Proof. It is clear from Proposition 3.1 that each set $J_\theta $ is closed, connected, and has non-empty intersection with M. It is also clear that $f(J_\theta )=J_{\tau (\theta )}$ . However, because the sets $f^i(D_k)\cap f^j(D_k)$ are allowed to intersect at a point, it is not clear if the sets $J_\theta $ are pairwise disjoint. We prove this fact now. Suppose there are $\theta \neq \theta '$ with $J_\theta \cap J_{\theta '}\neq \emptyset $ . Find k minimal such that $\theta _k\neq \theta ^{\prime }_k$ . Then clearly

$$ \begin{align*} J_\theta \cap J_{\theta'} = f^{\theta_k}(D_k) \cap f^{\theta^{\prime}_k}(D_k) = \{a\} \end{align*} $$

for some single point $a\in X$ . Taking the image by $f^{\alpha _k}$ , we have

$$ \begin{align*} f^{\alpha_k}(a) \in f^{\alpha_k}(J_\theta)\cap f^{\alpha_k}(J_{\theta'}) = J_{\tau^{\alpha_k}(\theta)} \cap J_{\tau^{\alpha_k}(\theta')} = f^{\theta_k+\alpha_k}(D_k) \cap f^{\theta^{\prime}_k+\alpha_k}(D_k) = \{a\}, \end{align*} $$

because $D_k$ is periodic with period $\alpha _k$ . This shows that a is periodic with period $\alpha _k$ . In particular, it does not belong to the infinite minimal set M. Now for $n\in \mathbb {N}$ , let $J_n=J_{\tau ^{n\alpha _k}(\theta )}$ . Then $a\in J_n$ and we can choose an additional point $m_n \in M \cap J_n$ for all n. Thus the $J_n$ sets are non-degenerate subdendrites and intersect pairwise only at a. In particular, the sets $(a,m_n]$ are pairwise disjoint connected subsets of X, so their diameters must converge to zero (see e.g. [Reference Misiurewicz and Marchioro19, Lemma 2.3]). However, because M is closed, this shows that $a\in M$ , a contradiction.

Now that the sets $J_\theta $ , $\theta \in \Omega $ have been shown to be pairwise disjoint, we see immediately that $\pi $ is well defined. It is also easy to see that $\pi $ is continuous and $\tau \circ \pi = \pi \circ f$ .

Again, because the sets $J_\theta $ , $\theta \in \Omega $ are pairwise disjoint connected sets in X, only countably many of them can have positive diameter. It follows that $\pi ^{-1}(\theta )$ is a singleton except for countably many $\theta $ . It remains to show that $M\cap J_\theta $ is countable when $J_\theta $ is non-degenerate. Because M is minimal and $J_\theta $ never returns to itself, we must have $M\cap J_\theta \subset \operatorname {\mathrm {Bd}}(J_\theta )$ , where $\operatorname {\mathrm {Bd}}(J_{\theta })$ stands for the boundary of $J_{\theta }$ . However, the boundary in X of the subdendrite $J_\theta $ is a subset of $E(J_\theta ) \cup B(X) \cup \overline {E(X)}$ , which is countable by the assumption that $E(X)$ has countable closure. Here we use the well-known facts that $B(X)$ is countable in any dendrite, and the cardinality of the endpoint set of a dendrite cannot increase when we pass to a subdendrite, see e.g. [Reference Naghmouchi21, Reference Scarpellini22].

Proof. Using Lemma 3.2, we know that there are uncountably many singleton sets $J_\theta $ . Now a dendrite whose endpoint set has countable closure is always the union of a countable sequence of free arcs and a countable set, see [Reference Askri5, Theorem 2.2]. It follows that we can find $\theta $ with the singleton $J_\theta $ in the interior of a free arc A in X. Because $J_\theta $ is the nested intersection $\bigcap _N f^{\theta _N}(D_N)$ , we can find N large enough that $f^{\theta _N}(D_N)$ is contained in A. Then $f^{\theta _k}(D_k)$ is a free arc for all $k\geq N$ .

Proof. Throughout the proof, we will use freely the following well-known property of $\omega $ -limit sets (e.g. [Reference Bartoš, Bobok, Pyrih, Roth and Vejnar6]): if for fixed $n\geq 2$ we write $W_i=\omega _{f^n}(f^i(x))$ for $0\leq i<n$ , then $\omega _f(x)=\bigcup _{i=0}^{n-1} W_i$ and $f(W_i)=W_{i+1 \; (\text {mod }n)}$ . In particular, if $\omega _f(x)$ is uncountable, then so is each $W_i$ , and if $\omega _f(x)$ contains a given fixed point, then each $W_i$ contains it as well. We continue to use the notation $[x,y]$ for the unique arc in X with endpoints $x,y\in X$ , and if $z\in (x,y)=[x,y]\setminus \{x,y\}$ , we will say for simplicity that z lies between x and y.

Now let $L=\omega _f(x)$ , where x is recurrent but not periodic. Then L is the closure of the orbit of x, and hence it is a perfect uncountable set.

Step 1: L does not contain a periodic point with a free arc neighborhood in X. Suppose to the contrary that $a\in L$ , $f^N(a)=a$ , and some free arc C is a neighborhood of a in X. Then, by the standard properties mentioned above, $a\in \omega _{f^N}(f^i(x))$ for some $0\leq i<N$ . Replacing f with its iterate and x with its image, we may safely assume that $N=1$ and $i=0$ , that is, a is a fixed point in $L=\omega _f(x)$ .

Because periodic points are never isolated in infinite $\omega $ -limit sets, we know that L accumulates on a from at least one side in the free arc C. So, choose an endpoint b of C such that $L\cap [a,b]$ accumulates on a. For convenience, we let C carry its natural order as an arc, oriented in such a way that $a<b$ . Choose five points $y_i\in L\cap [a,b]$ with $a<y_1<y_2<y_3<y_4<y_5<b$ . Choose three small arc neighborhoods $I_2, I_3, I_4$ containing $y_2,y_3,y_4$ , respectively, and let them be pairwise disjoint and lie between $y_1$ and $y_5$ . Because the orbit of x visits each of these neighborhoods $I_i$ infinitely often, there must be points in $I_2$ and $I_4$ which visit $I_3$ , so by [Reference Misiurewicz and Marchioro19, Theorem 2.13], f has a periodic point c between $y_1$ and $y_5$ . Replacing x with a point from its orbit near $y_1$ , we may assume that $a<x<c<y_5<b$ . Let r be the period of c and put $g=f^r$ . Because a was already fixed for f, we have $a\in \omega _g(x)$ as well. Note that because x is recurrent for f, it is also recurrent for g.

Now we may use the claim to finish Step 1. Because x belongs to the closed invariant set I, we have $\omega _g(x)=\omega _{g|_I}(x)$ . However, $g|_I$ is an interval map, and when an infinite $\omega $ -limit set for an interval map contains a periodic point, the topological entropy must be positive (see [Reference Nadler20]), a contradiction.

Step 2: L does not contain any periodic points. Suppose to the contrary that $a\in L$ is a periodic point. As in Step 1, we may assume that a is fixed. By [Reference Askri5, Theorem 2.2], the dendrite X is the union of a countable sequence of free arcs together with a countable set. In particular, we can find a free arc C not containing a with $L\cap C$ uncountable. Write $C=[u,v]$ with v between u and a and let $<$ denote the order in C with $u<v$ . Because $L\cap C$ is infinite, we may choose four points $x_i\in \mathrm {Orb}_f(x)$ with $u<x_1<x_2<x_3<x_4<v$ . As in Step 1, we can use small arc neighborhoods of $x_2,x_3,x_4$ to find a periodic point c with $u<x_1<c<v$ , and because $x_1$ is in the orbit of x, we may redefine $x=x_1$ without changing $\omega _f(x)$ . Let r denote the period of c and put $g=f^r\kern-3pt.$ Because x is recurrent also for g, we have $\mathrm {Orb}_g(x)\cap [u,c]$ infinite, so we can find two points $x_5,x_6\in \mathrm {Orb}_g(x)$ with $u<x_5<x_6<c$ and passing forward along the orbit, we can redefine $x=x_6$ without changing $\omega _g(x)$ . In particular, $x_5\in \omega _g(x)=\overline {\mathrm {Orb}_g(x)}$ , so we can choose $p\geq 1$ with $g^p(x)$ close to $x_5$ so that $u<g^p(x)<x<c$ .

Let $l=\omega _{g^p}(x)$ . We have $x\in l$ because x is recurrent and $a\in l$ because a is a fixed point in $\omega _f(x)$ . Moreover, $c\not \in l$ as a result of Step 1. So let $X_0, X_1$ denote the connected components of $X\setminus \{c\}$ containing x and a, respectively, and put $l_i=l\cap X_i$ . Then $l=l_0\cup~l_1$ expresses l as the disjoint union of two non-empty open subsets (in the topology induced from X to l). Recall that every $\omega $ -limit set $\omega _f(x)$ is weakly incompressible, that is, $f(\overline {U})\not \subset U$ for any set $U\subsetneq \omega _f(x)$ open in $\omega _f(x)$ (see, e.g. [Reference Sarnak25]). Thus we have $g^p(l_0)\cap l_1 \neq \emptyset $ . Therefore, we may choose $y\in l_0$ with $g^p(y)\in l_1$ , and because $\mathrm {Orb}_{g^p}(x)$ is dense in $l_0$ , we may choose y from the orbit of x. We finish the proof in two cases, depending on the location of y.

Suppose first that y is between x and c. In the ordering of the arc $[g^p(x),g^p(y)]$ , we have $g^p(x) < x < y < c < g^p(y)$ . Put $I=[x,y]$ and $J=[y,c]$ . Clearly $g^p(I)\supseteq I\cup~J$ . Because $y\in \mathrm {Orb}_g(x)$ , we have $\omega _{g}(y)=\omega _{g}(x)\supset \mathrm {Orb}_{g}(x)\ni g^p(x)$ . In particular, we may choose $n>p$ to make $g^{n}(y)$ as close to $g^p(x)$ as we like, so that $x,y \in [g^{n}(y),c]$ . However, then $g^{n}(J) \supseteq I\cup J$ . We conclude that g possesses an arc horseshoe and thus g has positive topological entropy, which is a contradiction with $h_{\mathrm {top}}(f)=0$ .

Suppose instead that x is between y and c. Then $c\in [g^p(x),g^p(y)]$ , so there is $c_{-1}\in (x,y)$ with $g^p(c_{-1})=c$ . In the ordering of the arc $[y,g^p(y)]$ , we have $y<c_{-1}<x<c$ and we also have $x\in (g^p(x),c)$ . Put $I=[c_{-1},x]$ and $J=[x,c]$ . Because $y\in \mathrm {Orb}_g(x)\subset \omega _g(x)$ , we can find $n>p$ with $g^n(x)$ as close to y as we like. In particular, we can get $x,c_{-1}\in [g^n(x),c]$ . However then $g^n(I) \cap g^n(J) \supseteq I\cup J$ . Again we conclude that g has positive topological entropy, which is a contradiction. This ends the proof.

Proof. Let $x\in \mathrm {Rec}(f)$ . If x is periodic, then it is minimal, so assume x is not periodic. Let $L=\omega (x)$ . Let $M\subset L$ be a minimal set. By Lemma 3.5, L contains no periodic orbits, so M is an infinite minimal set. Then Proposition 3.1 applies and we get a sequence of f-periodic subdendrites $(D_k)_{k\geq 1}$ and periods $(\alpha _k)$ satisfying properties (1)–(5) of that proposition. By Lemma 3.4 for all sufficiently large k, we have that $f^i(D_k)$ is a free arc for suitable i. Because M is infinite and $D_k$ is periodic, we have $M\cap \operatorname {\mathrm {int}} f^i(D_k)\neq \emptyset $ and as a consequence, $Orb_f(x)\cap D_k\neq \emptyset $ , for every sufficiently large k. Hence, $\bigcap _{k\geq 1}Orb_f(D_k)$ contains L, that is, property (4) still holds with L in the place of M.

We claim that property (5) also holds with L in the place of M. Fix k and denote $L_i=f^i(D_k)\cap L$ for $0\leq i<\alpha _k-1$ . Observe that L does not contain periodic points, and the set $\mathrm {Orb}(f^i(D_k)\cap f^j(D_k))$ is always finite and invariant for any $i\neq j$ (can be empty) and hence $L_i \cap f^j(D_k)=\emptyset $ for $i\neq j$ . This shows that the sets $L_i \cap L_j=\emptyset $ for $i\neq j$ . Clearly $f(L_i) \subseteq L_{i+1 (\text {mod }\alpha _k)}$ , and $f(L)=L$ because $\omega $ -limit sets are always mapped onto themselves. This shows that $f(L_i)=L_{i+1 (\text {mod } \alpha _k)}$ . In particular, we conclude that $L_i$ is uncountable for each i.

Again using Lemma 3.4, choose k large enough that $f^i(D_k)$ is a free arc for some $0\leq i<\alpha _k-1$ and let $A=f^i(D_k)$ denote that free arc. We have just shown that $L_i=A\cap L$ is uncountable, so because A is a free arc, there are points from $\omega (x)$ in its interior. Thus we can find a point $y=f^l(x)$ from the forward orbit of x in A. Then $\omega _f(x)=\omega _f(y)$ and y is also recurrent for f. Because $\mathrm {Rec}(f^{\alpha _k})=\mathrm {Rec}(f)$ , y is also recurrent for $f^{\alpha _k}$ . However, the restriction of $f^{\alpha _k}$ to A is an interval map with zero topological entropy. For such a map, all recurrent points are minimal points, see e.g. [Reference Bartoš, Bobok, Pyrih, Roth and Vejnar6, Ch. VI. Proposition 7]. Thus, y is a minimal point for $f^{\alpha _k}$ , and hence also for f. This shows that $\omega _f(y)=\omega _f(x)$ is a minimal set, and hence x itself is minimal.

Proof. Let $\mu $ be any finite invariant measure concentrated on $\bigcup _i A_i$ . Because each $A_i$ is invariant, that is, $f(A_i)\subset A_i$ , and f preserves $\mu $ , we may assume by throwing away a set in X of $\mu $ -measure zero that $f^{-1}(A_i)=A_i$ for each i.

We may take the index set for the variable i to be $\{1,\ldots ,n\}$ in the finite case or $\mathbb {N}$ in the countable case. Then putting $B_i = A_i \setminus \bigcup _{j<i} A_j$ for each i, we get a collection $\{B_i\}$ of pairwise disjoint invariant Borel sets. Now let $I=\{i~:~\mu (B_i)>0\}$ and write $\mu _i=\mu |_{B_i}$ for the (unnormalized) restriction of $\mu $ to $B_i$ . Then we get a direct sum decomposition of Hilbert spaces $L^2_{\mu }(X) = \bigoplus _{i\in I} L^2_{\mu _i}(B_i).$ (Here in a direct sum $\bigoplus H_i$ of Hilbert spaces, we include all $(v_i), v_i\in H_i$ such that $\sum ||v_i||^2<\infty $ . We do not require that all but finitely many $v_i$ vanish.) We may extend each function $\phi \in L^2_{\mu _i}(B_i)$ to an element of $L^2_{\mu _i}(X)$ by letting $\phi $ vanish outside of $B_i$ . Because $f^{-1}(B_i)=B_i$ , we see that if $\phi \circ f = \lambda \phi $ holds $\mu _i$ -almost everywhere (a.e) in $B_i$ , then by letting $\phi $ vanish outside $B_i$ , it continues to hold $\mu $ -a.e in X. Thus we have the equivalent direct sum decomposition

(4.1) $$ \begin{align} L^2_{\mu}(X) = \bigoplus_{i\in I} L^2_{\mu_i}(X), \end{align} $$

and an eigenfunction in a coordinate space is still an eigenfunction in the whole space. For each $i\in I$ , the normalized measure $\mu _i/\mu (B_i)$ is an invariant probability measure for f concentrated on $B_i \subset A_i$ , so by hypothesis, the eigenfunctions of the Koopman operator on the space $L^2_{\mu _i/\mu (B_i)}(X)$ have dense linear span. Dropping the normalizing constant, the same holds for $L^2_{\mu _i}(X)$ . Passing through the direct sum decomposition, it follows that the eigenfunctions of the Koopman operator on the space $L^2_{\mu }(X)$ have dense linear span, that is, $\mu $ has discrete spectrum.

The last statement of the lemma follows by the Poincaré recurrence theorem, whereby if $\mathrm {Rec}(f) \subseteq \bigcup A_i$ , then every measure $\mu \in M_f(X)$ is concentrated on $\bigcup A_i$ .

Proof. Suppose that F has positive entropy. Then by [Reference Li, Oprocha and Zhang15], there exists an arc horseshoe $I_1,I_2$ with $F^{n_1}(I_1)\cap F^{n_2} (I_2) \supset I_1 \cup I_2$ for some $n_1,n_2\in \mathbb N$ . Then $F^{i}(I_j)$ is not a single point for any $i=1,\ldots ,n_j$ and $j=1,2$ . However, if $F(J)$ is non-degenerate for an arc J, then $f(J)\supset F(J)$ which implies that $f^{n_1}(I_1)\cap f^{n_2}(I_2)\supset I_1\cup I_2$ which implies that f has positive topological entropy. A contradiction.

Proof. Let $Z=\{z\in \overline {E(X)}~:~\omega _f(z)\text { is an infinite minimal set}\}.$ Following arguments in [Reference Naghmouchi21, Theorem 10.27], let $(T_n)_{n \in \mathbb {N}} \subset X$ be an increasing sequence of topological trees with endpoints in $E(X)$ defined as follows. We inductively construct the sequence $(T_n)_{n\in \mathbb {N}}$ starting with $T_1 = \{e_1\}$ for some $e_1 \in E(X)$ . Then for $n\geq 1$ , we attach to $T_n$ an arc $[e,e_{n+1}]$ whose one endpoint $e_{n+1}$ belongs to $E(X)\setminus T_n$ and $e\in T_n$ . Because $E(X)$ is countable, we can put every endpoint into one of the trees, that is, we let the sequence $(e_n)_{n \in \mathbb {N}}$ be an enumeration of $E(X)$ , and then $\bigcup _{n\geq 1} T_n$ being a connected set must coincide with the whole dendrite X.

Let $\hat {T}_n=\bigcap _{i=0}^{\infty } f^{-i}(T_n)$ be the maximal invariant set completely contained in $T_n$ . Let $\mathrm {Per}(f)$ be the set of periodic points of f. We claim that

(4.2) $$ \begin{align} \mathrm{Rec}(f) \subset \mathrm{Per}(f) \cup \bigg(\bigcup_{z\in Z}\omega_f(z)\bigg) \cup \bigg(\bigcup_{n} \hat{T}_n\bigg). \end{align} $$

To see this, let x be a non-periodic recurrent point whose orbit is not contained in any of the trees $T_n$ . This means that there are points $f^{n_i}(x)$ which belong to $T_{m_i}\setminus T_{m_i-1}$ for some strictly increasing sequences $m_i$ , $n_i \to \infty $ . Then the arcs $[f^{n_i}(x), e_{m_i}]$ in X are pairwise disjoint, so by [Reference Misiurewicz and Marchioro19, Lemma 2.3], their diameters tend to zero. This shows that $\liminf _{n\to \infty } d(f^n(x),E(X))=0$ . Therefore, $\omega _f(x)\cap \overline {E(X)}\neq \emptyset $ . By Theorem 3.6, $\omega _f(x)$ is a minimal set, so choosing $z\in \omega _f(x)\cap \overline {E(X)}$ , we have $\omega _f(x)=\omega _f(z)$ . This establishes equation (4.2).

Now observe that any finite invariant measure concentrated on $\mathrm {Per}(f)$ has discrete spectrum, see e.g. [Reference Li, Tu and Ye17, Theorem 2.3]. As for the sets $\hat {T}_n$ , note that for each $n\in \mathbb {N}$ , the map $F=R\circ f$ , where $R\colon X\to T_n$ is a retraction, satisfies $F|_{\hat {T}_n}=f|_{\hat {T}_n}$ by the definition and therefore each f-invariant measure concentrated on $\hat {T}_n$ (a subset of a tree) has discrete spectrum, as, by [Reference Li, Tu and Ye17], all invariant measures of F have discrete spectrum.

Finally, we claim that any invariant measure concentrated on $\omega _f(z)$ , $z\in Z$ , has discrete spectrum. Let $(D_k)$ be the periodic subdendrites with periods $(\alpha _k)$ described in Proposition 3.1 and let $\pi :(\omega _f(z),f)\to (\Omega ,\tau )$ be the factor map onto the odometer described in Lemma 3.2. Let $\mu \in M_f(X)$ be any invariant measure concentrated on $\omega _f(z)$ . Then the pushforward measure $\pi _*(\mu )$ is invariant for the odometer, so by unique ergodicity, it is the Haar measure on $\Omega $ and it has discrete spectrum as a consequence of [Reference Walters26, Theorem 3.5]. Now because $\omega _f(z)$ contains no periodic points, we know that $\mu $ is non-atomic and therefore countable sets have measure zero. Then in the category of measure preserving transformations, the factor map $\pi :(\omega _f(z),\mu ,f)\to (\Omega ,\pi _*(\mu ),\tau )$ is in fact an isomorphism, because by Lemma 3.2, it is invertible except on a set of $\mu $ -measure zero. This implies that $\mu $ has discrete spectrum.

We have shown that an invariant measure concentrated on any of the countably many invariant sets in equation (4.2) has discrete spectrum. By Lemma 4.1, this completes the proof.