[proofplan]
The Smale-Birkhoff horseshoe construction supplies a return iterate $f^m$, a small rectangle near the transverse homoclinic orbit, and two Markov strips crossing that rectangle. We define $K$ as the maximal set whose full $f^m$-orbit remains in those two strips, and we encode each point by the bi-infinite sequence of strips it visits. The Markov crossing and hyperbolicity properties make this itinerary map a continuous bijection onto the full two-shift, hence a topological conjugacy by compactness. Periodic symbol sequences then give infinitely many periodic orbits, and the entropy of the full shift gives positive entropy for $f$ after using the iterate formula for topological entropy.
[/proofplan]
[step:Extract two Markov strips from the Smale-Birkhoff horseshoe construction]
By the local horseshoe form of the [Smale-Birkhoff Homoclinic Theorem](/theorems/7752), applied to the transverse homoclinic point $q$ of the hyperbolic periodic point $p$ (citing a result not yet in the wiki: Smale-Birkhoff Homoclinic Theorem), there exist an integer $m \in \mathbb{N}$, a compact topological rectangle $R \subset M$ contained in an arbitrarily small neighbourhood of the homoclinic orbit, and two pairwise disjoint compact vertical strips $R_0,R_1 \subset R$. Moreover, when a neighbourhood $U \subset M$ of the homoclinic orbit is prescribed, the construction may be arranged so that the finite tower
\begin{align*}
\bigcup_{j=0}^{m-1} f^j(R)
\end{align*}
is contained in $U$. Fix a metric $d$ inducing the topology on the compact rectangle $R$. Define the global diffeomorphism
\begin{align*}
F: M \to M
\end{align*}
by $F(x) := f^m(x)$ for every $x \in M$. The strips have the following Markov property for this return iterate.
For each $a \in \{0,1\}$, the image $F(R_a)$ crosses $R$ as a horizontal strip. The statement also includes the cylinder property supplied by the horseshoe construction: every two-sided finite itinerary determines a nonempty compact cylinder, these cylinders are nested as the itinerary window grows, and their diameters tend to $0$. The present argument uses these geometric conclusions as input rather than reproving the hyperbolic strip estimates.
[/step]
[step:Define the maximal invariant set for the return map]
Define the two-strip domain
\begin{align*}
R_* := R_0 \cup R_1.
\end{align*}
Since $F=f^m:M \to M$ is a diffeomorphism, $F^n:M \to M$ is defined and continuous for every $n \in \mathbb{Z}$. Define the maximal $F$-invariant set in $R_*$ by
\begin{align*}
K := \{x \in R_* : F^n(x) \in R_* \text{ for every } n \in \mathbb{Z}\}.
\end{align*}
Equivalently,
\begin{align*}
K = \{x \in R_* : f^{mn}(x) \in R_* \text{ for every } n \in \mathbb{Z}\}.
\end{align*}
The set $K$ is compact. Indeed, for each $n \in \mathbb{Z}$, define
\begin{align*}
K_n := R_* \cap (F^n)^{-1}(R_*).
\end{align*}
The set $R_*$ is compact and hence closed in the Hausdorff manifold $M$, and $F^n$ is continuous. Therefore $K_n$ is closed in $R_*$. Since
\begin{align*}
K = \bigcap_{n \in \mathbb{Z}} K_n,
\end{align*}
the set $K$ is closed in the compact set $R_*$, hence compact.
The same definition gives $F(K)=K$. If $x \in K$, then for every $n \in \mathbb{Z}$ one has
\begin{align*}
F^n(F(x)) = F^{n+1}(x) \in R_*,
\end{align*}
so $F(x) \in K$. Conversely, if $y \in K$, set $x := F^{-1}(y)$. Then for every $n \in \mathbb{Z}$,
\begin{align*}
F^n(x) = F^{n-1}(y) \in R_*,
\end{align*}
because $n-1 \in \mathbb{Z}$ and $y \in K$. Thus $x \in K$ and $F(x)=y$. Hence $F(K)=K$.
[/step]
[step:Encode each orbit by its strip itinerary]
Let
\begin{align*}
\Sigma_2 := \{0,1\}^{\mathbb{Z}}
\end{align*}
with the [product topology](/page/Product%20Topology), and define the full two-sided shift
\begin{align*}
\sigma: \Sigma_2 \to \Sigma_2
\end{align*}
by
\begin{align*}
(\sigma s)_n := s_{n+1}
\end{align*}
for every $s=(s_n)_{n \in \mathbb{Z}} \in \Sigma_2$ and every $n \in \mathbb{Z}$.
Define the itinerary map
\begin{align*}
\pi: K \to \Sigma_2
\end{align*}
as follows: for $x \in K$, the sequence $\pi(x)=(\pi(x)_n)_{n \in \mathbb{Z}}$ is determined by the condition
\begin{align*}
F^n(x) \in R_{\pi(x)_n}
\end{align*}
for every $n \in \mathbb{Z}$. This is well-defined because $R_0$ and $R_1$ are disjoint and every $F^n(x)$ lies in $R_*$.
For every $x \in K$ and every $n \in \mathbb{Z}$,
\begin{align*}
F^n(F(x)) = F^{n+1}(x).
\end{align*}
Therefore the strip visited by $F^n(F(x))$ is the strip visited by $F^{n+1}(x)$, so
\begin{align*}
\pi(F(x)) = \sigma(\pi(x)).
\end{align*}
Thus $\pi$ semiconjugates $F|_K$ to $\sigma$.
[guided]
The purpose of the itinerary map is to replace the geometric orbit of a point by a symbolic record of which of the two strips it visits. We first declare the target space precisely. Let
\begin{align*}
\Sigma_2 := \{0,1\}^{\mathbb{Z}}
\end{align*}
with the product topology. Define the full two-sided shift
\begin{align*}
\sigma: \Sigma_2 \to \Sigma_2
\end{align*}
by the rule
\begin{align*}
(\sigma s)_n := s_{n+1}
\end{align*}
for every bi-infinite sequence $s=(s_n)_{n \in \mathbb{Z}}$ and every integer $n$.
Now define the coding map
\begin{align*}
\pi: K \to \Sigma_2.
\end{align*}
For a point $x \in K$, every iterate $F^n(x)$ lies in $R_* = R_0 \cup R_1$. Since the two strips $R_0$ and $R_1$ are disjoint, there is a unique symbol $\pi(x)_n \in \{0,1\}$ such that
\begin{align*}
F^n(x) \in R_{\pi(x)_n}.
\end{align*}
This defines a unique sequence $\pi(x)=(\pi(x)_n)_{n \in \mathbb{Z}} \in \Sigma_2$.
The coding automatically intertwines the dynamics with the shift. Indeed, for each $n \in \mathbb{Z}$,
\begin{align*}
F^n(F(x)) = F^{n+1}(x).
\end{align*}
Thus the $n$th symbol of the itinerary of $F(x)$ is the strip containing $F^{n+1}(x)$, which is the $(n+1)$st symbol of the itinerary of $x$. Therefore
\begin{align*}
\pi(F(x)) = \sigma(\pi(x)).
\end{align*}
This is the semiconjugacy relation. The remaining work is to prove that no symbolic information is lost and that every symbolic sequence occurs.
[/guided]
[/step]
[step:Use the strip geometry to prove that the itinerary map is bijective]
We prove surjectivity first. Fix a sequence $s=(s_n)_{n \in \mathbb{Z}} \in \Sigma_2$. For integers $N \geq 1$, define the compact finite-itinerary set
\begin{align*}
K_N(s) := \{x \in R_* : F^n(x) \in R_{s_n} \text{ for every } -N \leq n \leq N\}.
\end{align*}
This definition uses the globally defined iterates $F^n:M \to M$ for both positive and negative integers $n$.
The Markov property of the two strips implies that each $K_N(s)$ is nonempty, and the sets are nested:
\begin{align*}
K_{N+1}(s) \subset K_N(s).
\end{align*}
Since $R_*$ is compact, the intersection
\begin{align*}
\bigcap_{N=1}^{\infty} K_N(s)
\end{align*}
is nonempty. Every point in this intersection belongs to $K$ and has itinerary $s$. Hence $\pi$ is surjective.
We prove injectivity. Suppose $x,y \in K$ and $\pi(x)=\pi(y)=s$. Then for every $N \geq 1$, both $x$ and $y$ lie in $K_N(s)$. The hyperbolic contraction estimates from the horseshoe construction say that the diameter of $K_N(s)$ tends to $0$ as $N \to \infty$. Therefore
\begin{align*}
d(x,y) \leq \operatorname{diam}(K_N(s))
\end{align*}
for every $N \geq 1$, where diameters are computed with respect to the previously fixed metric $d$ on $R$. Passing to the limit gives $d(x,y)=0$, hence $x=y$. Thus $\pi$ is injective.
[guided]
Fix a bi-infinite sequence
\begin{align*}
s=(s_n)_{n\in\mathbb Z}\in\Sigma_2.
\end{align*}
For each $N\geq 1$, the cylinder hypothesis gives a nonempty compact set
\begin{align*}
K_N(s)=\{x\in R_*:F^n(x)\in R_{s_n}\text{ for every }-N\leq n\leq N\}.
\end{align*}
The sets $K_N(s)$ are nested. Since $K_1(s)$ is compact, the nested intersection theorem gives a point
\begin{align*}
x\in\bigcap_{N=1}^{\infty}K_N(s).
\end{align*}
For every integer $n$, choosing $N\geq |n|$ gives $F^n(x)\in R_{s_n}\subset R_*$. Thus $x\in K$ and $\pi(x)=s$, proving surjectivity.
For injectivity, suppose $x,y\in K$ and $\pi(x)=\pi(y)=s$. Then $x$ and $y$ lie in $K_N(s)$ for every $N\geq 1$, so
\begin{align*}
d(x,y)\leq\operatorname{diam}(K_N(s)).
\end{align*}
The cylinder hypothesis gives $\operatorname{diam}(K_N(s))\to 0$, hence $d(x,y)=0$ and $x=y$. Therefore the itinerary map is bijective.
[/guided]
[/step]
[step:Upgrade the itinerary bijection to a topological conjugacy]
The map $\pi:K \to \Sigma_2$ is continuous. Indeed, to determine the coordinates $\pi(x)_n$ for all $|n| \leq N$, it is enough to know in which of the finitely many closed strips the points $F^n(x)$ lie. On $K$, points with a fixed finite itinerary form a closed-and-open cylinder subset relative to $K$, because the Markov strips are separated and the branch maps are continuous. Therefore preimages under $\pi$ of cylinder sets in $\Sigma_2$ are open in $K$, which proves continuity.
The space $K$ is compact, and $\Sigma_2$ is Hausdorff because it is a product of finite discrete Hausdorff spaces. Since $\pi$ is a continuous bijection from a [compact space](/page/Compact%20Space) to a [Hausdorff space](/page/Hausdorff%20Space), $\pi$ is a homeomorphism. Together with
\begin{align*}
\pi \circ F|_K = \sigma \circ \pi,
\end{align*}
this proves that
\begin{align*}
F|_K = f^m|_K: K \to K
\end{align*}
is topologically conjugate to the full two-sided shift on two symbols.
[guided]
The previous step proved that $\pi$ is bijective. To prove continuity, take a cylinder set in $\Sigma_2$ that fixes the coordinates with $|n|\leq N$. Its preimage under $\pi$ is the subset of $K$ whose iterates $F^n(x)$ lie in the prescribed strips for those finitely many $n$. By the Markov strip construction, these finite-itinerary subsets are relatively open in $K$, so preimages of cylinder sets are open. Since cylinder sets form a basis for the product topology on $\Sigma_2$, $\pi$ is continuous.
Now $K$ is compact and $\Sigma_2$ is Hausdorff. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so $\pi^{-1}$ is continuous. The itinerary computation gave
\begin{align*}
\pi(F(x))=\sigma(\pi(x))
\end{align*}
for every $x\in K$. Because $\pi$ is a homeomorphism, this identity is exactly a topological conjugacy between $F|_K=f^m|_K$ and the full two-sided shift $\sigma$.
[/guided]
[/step]
[step:Build an invariant set for the original map]
Define
\begin{align*}
\widehat K := \bigcup_{j=0}^{m-1} f^j(K).
\end{align*}
Each set $f^j(K)$ is compact because $K$ is compact and $f^j: M \to M$ is continuous. Hence $\widehat K$ is compact as a finite union of compact sets.
We now prove $f(\widehat K)=\widehat K$. If $0 \leq j \leq m-2$, then
\begin{align*}
f(f^j(K)) = f^{j+1}(K) \subset \widehat K.
\end{align*}
For the last term,
\begin{align*}
f(f^{m-1}(K)) = f^m(K) = K,
\end{align*}
because $K$ is invariant under $f^m$. Thus $f(\widehat K) \subset \widehat K$. Applying the same argument to $f^{-1}$, or using that $f$ is a diffeomorphism and maps the finite cycle of compact sets onto itself, gives the reverse inclusion. Therefore $f(\widehat K)=\widehat K$.
[/step]
[step:Derive periodic orbits and positive topological entropy]
For each integer $r \geq 1$, choose a periodic sequence $s_r \in \Sigma_2$ of least shift period $r$. Since $\pi:K \to \Sigma_2$ is a conjugacy, the point
\begin{align*}
x_r := \pi^{-1}(s_r)
\end{align*}
satisfies
\begin{align*}
F^r(x_r)=x_r.
\end{align*}
Equivalently,
\begin{align*}
f^{mr}(x_r)=x_r.
\end{align*}
The full two-shift has periodic sequences of arbitrarily large least period, so this gives infinitely many distinct periodic orbits of $f$ inside the invariant set $\widehat K$.
We now justify the stated accumulation on the homoclinic orbit. Let $U \subset M$ be an arbitrary neighbourhood of the homoclinic orbit $\mathcal{O}(q)$. Use the local horseshoe construction with the full finite tower contained in $U$, so that
\begin{align*}
\bigcup_{j=0}^{m-1} f^j(R) \subset U.
\end{align*}
Since $K \subset R$, the corresponding invariant set satisfies
\begin{align*}
\widehat K = \bigcup_{j=0}^{m-1} f^j(K) \subset \bigcup_{j=0}^{m-1} f^j(R) \subset U.
\end{align*}
The preceding periodic-sequence argument produces a periodic orbit of $f$ contained in this corresponding finite union $\widehat K$. Since $U$ was arbitrary, periodic orbits of $f$ accumulate on $\mathcal{O}(q)$.
Finally, [topological entropy is invariant under topological conjugacy](/theorems/6805) and the full two-shift has entropy $\log 2$ (citing a result not yet in the wiki: entropy of the full shift). Hence
\begin{align*}
h_{\mathrm{top}}(f^m|_K)=\log 2.
\end{align*}
The compact set $\widehat K$ is $f$-invariant, so $f|_{\widehat K}:\widehat K \to \widehat K$ is a continuous map on a compact space. The iterate $(f|_{\widehat K})^m$ restricts on the compact invariant subset $K \subset \widehat K$ to $f^m|_K$. Monotonicity of entropy under restriction gives
\begin{align*}
h_{\mathrm{top}}((f|_{\widehat K})^m) \geq h_{\mathrm{top}}(f^m|_K)=\log 2.
\end{align*}
The standard iterate identity for topological entropy applies to the compact dynamical system $f|_{\widehat K}:\widehat K \to \widehat K$, so
\begin{align*}
h_{\mathrm{top}}((f|_{\widehat K})^m)=m h_{\mathrm{top}}(f|_{\widehat K}).
\end{align*}
Therefore
\begin{align*}
h_{\mathrm{top}}(f|_{\widehat K}) \geq \frac{\log 2}{m} > 0.
\end{align*}
By the stated noncompact-space convention, $h_{\mathrm{top}}(f)$ is the supremum of the entropies over compact $f$-invariant subsets of $M$, and $\widehat K$ is one such subset. Hence
\begin{align*}
h_{\mathrm{top}}(f) \geq h_{\mathrm{top}}(f|_{\widehat K}) \geq \frac{\log 2}{m} > 0.
\end{align*}
This completes the proof.
[/step]
