimal under g and thus since $Z' \subset \pi _k^{-1}(Z)$ and minimal sets are always disjoint or equal, using (2.1) we have that $Z' = Z_j'$ for some j. Similarly, $Z" = Z_{j'}^{\prime }$ for some $j'$ , and thus $Z' = T^p{Z_1'}$ and $Z" = T^{p'}{Z_1'}$ , and so $Z" = T^{p'-p}(Z')$ as required.

Turning to (c), since the deck group of $S_k$ is ${\mathbb Z}_k$ there is a natural identification of $\pi _k^{-1}(x)$ with ${\mathbb Z}_k$ , with $x_1'$ identified with zero. Since $\tilde {g}_k^q({\tilde {x}}) = T^p{\tilde {x}}$ in ${\mathbb R}$ the induced action of $\tilde {g}_k^q$ on ${\mathbb Z}_k$ is $n \mapsto n+p\,\mathrm {mod}\ k$ . An easy elementary number theory argument yields that this action has exactly $\gcd (p,k)$ distinct orbits. Thus $\tilde {g}_k^q$ has exactly $\gcd (p,k)$ distinct orbits when acting on $\pi ^{-1}_k(x)$ . But $x_i', x_j'\in \pi ^{-1}_k(x)$ are on the same $\tilde {g}_k^q$ orbit if and only if they are on the same $\tilde {g}_k$ orbit and each orbit in $\pi ^{-1}_k(o(x,g))$ contains at least one point from $\pi ^{-1}_k(x)$ . Thus $\tilde {g}_k$ acting on $\pi ^{-1}_k(o(x,g))$ has exactly $\gcd (p,k)$ orbits. The rest of the form of (2.2) follows from part (a).

Proof. For part (a), since $\rho (h) = p/q$ in lowest terms, every periodic point has period q. By the conditions on the derivatives of $h\in {\mathcal {H}}_\ell $ , there are four classes of periodic points.

1. (1) x is unstable with $Dh^q(x)>1$ and $o(x,h)\subset \operatorname {\mathrm {Int}}(P(h))$ .

2. (2) x is superstable with $Dh^q(x) =0$ and $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Int}}(J_i)) \not = \emptyset $ while $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Fr}}(J_i)) = \emptyset $ .

3. (3) x is superstable with $Dh^q(x) =0$ and $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Int}}(J_i)) \not = \emptyset $ while $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Fr}}(J_i)) \not = \emptyset $ and $o(x,h)$ contains both left and right endpoints of flat spots.

4. (4) x is semi-stable with $Dh^q(x) =0$ from one side and $Dh^q(x)>1$ from the other and $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Int}}(J_i)) = \emptyset $ while $o(x, h) \cap (\bigcup _{i=1}^\ell \operatorname {\mathrm {Fr}}(J_i)) \not = \emptyset $ and $o(x,h)$ contains only left or only right endpoints of flat spots.

This implies that all periodic points are isolated so there are finitely many of them.

For $i=1, \ldots , 4$ let $n_i$ be the number of periodic orbits of type (1). Using the fixed point index on $h^q$ , $n_1 = n_2 + n_3$ . Each orbit of type (3) hits two flat spots and each of type (2) and (4) at least one and a flat spot cannot contain multiple periodic orbits, and so $n_1 + 2 n_3 + n_4 \leq \ell $ . Thus the total number of periodic orbits wholly contained on $P(h)$ is $n_1 + n_3 + n_4 = n_2 + 2 n_3 + n_4 \leq \ell $

For part (b) assume first that $h^n(J_i) \cap J_{i'} = \emptyset $ for all $n>0$ and $i'$ . If $J_i$ was loose, there would exist $z_1, z_2\in Z$ with $z_1\leq J_i \leq z_2$ with at least one inequality strict and $Z\cap (z_1, z_2) = \emptyset $ . Thus $h((z_1, z_2))$ is a non-trivial interval with $h^n((z_1, z_2)) \subset P(h)$ for all $n>0$ . This is impossible since h is expanding in $P(h)$ and so $J_i$ must not be loose.

For the converse, say $h^n(J_i) \in J_{i'}$ for some $n>0$ and first note $i=i'$ is impossible since that would imply h has a periodic point. Since $h^n(J_i)$ is a point there exists a non-trivial interval $[z_1, z_2]$ properly containing $J_i$ with $h^n([z_1, z_2]) = J_{i'}$ and so $(z_1, z_2)\cap Z = \emptyset $ , and so $J_i$ is a loose flat spot.

Finally, since $h^n(J_i) \cap J_{i} = \emptyset $ and there are finitely many flat spots there is at least one $J_{i}$ with $h^n(J_i) \cap J_{i'} = \emptyset $ for all $n>0$ and $i'$ .

For (c), assume y is such that $o(y,h)\subset P(h)$ and $o(y,h)\cap Z = \emptyset $ . Let $x,x'\in Z$ with $y\in (x, x')$ and $(x, x')\cap Z = \emptyset $ . Because of the uniform expansion in $P(h)$ there are a flat spot J and an $n\geq 0$ so that $J\subset h^n([x, x'])$ . If $\rho (h)\not \in {\mathbb Q}$ , then by (c) for some $n'$ , $h^{n+n'}([x, x'])$ is a tight flat spot and so $h^{n + n' + 1}(y) \in Z$ , where Z is the semi-Denjoy minimal set given in Lemma 3.1(c).

Now assume $\rho (h) = p/q$ . In this case x and $x'$ are periodic orbits and so $J \subset h^{n+w q}([x, x'])$ for all $w\geq 0$ , and so $ h^{n + wq} (y) \in h^{n+w q}([x, x'])\setminus \operatorname {\mathrm {Int}}(J)$ using the assumption that $o(y,h)\subset P(h)$ . But from (a), $ h^{n + wq} (J) \subset J$ . Thus by monotonicity, $ h^{n + wq} (y)$ is either always in the left component of $[x, x'] \setminus \operatorname {\mathrm {Int}}(J)$ or in the right component. This violates the expansion in $P(h)$ and so, for some $j> 0$ , $h^j(y)\in Z$ , which yields (3.1).

Proof. Parts (a) and (b) are easy to verify. For (c) we prove the case $k=\infty $ which implies the $k < \infty $ cases. Assume ${\hat {\pi }}_\infty ({\underline s}) = {\underline w}$ . The transitions in (5.2) coupled with the structure of ${\underline w}$ imply that once $s_0$ is determined the parity structure of ${\underline s}$ determines all of ${\underline w}$ . Similarly, once $t_0$ is determined all of ${\underline t}$ is determined. Once again (5.2) implies that if $s_0 - t_0 = 2n$ then for all i, $s_i = t_i + 2n$ .

Proof. We prove the $k=\infty $ case; the $k<\infty $ case follows. Theorem 6.1 shows that $\iota _\infty $ is an order-preserving bijection. Since $\iota _\infty T_\infty ^k = {\hat {T}}_\infty ^k \iota _\infty $ , we have that $T_\infty ^k(Z) = Z$ if and only if ${\hat {T}}^k_\infty \iota _\infty (Z) = \iota _\infty (Z)$ . Using the additional fact that $\iota _\infty \tilde {g} = \sigma _\infty \iota _\infty $ we have that $\tilde {g}$ is weakly order-preserving on Z if and only if $\sigma _\infty $ is weakly order-preserving on $\iota _\infty (Z)$

Proof. For part (a) we know that $\iota _k$ is a conjugacy that takes kfsm sets to kfsm sets, which yields that ${\mathcal {N}}_k(g)\rightarrow {\hat {{\mathcal {N}}}}_k(g)$ is a homeomorphism. By hypothesis any $g\in {\mathcal {G}}$ is $C^2$ in $P(g)$ and so there is some $M>1$ with $g'< M$ on $P(g)$ and thus on all address intervals $I_j$ . It is standard that this implies that $\iota _k$ is Hölder with exponent $\nu = \log 2k/\log M$ . This then implies that $\iota _k$ preserves Hausdorff convergence and so ${\mathcal {B}}_k(g)\rightarrow {\hat {{\mathcal {B}}}}_k(g)$ is a homeomorphism.

For part (b), since the space of all $\tilde {g}_k$ -invariant Borel probability measures is compact metric, it suffices to show that ${\mathcal {N}}_k(g)$ is closed, and so assume $\mu _n\in {\mathcal {N}}_k(g)$ and $\mu _n\rightarrow \mu $ weakly with $X_n := \operatorname {\mathrm {supp}}(\mu _n)$ a recurrent kfsm set.

As noted in Remark 7.4 the collection of all kfsm sets in $\Lambda _k$ is compact in the Hausdorff topology and so there exist a kfsm set X and $n_i\rightarrow \infty $ with $X_{n_i}\rightarrow X$ . A standard argument which we give here shows that $\operatorname {\mathrm {supp}}(\mu ) \subset X$ . If this inclusion does not hold, there exists an $x\in \operatorname {\mathrm {supp}}(\mu ) \cap X^c$ , then let $\epsilon = d(x, X)$ . Since the atoms of $\mu $ are countable, we may find an $\epsilon _1 < \epsilon /4$ so that, letting $U = N_{\epsilon _1}(x)$ , we have that $\mu (\operatorname {\mathrm {Fr}}(U)) = 0$ . Thus U is a continuity set for $\mu $ . It then follows, using the fact that $x\in \operatorname {\mathrm {supp}}(\mu )$ and a standard result (pp. 16–17 of [Reference Billingsley7]), that $\mu _{n_i}(U) \rightarrow \mu (U)>0$ . Thus for large enough i, with $m = n_i$ we have $X_m\subset N_{\epsilon /4}(X)$ and so $\emptyset = U \cap X_m = U\cap \operatorname {\mathrm {supp}}(\mu _m)$ with $\mu _m(U)>0$ a contradiction. Thus $\operatorname {\mathrm {supp}}(\mu ) \subset X$ . Now any invariant measure supported on X must be supported on its recurrent set and so $\mu \in {\mathcal {N}}_k(g)$ , as required. The compactness of ${\hat {{\mathcal {N}}}}_k(g)$ follows from part(a).

Proof. By Theorem 7.2 there exists a continuous, semi-monotone $H:S_k\rightarrow S_k$ which interpolates the action of $\tilde {g}_k$ on Z. Rescaling to the standard circle, let $H_k:S^1\rightarrow S^1$ be defined as $H_k:= D_k\circ H\circ D_k^{-1}$ . By Lemma 3.1(a), $\rho (H_k) = \omega $ is a single number and, since $\rho _k(Z) = k \rho (D Z, H_k)$ , (a) follows. If $\rho _k(Z)\not \in {\mathbb Q}$ then $\rho (D Z, H_k)\not \in {\mathbb Q}$ and so, by Lemma 3.1(e), $D_kZ$ and thus Z is a semi-Denjoy minimal set, yielding (b).

Now assume $\rho _k(Z) = p/q$ in lowest terms and so $\rho (D_k Z, H_k) = p/(qk)$ . Written in lowest terms,

$$ \begin{align*}\frac{p}{qk} = \frac{p/\gcd(p,k)}{kq/\gcd(p,k)}.\end{align*} $$

But since $H_k$ is semi-monotone, its recurrent set is a collection of periodic orbits and its rotation number in lowest terms has their period as its denominator which is thus $q k{/}\kern-1pt\gcd (p,k)$ . Since by assumption, $Z\kern1pt{\subset}\kern1pt \Lambda _k\kern-0.5pt(g)$ we may choose H to have k flat spots. Thus, using Lemma 3.5, Z consists of at least one and at most k periodic orbits, finishing (c).

It is standard from Aubry–Mather theory that $\rho $ is continuous on the collection of all kfsm sets and thus it is continuous restricted to the recurrent kfsm sets. As for measures, since $\rho (\mu ) = \int \Delta _g\; d\mu $ using definition (2.5) with $\Delta _g$ continuous, continuity of ${\hat {\rho }}_k$ follows from the definition of weak convergence.

Proof. We begin by proving portions of (a) and (b). For part (a) we first show that $B_k(\alpha , \vec {\nu })$ is minimal using a characterization usually attributed to Birkhoff. If $f:X\rightarrow X$ is a continuous function of a compact metric space and $x\in X$ , then $\operatorname {\mathrm {Cl}}(o(x,f))$ is a minimal set if and only if for all $\epsilon>0$ there exists an N so that for all $n\in {\mathbb N}$ there is a $0<i\leq N$ with $d(f^{n+i}, x) < \epsilon $ . Pick x in the good set G. Since $(S^1, R_\alpha )$ is minimal, $o(x, R_\alpha )$ has the given property. Since $\zeta _k$ restricted to G is a homeomorphism and $\zeta _k R_\alpha = \sigma _k \zeta _k$ , $o(\zeta _k(x), R_\alpha )$ has the desired property and, further, $o(\zeta _k(x), R_\alpha )$ is dense in $\zeta _k(G)$ and thus in $B_k(\alpha , \vec {\nu }) = \operatorname {\mathrm {Cl}}(\zeta _k(G))$ . Thus $B_k(\alpha , \vec {\nu })$ is minimal under $\sigma _k$ .

For part (b) note first that since $R_{p/q}$ is of finite order and there are finitely many address intervals, $B_k(p/q, \vec {\nu })$ must consist of finitely many periodic orbits. The other properties in (a) and (b) will follow from (c) (proved using just these two partial results on (a) and (b)) and Theorem 7.9 using Remark 7.11(a).

For part (c), we first show that $B_k(\omega , \vec {\nu })$ is a recurrent symbolic kfsm set. By parts (a) and (b) we know that $B_k(\omega , \vec {\nu })$ is recurrent, and by Remark 8.4 that $B_k(\omega , \vec {\nu })\subset \Omega _k$ and $\rho _k(B_k(\omega , \vec {\nu }) = \omega $ . We show that $B_k(\omega , \vec {\nu })$ is a symbolic kfsm set by showing that its full lift $A_k(\omega , \vec {\nu })$ to $\Omega _\infty $ is as required by Definition 8.1. As noted in Remark 2.2, ${\hat {T}}^k_\infty (A_{k}(\omega ,\vec {\nu })) = A_{k}(\omega ,\vec {\nu })$ so we need to show that ${\hat {\sigma }}_\infty $ is semi-monotone on $A_k(\omega , \vec {\nu })$ .

The first step is to show that $\zeta _\infty $ is weakly order-preserving. Assume $x_1', x_2'\in G$ with $x_1' < x_2'$ . It could happen (when $\omega $ is rational) that $\zeta _\infty (x_1') = \zeta _\infty (x_2')$ , but if there exists a least n with $(\zeta _\infty (x_1'))_n \not = (\zeta _\infty (x_2'))_n$ , then since $I_m < I_{m+1}$ for all m and $R_\omega $ is order-preserving, certainly $(\zeta _\infty (x_1'))_n < (\zeta _\infty (x_2'))_n$ , and so $\zeta _\infty $ is weakly order-preserving.

We now show that ${\hat {\sigma }}_\infty $ is semi-monotone on $A_k(\omega , \vec {\nu })$ . Let G be the good set for $\zeta _\infty $ and assume ${\underline s}, {\underline t}\in \zeta _\infty (G)$ with ${\underline s} < {\underline t}$ . Then there exist $x_1', x_2'\in G$ with $\zeta _\infty (x_1') = {\underline s}$ and $\zeta _\infty (x_2') = {\underline t}$ and, of necessity, $x_1' < x_2'$ and so $R_\omega (x_1') < R_\omega (x_2')$ . Since $\zeta _\infty R_\omega = \sigma _\infty \zeta _\infty $ , we have

$$ \begin{align*} \sigma_\infty({\underline s}) = \sigma_\infty \zeta_\infty(x_1') = \zeta_\infty R_\omega(x_1) \leq \zeta_\infty R_\omega(x_2) = \sigma_\infty \zeta_\infty(x_2') = \sigma_\infty({\underline t}). \end{align*} $$

Thus $\sigma _\infty $ is weakly order-preserving on $\zeta _\infty (G)$ and so on $A_k(\omega , \vec {\nu })$ . We have that $A_k(\omega , \vec {\nu })$ satisfies all the conditions of the lift in Definition 7.5 and thus $B_k(\omega , \vec {\nu })$ is symbolic kfsm.

Now for the converse assume that ${\hat {Z}}\subset \Omega _k$ is symbolic recurrent kfsm with ${\hat {\rho }}({\hat {Z}}) = \omega $ . Let ${\hat {Z}}'\subset \Omega _\infty $ be the lift that satisfies Definition 7.5. The proof splits into the two cases when $\omega $ is rational and irrational.

First assume $\omega = p/q$ with $\gcd (p,q) = 1$ . We know from Lemma 7.9 and Remark 7.11 that ${\hat {Z}}$ consists of at most k distinct periodic orbits each with period $kq/d$ with $d=\gcd (p,k)$ . We assume for simplicity that ${\hat {Z}}$ is a single periodic orbit. The case of multiple periodic orbits is similar but with more elaborate indexing.

For $i = 0, \ldots , kq/d-1$ let $P_i = (2 i + 1)d/2q\subset S_k$ and ${\mathcal {P}}= \{P_i\}$ . Since ${\hat {Z}}$ is a kfsm periodic orbit with $\rho _k$ -rotation number $p/q$ we may find an order-preserving bijection $\phi :{\hat {Z}}\rightarrow {\mathcal {P}}$ with $\phi \sigma _k = R_{p/q} \phi $ on ${\hat {Z}}$ . Thus $\phi \sigma _k\phi ^{-1}$ acts on ${\mathcal {P}}$ as $P_i \mapsto P_{i+p/d}$ reducing indices mod $kq/d$ .

For $j = 0, \ldots , k-1$ , let $X^{\prime }_j = \phi ({\hat {Z}} \cap [j])$ where recall that $[j]$ is the length-one cylinder set in $\Sigma ^+_{2k}$ . Since $\phi $ is order-preserving, each $X_j'$ consists of a collection of adjacent points from ${\mathcal {P}}$ . If $X_j' = \{ P_{n(j)}, \ldots , P_{m(j)}\} \not = \emptyset $ , let $X_j = [P_{n(j)} - d/(2q), P_{m(j)} + d/(2q)]$ and when $X_j'=\emptyset $ let $X_j=\emptyset $ . We now claim that $\{X_j\}$ is an address system as used in the HM construction where $\vec {\nu }$ is defined by $\nu _{j+1} = |X_{2j}|$ and that $|X_{2j+1}| = p/q$ for $j = 0, \ldots , k-1$ , yielding ${\hat {Z}}' = B_k(p/q, \vec {\nu })$ .

Letting $\zeta $ be the itinerary map for the address system $\{X_j\}$ , for ${\underline s}\in {\hat {Z}}$ we have by construction that $\zeta \phi ({\underline s}) = {\underline s}$ . In addition, for all $x\in [\phi ({\underline s})-d/(2q), \phi ({\underline s}) + d/(2q)]$ we also have $\zeta (x) = \zeta \phi ({\underline s}) = {\underline s}$ . Thus for any point x in the good set G, $\zeta (x) = {\underline s}$ for some ${\underline s}\in {\hat {Z}}$ . This shows that ${\hat {Z}} = \operatorname {\mathrm {Cl}}(\zeta (G))$ . The last step needed to show that ${\hat {Z}} = B_k(p/q, \vec {\nu })$ is to check that the address system is of the type used in the HM construction.

We need only check that $|X_{2j+1}| = p/q$ , and for this it suffices to show that $\# X_{2j+1} = p/d$ . Assume first that $\# X_{2j+1} < p/d$ . Recalling that $\phi \sigma _k \phi ^{-1}$ acts on the $X_i'$ like $i \mapsto i + p/d$ , we see that there will be some $P_m\in X_{2j}'$ and $P_{m+p/d}\in X_{2j+2}'$ . Thus using $\phi ^{-1}$ there is a ${\underline s}\in {\hat {Z}}$ with $s_0 = 2j$ and $s_1 = 2j+2$ , a contradiction to the fact that ${\hat {Z}}\subset \Omega _k$ and thus its allowable transitions are given by (5.1). On the other hand, if $\# X_{2j+1}> p/d$ we have some $P_m\in X_{2j+1}'$ and $P_{m+p}\in X_{2j+1}'$ , again yielding a contradiction to ${\hat {Z}}\subset \Omega _k$ .

The irrational case is basically a continuous version of the rational one. By Remark 7.11(b) we have a continuous, onto $\phi :{\hat {Z}}\rightarrow S_k$ which is weakly order-preserving, $\phi \sigma _\infty = R_\alpha \phi $ , and $\#\phi ^{-1}(x) = 1$ for all but a countable number of $R_\alpha $ -orbits on which $\#\phi ^{-1}(x) = 2$ .

For $j= 0, \ldots , k-1$ , let $X_j = \phi ([j])$ . Thus $X_j$ is a closed interval (perhaps empty) with $\cup X_j = {\mathbb R}$ , $X_j \leq X_{j+1}$ and adjacent intervals intersect only in their single common boundary point. We use $\{ X_j\}$ as an address system with dynamics $R_\alpha $ , good set G, and itinerary map $\zeta $ . By construction, if ${\underline s}\in {\hat {Z}}$ with $\phi ({\underline s})\in G$ , then ${\underline s} = \zeta \phi ({\underline s})$ and so $\phi ^{-1}(G) = \zeta \phi (\phi ^{-1}(G)) = \zeta (G)$ . Since ${\hat {Z}}$ is a Cantor set and $\phi ^{-1}(G)$ is ${\hat {Z}}$ minus a countable set of $\sigma _k$ -orbits we have that $\phi ^{-1}(G)$ is dense ${\hat {Z}}$ . Thus, taking closures, ${\hat {Z}} = \operatorname {\mathrm {Cl}}(\zeta (G))$ .

To finish we must show that $\{ X_j\}$ is the type of address system allowable in the HM construction. We just need $|X_{2j+1}| = \alpha $ for all j. The proof is similar to the rational case. If $|X_{2j+1}| < \alpha $ then ${\hat {Z}}$ has a transition $2j \rightarrow 2j+2$ , and if $|X_{2j+1}|> \alpha $ then ${\hat {Z}}$ has a transition $2j+1 \rightarrow 2j+1$ . Either is a contradiction to ${\hat {Z}}\subset \Omega _k$ . Thus letting $\nu _{j+1} = |X_{2j}|$ for $j = 0, \ldots , k-1$ , we have ${\hat {Z}} = \operatorname {\mathrm {Cl}}(\zeta (G)) = B_k(\alpha , \vec {\nu })$ .

The last sentence in (c) follows from the construction of $\unicode{x3bb} _k(\omega , \vec {\nu })$ .

Proof. We suppress the dependence on k to simplify notation. For (a), the resonance hypothesis implies that there are odd a and b with $R_{\alpha }^n(X_{a\vec {\nu }}) = X_{b\vec {\nu }}$ for some $n>0$ where we may assume $a<b$ . Since $R_\alpha ^n(r_{a\vec {\nu }}) = r_{b\vec {\nu }}$ , by shrinking some $\nu _j$ for $a < j < b$ we obtain a $\vec {\nu }^{\;\prime }$ and $x < r_{a\vec {\nu }}$ and arbitrarily close to it with $x\in G_{\alpha \vec {\nu }^{\;\prime }}$ and $R_{\alpha }^n(x) \in X_{b+1, \vec {\nu }^{\;\prime }}$ . In this way we can obtain sequences $\vec {\nu }^{(i)}\rightarrow \vec {\nu }$ and $x_i\nearrow r_{a\vec {\nu }}$ with $x_i\in G_{\alpha , \vec {\nu }^{(i)}}$ and $R_{\alpha }^n(x_i) \in X_{b+1, \vec {\nu }^{(i)}}$ . Thus

$$ \begin{align*} \zeta_{\alpha,\vec{\nu}^{(i)}}(x_i) = .a\ldots (b+1) \zeta_{\alpha,\vec{\nu}^{(i)}}(R_{\alpha}^{n+1}(x_i). \end{align*} $$

To simplify matters, assume that $R_{\alpha }(r_{b\vec {\nu }}) \in G_{\alpha , \vec {\nu }}$ ; more complicated resonances are similar. Since $R_\alpha ^{n+1}(x_i)\rightarrow R_\alpha ^{n+1}(r_{a\vec {\nu }}) = R_{\alpha }(r_{b\vec {\nu }})$ using Remark 8.4(b),

$$ \begin{align*} \zeta_{\alpha,\vec{\nu}^{(i)}}(x_i) \rightarrow .a\ldots (b+1) \zeta_{\alpha,\vec{\nu}}(R_{\alpha}(r_{b\vec{\nu}}) := {\underline s}. \end{align*} $$

Passing to a subsequence if necessary, by the compactness of the collection of symbolic kfsm sets there is a kfsm Z with $B_k(\alpha , \vec {\nu }^{(i)})\rightarrow Z$ in the Hausdorff topology, and by its construction ${\underline s}\in Z$ . But ${\underline s}$ cannot be recurrent since by the resonance any length- $(n+1)$ block in $\zeta _{\alpha ,\vec {\nu }}(R_{\alpha } (r_{b\vec {\nu }})$ must start with a and end in b.

The argument for (b) is similar, but now the perturbation must be in the parameter $\omega $ . This is because if $\omega = p/q$ is fixed, then $R_{\omega }^n(X_a) = X_a$ with $n = qk/\gcd (p,k)$ for all $\vec {\nu }$ . Fix an a and so $R_{p/q}^n(r_a) = r_a$ . By increasing $\omega $ incrementally we may find sequences $\omega ^{(i)}\searrow p/q$ and $x_i\nearrow r_a$ with $x_i\in G_{\omega ^{(i)}, \vec {\nu }}$ so that the initial length- $(n+1)$ block of $\zeta _{\omega ^{(i)} \vec {\nu }}(x_i)$ is $a \ldots a+1$ . Thus if $\zeta _{p/q\vec {\nu }}(r_a + \epsilon ) = P^\infty $ for small $\epsilon $ then

$$ \begin{align*} \zeta_{\omega^{(i)},\vec{\nu}}(x_i) \rightarrow .a\ldots (a+1) P_2 P_3 \ldots P_{n-1} P^\infty := {\underline t} \end{align*} $$

where $P = (a+1) P_2 P_3 \ldots P_{n-1}$ . As in the proof of (a) passing to a subsequence if necessary, there is a kfsm Z with $B_k(\omega ^{(i)}, \vec {\nu })\rightarrow Z$ in the Hausdorff topology, and by its construction ${\underline t}\in Z$ . But ${\underline t}$ cannot be recurrent since any length- $(n+1)$ block in $P^\infty $ must start and end with a.

Proof. Since we know from Lemma 7.8 that $\iota _k$ and $(\iota _k)_*$ are homeomorphisms we only consider $B_k$ and $\unicode{x3bb} _k$ . While these are functions of $\vec {\nu }$ alone, for the proof it is clearer to resort to the two-variable versions with the proviso that $\omega = 1 - \sum \nu _i/k$ . Note that we have already shown in Theorem 8.5 that $\unicode{x3bb} _k$ and $B_k$ are onto ${\hat {{\mathcal {N}}}}_k(g)$ and ${\hat {{\mathcal {B}}}}_k(g)$ , respectively. We will often need to include the explicit dependence of various objects on the variables, for example, $\ell _j(\omega , \vec {\nu })$ , and we often suppress the dependence on k.

We prove (b) first. We first show that $\unicode{x3bb} _k$ is continuous. For each $j = 1, \ldots , 2k-1$ and $i\in {\mathbb N}$ , let $\ell _j^{(i)}(\omega , \vec {\nu }) = R_{\omega }^{-i}(\ell _j(\omega , \vec {\nu }))$ . The first observation from the HM construction is that

(9.2) $$ \begin{align} |\ell_j^{(i)}(\omega, \vec{\nu}) - \ell_j^{(i)}(\omega_0, \vec{\nu}_0)| \leq \|(\omega, \vec{\nu}) - (\omega_0, \vec{\nu}_0)\|_1. \end{align} $$

For a length-N block $B = b_0 \ldots b_{N-1}$ in $\Omega _k$ , let

$$ \begin{align*} Y_B(\omega, \vec{\nu}) = \bigcap_{i=0}^{N-1} R^{-i}_{\omega} (\operatorname{\mathrm{Int}}(X_{b_i}(\vec{\nu}))), \end{align*} $$

and so $x\in G_{\omega , \vec {\nu }} \cap Y_B(\omega , \vec {\nu })$ implies that $\zeta _{\omega \vec {\nu }} (x)$ begins with the block B. Also by the HM construction, $\unicode{x3bb} _k(\omega , \vec {\nu }) ([B]) = m( Y_B(\omega , \vec {\nu }))$ , with m being Lebesgue measure on the circle.

Recall that the weak topology on $\Sigma _{2k}^+$ is generated by the metric

$$ \begin{align*} d(\mu, \mu') = \sum_{i=1}^\infty \frac{|\mu([B_i]) - \mu'([B_i])|}{2^i} \end{align*} $$

where $\{B_i\}$ is some enumeration of the blocks in $\Sigma _{2k}^+$ . Since each $Y_B(\omega , \vec {\nu })$ is a (perhaps empty) interval with endpoints some $\ell _i^{(j)}(\omega , \vec {\nu })$ , inequality (9.2) implies that

$$ \begin{align*} |m(Y_B(\omega, \vec{\nu})) - m(Y_B(\omega_0, \vec{\nu}_0))| \leq 2 \|(\omega, \vec{\nu}) - (\omega_0, \vec{\nu}_0)\|_1. \end{align*} $$

Thus, summing over blocks,

$$ \begin{align*} d(\unicode{x3bb}_k(\omega, \vec{\nu}), \unicode{x3bb}_k(\omega_0, \vec{\nu}_0)| \leq 2 \|(\omega, \vec{\nu}) - (\omega_0, \vec{\nu}_0)\|_1, \end{align*} $$

so $\unicode{x3bb} _k$ is continuous.

Since by definition in the HM construction, $\unicode{x3bb} _k(\omega , \vec {\nu })([2j]) = \nu _{j+1}$ , $\unicode{x3bb} _k$ is injective. Recall now that for the model map, $\operatorname {\mathrm {HM}}(f_m) = {\mathcal {D}}_k$ which is compact. So $\unicode{x3bb} _k:\operatorname {\mathrm {HM}}(f_m)\rightarrow {\hat {{\mathcal {N}}}}_k(f_m)$ is a homeomorphism whose image is the set of all measures on recurrent symbolic kfsm sets in $\Omega _k$ . Thus, since $\operatorname {\mathrm {HM}}_k(g)\subset {\mathcal {D}}_k$ we have that $\unicode{x3bb} _k:\operatorname {\mathrm {HM}}_k(g)\rightarrow {\hat {{\mathcal {N}}}}_k(g)$ is also a homeomorphism. The compactness of ${\hat {{\mathcal {N}}}}_k(g)$ was proved in Lemma 7.8.

The proof of (a) is based on the following claim: $B_k$ is continuous at $(\omega _0, \vec {\nu }_0)$ if and only if for all N there exists $\delta>0$ so that $\|(\omega , \vec {\nu }) - (\omega _0, \vec {\nu }_0)\| < \delta $ implies that for all blocks B of length up to N we have $Y_B(\omega _0, \vec {\nu }_0)$ non-empty exactly when $Y_B(\omega , \vec {\nu })$ is non-empty.

To prove the claim, first note that continuity is equivalent to the following: given $\epsilon> 0$ , there exists $\delta>0$ so that $\| (\omega , \vec {\nu }) - (\omega _0, \nu _0)\| < \delta $ implies that for each ${\underline s}\in \zeta _{\omega \vec {\nu }}(G_{\omega \vec {\nu }})$ there is a ${\underline t}\in \zeta _{\omega _0\vec {\nu }_0}(G_{\omega _0 \vec {\nu }_0})$ with $d({\underline s}, {\underline t}) < \epsilon /2$ and for each ${\underline t}\in \zeta _{\omega _0 \vec {\nu }_0}(G_{\omega _0 \vec {\nu }_0})$ there is an ${\underline s}\in \zeta _{\omega \vec {\nu }}(G_{\omega \vec {\nu }})$ with $d({\underline s}, {\underline t}) < \epsilon /2$ . This implies that $\operatorname {\mathrm {HD}}(\zeta _{\omega \vec {\nu }}(G_{\omega \vec {\nu }}), \zeta _{\omega _0 \vec {\nu }_0}(G_{\omega _0 \vec {\nu }_0})) < \epsilon /2$ and thus $\operatorname {\mathrm {HD}}(B_k(\omega ,\vec {\nu }), B_k(\omega _0,\vec {\nu }_0)) < \epsilon $ . Since $d({\underline s}, {\underline t})$ is small exactly when ${\underline s}$ and ${\underline t}$ agree in a long prefix block B and $Y_B(\omega , \vec {\nu }) = \zeta _{\omega \vec {\nu }}^{-1}([B])$ , the claim follows.

We show that $B_k$ satisfies the condition in the claim when $(\omega , \vec {\nu })$ is non-resonant. Given N for $j = 1, \ldots , 2k-1$ and $i = 0, \ldots , N$ , consider again $\ell _j^{(i)}(\omega , \vec {\nu }) = R_{\omega }^{-i}(\ell _j(\omega , \vec {\nu })$ . By the HM construction we have $\ell _{2m}^{(n+1)}(\omega , \vec {\nu }) = \ell _{2m-1}^{(n)}(\omega , \vec {\nu })$ for all n, m, and $(\omega , \vec {\nu })$ . By non-resonance at $(\omega _0, \vec {\nu }_0)$ , all the other $\ell _j^{(i)}(\omega _0, \vec {\nu }_0)$ are disjoint. Since by (9.2) each $\ell _j^{(i)}(\omega , \vec {\nu })$ depends continuously on $(\omega , \vec {\nu })$ and the endpoints of each $Y_B(\omega , \vec {\nu })$ are some $\ell _j^{(i)}(\omega , \vec {\nu })$ , we may find a $\delta $ so that $\| (\omega , \vec {\nu }) - (\omega _0, \vec {\nu }_0)\| < \delta $ implies that the $\ell _j^{(i)}(\omega , \vec {\nu })$ are ordered around $S_k$ in the same way and with the same gaps between them as the $\ell _j^{(i)}(\omega _0, \vec {\nu }_0)$ . This implies that for each block B of length $B\leq N$ , $Y_B(\omega _0, \vec {\nu }_0)$ is non-empty exactly when $Y_B(\omega , \vec {\nu })$ is non-empty and so $B_k$ is continuous.

For the discontinuity, since the sets Z in Lemma 9.3(ab) are not recurrent, they are not equal to $B_k(\omega _0, \vec {\nu }_0)$ .