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For a continuous map on a compact metric space, the Bowen metric compares the first $n$ points of two forward orbits. Its radius-$ϵ$ ball is $B_n(x,ϵ)=\{y:d_n(x,y)<ϵ\}$. If $s_n(ϵ)$ is the largest cardinality of an $ϵ$-separated set in this metric, then topological entropy is ${\lim }_{ϵ\downarrow 0}{\mathrm{limsup}}_{n\to \infty }{n}^{-1}\log s_n(ϵ)$. The uniform continuity of the identity between two compatible metrics gives the comparison of separated-set counts needed for independence of the original metric.

Use the separated-set definition of topological entropy. For a compatible metric $d$ and $n\ge 1$, define the Bowen metric and separated-set count byFor fixed $n$, the metric $d_n$ generates the original topology: it includes the $k=0$ term and all finitely many iterations of a map are continuous. Thus its space is compact and $s_n(d,ϵ)$ is finite. With natural logarithms, setThe last limit exists, possibly with value infinity, because decreasing $ϵ$ can only increase separated-set counts.

Let $d^{\prime }$ be another compatible metric. The identity $(X,d)\to (X,d^{\prime })$ is a continuous function on a compact domain, hence uniformly continuous. For every $ϵ>0$, choose $\delta >0$ such that $d(x,y)<\delta$ implies $d^{\prime }(x,y)<ϵ$. The same implication at every iterate givesBy its contrapositive, every $ϵ$-separated set for $d_n^{\prime }$ is $\delta$-separated for $d_n$. Consequently $s_n(d^{\prime },ϵ)\le s_n(d,\delta )$ for every $n$, andTaking $ϵ\downarrow 0$ gives $h_{d^{\prime }}(f)\le h_d(f)$. Interchanging the two compatible metrics gives the reverse inequality. ThereforeCompactness is used for uniform continuity, not merely for pointwise equivalence of the two metrics.

An interval covering relation $I\to J$ means $F(I)\supseteq J$. The associated directed graph has one vertex for each interval and an edge $i\to j$ for each covering. Its transition matrix is $A_{ij}=1$ when that edge exists, and zero otherwise. Matrix multiplication counts walks; thus $\boxed{\mathrm{tr}(A^n)}$ counts closed length-$n$ itineraries, with a distinguished starting vertex.

To see how itineraries force periodic orbits, a continuous map covering a closed interval has a closed subinterval mapping onto it: choose successive first/last crossings of its endpoints. Pull these subintervals back along a closed walk. This yields $K\subseteq I_{i_0}$ with $F^n(K)\supseteq I_{i_0}\supseteq K$. At the two preimages of the ends of $K$, $F^n(x)-x$ has opposite weak signs; the intermediate value theorem gives a fixed point of $F^n$.

A graph alone does not give an exact count of periodic points for a general continuous map: extra oscillations or a fixed interval can create additional points. For example on $I=[0,1]$, both $F(x)=x$ and $F(x)=1-x$ have matrix $A=(1)$; the first has infinitely many fixed points and the second only one. Distinct itineraries give distinct points when their interval interiors are disjoint and the periodic orbit avoids shared endpoints. In that setting, $\mathrm{tr}(A^n)$ is a guaranteed lower bound, not a universal equality. Likewise, the primitive-word count $(1/n){\sum }_{d\mid n}\mu (d)\mathrm{tr}(A^{n/d})$ counts guaranteed least-period itineraries under this separation condition. With overlapping intervals, even distinctness needs qualification. This is the scope of counting cycles in an interval covering graph.

For the first five-cycle, put $I_j=[x_j,x_{j+1}]$, $0\le j\le 3$. The graph and matrix areThe four traces are $1,3,7,15$. Shared endpoints have least period five, so they do not spoil the distinctness argument for the requested periods. Primitive cyclic itineraries are: $3$ for period one; $(3,2)$ for period two; $(3,1,2)$ and $(3,3,2)$ for period three; $(0,1,2,3)$, $(3,3,1,2)$ and $(3,3,3,2)$ for period four. Consequently the monotone five-cycle interval covering pattern guaranteesThese words specify the successive intervals containing their points; cyclic rotations denote the same orbit.

There is a horseshoe for an interval map for $F^2$: both $I_2$ and $I_3$ cover $I_2\cup I_3$ under that iterate, and their interiors are disjoint. Repeated binary choices produce at least $2^n$ separated length-$n$ itineraries for the second iterate, hence positive topological entropy, at least $(\log 2)/2$ for $F$. Thus the dynamics are chaotic in this standard interval-map sense. No conclusion of global transitivity on all of $\mathbb{R}$ follows.

A one-step horseshoe is not guaranteed if that is the intended definition. For example the unimodal continuous map $F(x)=x+1$ for $x\le 3$ and $F(x)=16-4x$ for $x\ge 3$ has cycle $0,1,2,3,4$. A left-branch interval cannot cover itself, since its image is shifted to the right. Two right-branch intervals cannot each cover their union because that branch is injective. If one interval straddles the maximum and covers itself, its right endpoint must map below its left endpoint; every disjoint interval farther to the right then maps below it and cannot self-cover. A disjoint interval to the left also cannot self-cover. So this example has no one-step two-branch horseshoe, although its second iterate does. It also proves that $F$ can be unimodal.

For the second order, write $J_0=[x_3,x_1]$, $J_1=[x_1,x_0]$, $J_2=[x_0,x_2]$, $J_3=[x_2,x_4]$. Endpoint images giveThe traces are $1,3,1,7$. The guaranteed primitive words are $1$, $(0,3)$ and $(1,2,0,3)$, giving $\boxed{\text{periods }1,2,4\text{ but not necessarily }3}$. To prove the last qualification, take ordered points $0,1,2,3,4$ and linearly interpolate the map values $4,3,1,0,2$ on $[0,4]$, extending by $G(x)=4-x$ for $x<0$ and $G(x)=2$ for $x>4$. Both exterior regions enter $[0,4]$ after at most two steps, so they introduce no extra periodic points. Its only closed length-three itinerary is $111$; on $J_1$ it is $x\mapsto 5-2x$, whose third iterate has only the same fixed point. Endpoints have least period five, so this example has no three-cycle.

The distinct fourth-iterate return walks $1111$ and $1203$ yield two subintervals of $J_1$ that return over $J_1$, giving a horseshoe for $G^4$ and positive entropy at least $(\log 2)/4$. Thus $G$ too must have chaotic interval dynamics and a horseshoe in some iterate. The affine example has no one-step horseshoe: its increasing branch lies strictly below the diagonal, while on the decreasing branch injectivity prevents two self-covering intervals; a minimum-straddling self-covering interval excludes a self-covering interval on either side by the same endpoint argument. The distinction between a map and its iterate is essential for both patterns.

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.

Two continuous maps are topologically conjugate if a homeomorphism $H$ between their state spaces satisfies the displayed intertwining identity. It is conjugate maps with a topology-preserving conjugating map. On compact metric spaces, pulling back a compatible metric through $H$ preserves every Bowen metric, and hence topological entropy.