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For $1<s\le \sqrt{2}$, set $h_s(x)=-(s-1)x$ and $J_s=[-(s-1),s-1]$. The tent map sends $J_s$ to $[1+s-s^2,1]$, with disjoint interiors, and $T_s^2(J_s)\subseteq J_s$. Directly,Successive squaring eventually brings every $s>1$ into $(\sqrt{2},2]$. A horseshoe for an interval map of an iterate in the renormalized coordinate lifts to a horseshoe of an iterate of the original map.

// Target: dynamical-systems.bigb

A continuous interval map is chaotic in Glendinning's sense when some positive iterate has a horseshoe for an interval map.

If a directed covering graph of an interval map has two distinct closed walks of the same length $n$, based at the same interval and with different first edges, successive use of the interval covering relation produces two subintervals with disjoint interiors that $F^n$ maps across the base interval. Hence $F^n$ has a horseshoe for an interval map.

Suppose a interval map $F:A\to A$ maps every nondegenerate subinterval onto $A$ after finitely many iterates. Choose two separated closed intervals inside the interior of $A$ and a common iterate mapping each onto $A$. Between first-passage endpoint levels in each interval, take an open interval that maps exactly onto the interior of $A$. These two disjoint intervals are the branches of a horseshoe for an interval map for that iterate.

// Target: dynamical-systems.bigb

A continuous interval map has positive topological entropy if and only if some positive iterate has a horseshoe for an interval map.

A five-cycle whose points are in orbit order along the real line gives four interval vertices with arrows $0\to 1\to 2\to 3$ and $3\to 0,1,2,3$. The transition-matrix traces for lengths one through four are $1,3,7,15$. Primitive closed words force at least one fixed point, one two-cycle, two three-cycles and three four-cycles. The second iterate has a two-branch horseshoe for an interval map on vertices two and three. The map itself need not have a one-step horseshoe: the unimodal affine example $F(x)=x+1$ for $x\le 3$, $F(x)=16-4x$ for $x\ge 3$ has precisely this five-cycle and has no two disjoint self-covering branches.

Use the interval exactness of a tent-map core from the preceding part. Choose two separated nondegenerate closed intervals $J_0,J_1$ inside $J=(1-s,1)$. There are integers $n_i$ with $T_s^{n_i}(J_i)=A$. Taking $N=\max (n_0,n_1)$ works for both, because $T_s(A)=A$:Within each $J_i$, the continuous function $F$ attains both endpoint levels $1-s$ and $1$. Choose a segment between those levels, reversing its orientation if necessary. Starting from a point at level $1-s$, take the first point at level one, and then the last point at level $1-s$ before it. Between these two points, continuity and the first-passage choice give $1-s<F(x)<1$, with all intermediate levels attained. The interior $K_i$ of that segment therefore satisfies $F(K_i)=J$. The intervals $K_0,K_1$ are disjoint and lie inside $J$.

Thus interval exactness produces a horseshoe:This is a horseshoe for an interval map for an iterate of $T_s$, proving Glendinning chaos for every $\sqrt{2}<s\le 2$. The proof does not require the same horseshoe iterate $N$ to work for every parameter.

For $1<s\le \sqrt{2}$, use centered tent-map period-doubling renormalization. LetThe image $T_s(J_s)=[1+s-s^2,1]$ lies to the right of $J_s$, with disjoint interiors, because $1+s-s^2\ge s-1$. It lies in the positive branch. Consequently,and its range $[-c,(s^2-1)c]$ is contained in $J_s$. The orientation-reversing affine map $h_s$ gives the exact topological conjugacyThus renormalization squares the slope and halves the return period. Since $s>1$, there is a least $n\ge 0$ for which $s^{2^n}>\sqrt{2}$. Minimality implies $s^{2^n}\le 2$, and all previous slopes are in the renormalizable range. Successive affine maps therefore conjugate the restriction of $T_s^{2^n}$ on a nested central interval to $T_{s^{2^n}}$.

By the preceding part, an iterate $T_{s^{2^n}}^m$ has a horseshoe for an interval map. Pull its two branch intervals and its range interval back through the composed conjugacy. They give a horseshoe for $T_s^{2^{n}m}$. ThereforeAt the endpoint $s=\sqrt{2}$, one renormalization gives the full slope-two tent map, so the argument includes that endpoint without assuming the strict inequality from part (a).

A horseshoe for an interval map consists of an open interval $J$ containing two disjoint open subintervals each mapped onto $J$. Glendinning chaos means that some positive iterate has a horseshoe.

Solving on the appropriate branches of the tent map givesHere $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate isBoth endpoints map to $x_0$, and the minimum is $v=\mu -{\mu }^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu -1)({\mu }^2-2)\ge 0$. Restricting them to preimages of the interior proves a horseshoe for $F^2$ when $\mu \ge \sqrt{2}$.

For $1<\mu \le \sqrt{2}$, $G(J)\subseteq J$. Set $d=(\mu -1)/(\mu +1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hG{h}^{-1}=F_{{\mu }^2}$. The renormalization of the tent map near its fixed point therefore makes $F^4$ conjugate on $J$ to $F_{{\mu }^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le \mu <\sqrt{2}$.

For $\mu \ge \sqrt{2}$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by ${\mu }^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a fixed point with a periodic itinerary, distinct from $x_0$ and at distance $O({\mu }^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.

For $\mu <\sqrt{2}$, the conjugate tent parameter $\nu ={\mu }^2<2$ has every nonzero periodic point in $[\nu -{\nu }^2/2,\nu /2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus $x_0$ is isolated among periodic points, even though a fourth-iterate horseshoe exists elsewhere.

In Glendinning chaos, a continuous interval map is chaotic when some positive iterate has a horseshoe for an interval map. In the interval version, one may take an open interval $J$ containing two disjoint open intervals $K_0,K_1$ with $F^m(K_0)=F^m(K_1)=J$.

Let a three-cycle be ordered $a<b<c$. There are only two possible cyclic orderings; reflection of the coordinate interchanges them, so assume $F(a)=b$, $F(b)=c$, $F(c)=a$. ThenAlso $F(b)=c$ and $F(c)=a$, so the intermediate value theorem gives $d\in (b,c)$ with $F(d)=b$, whence $F^2(d)=c$. Both $[a,b]$ and $[b,d]$ therefore map across $[a,c]$ under $F^2$.

To obtain exact open-interval images, use the following elementary first-passage observation. A continuous function on $[r,s]$ with endpoint values $a,c$ has a subinterval on which its endpoint values are $a,c$ and all interior values lie strictly between them: take the first hit of the second level, then the last hit of the first level before it. The intermediate value theorem makes its interior image exactly $(a,c)$. Apply this to the two intervals, reversing endpoint order on the first. Their selected interiors $K_0\subset (a,b)$ and $K_1\subset (b,d)$ are disjoint and lie in $J=(a,c)$, withThis proves the three-cycle forces a two-iterate interval horseshoe result and establishes the required horseshoe for $F^2$.

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.

A horseshoe for an interval map consists of two branches covering a common interval: there are an open interval $J$ and disjoint open subintervals $K_0,K_1\subseteq J$ with $f(K_0)=f(K_1)=J$. Equivalently one can work with two closed intervals with disjoint interiors whose images cover their union, and then restrict the branches. Glendinning chaos means that some positive iterate $f^m$ has such a horseshoe, rather than requiring one for $f$ itself.

For the three-cycle, use the intermediate value theorem on endpoint images. The interval covering relations are $I_a\to I_b$ and $I_b\to I_a,I_b$. Consequently both $f^2(I_a)$ and $f^2(I_b)$ cover $I_a\cup I_b$. This proves $f^2$ has a horseshoe. The directed covering graph of an interval map has exactly these forced arrows; a closed walk $a\to b\to b\to a$ has length three and a primitive itinerary. Pulling back the intervals successively and using the fixed-point property produces a point fixed by $f^3$ with this itinerary. It cannot be fixed by $f$, and a shared endpoint belongs to the already given three-cycle, so a genuine period-three orbit results.

For the increasing spatial order of the five-cycle, let $I_j=[x_{j-1},x_j]$, $1\le j\le 4$. The forced graph isThe primitive loops $I_4$, $I_3{I}_4$, $I_2{I}_3{I}_4$, and $I_1{I}_2{I}_3{I}_4$ force exact periods $\boxed{1,2,3,4}$. For periods below five the known five-cycle endpoints cannot be responsible, so the interval itineraries give distinct interior orbits. Both $g^2(I_3)$ and $g^2(I_4)$ cover $I_3\cup I_4$, so $g$ is guaranteed to be Glendinning-chaotic.

A horseshoe for $g$ itself is not guaranteed. As a counterexample put $x_j=j$ on $[0,4]$ and interpolate by $g(x)=x+1$ for $0\le x\le 3$, $g(x)=16-4x$ for $3\le x\le 4$. This has the required five-cycle. The unique fixed point is $16/5$, with $g(x)>x$ to its left and $g(x)<x$ to its right. Any nondegenerate closed interval whose image covers itself must contain this fixed point in its interior: an interval entirely on either side, or with the fixed point as an endpoint, fails to cover its extreme endpoint. Two intervals with disjoint interiors therefore cannot both cover themselves, as a horseshoe would require. Hence this example has no one-iterate horseshoe.

For the alternative spatial order, set $y_0=x_4,y_1=x_2,y_2=x_1,y_3=x_3,y_4=x_0$ and $J_j=[y_{j-1},y_j]$. Endpoint images giveThe self-loop at $J_2$, the two-cycle $J_1{J}_4$, and the four-cycle $J_2{J}_3{J}_1{J}_4$ force $\boxed{1,2,4}$; period three is not guaranteed. To see this last claim, take the connect-the-dots map at $y_j=j$ with values $(4,3,1,0,2)$. Its exact interval transitions are those displayed. Any length-three closed itinerary is the repeated self-loop in $J_2$; there the map is the affine decreasing function $5-2x$, whose third iterate has only its original fixed point. Endpoints belong to the five-cycle. Thus no exact three-cycle exists in this example.

Chaos is still guaranteed: from $J_2$ there are two different length-four closed walks, four self-loops and $J_2{J}_3{J}_1{J}_4{J}_2$. Pullbacks give disjoint branches of $g^4$ over a common interval in $J_2$, hence a horseshoe for an interval map for an iterate. A horseshoe for $g$ itself is again not guaranteed: the same connect-the-dots example is strictly decreasing on $[0,3]$ and increasing on $[3,4]$ with image $[0,2]$. Its unique fixed point is $5/3$, again with $g(x)>x$ to the left and $g(x)<x$ to the right. The same fixed-point argument rules out two self-covering intervals with disjoint interiors. These counterexamples distinguish the three separate questions about periods, chaos and a one-iterate horseshoe.

An interval map has a horseshoe for an interval map when two disjoint subintervals are each mapped across a common interval. It is Glendinning-chaotic when one of its positive iterates has such a horseshoe.

Let the points of the three-cycle be $x_1<x_2<x_3$ and put $I_0=[x_1,x_2]$, $I_1=[x_2,x_3]$. There are two possible cyclic orders. If $x_1\mapsto x_2\mapsto x_3\mapsto x_1$, the intermediate value theorem givesfor the interval covering relation. Thus there are two closed covering walks of length two based at $I_1$. In the reverse cyclic order there are instead the arrows $I_0\to I_0$, $I_0\to I_1$ and $I_1\to I_0$, again giving two closed walks of length two. The horseshoe from two closed covering walks therefore shows that $F^2$ has a horseshoe in either case, so $F$ is chaotic.

The Sharkovsky theorem orders the positive integers asand says that a cycle of a given period forces cycles of every period to its right.

Suppose $N=2^{r}q$ with $q>1$ odd. Then $F^{2^r}$ has a $q$-cycle. The Sharkovsky theorem gives that map a six-cycle, and squaring it produces a three-cycle. Hence $F^{2^{r+1}}$ has a three-cycle and an iterate of $F$ has a horseshoe. Thus

A horseshoe contains the symbolic dynamics of the Bernoulli shift. Periodic binary words give cycles of every symbolic period; after translating from an iterate back to $F$, this supplies cycles of many periods that are not powers of two.

For the logistic map, a ten-cycle would imply chaos because $10$ is not a power of two. Hence there is no ten-cycle for $\mu <{\mu }_{\infty }$. A three-cycle forces every period, so a ten-cycle exists for $\mu >1+\sqrt{8}$. The stated facts alone give no conclusion for the intermediate range:

A map is Devaney-chaotic on $\Lambda$ when it is topologically transitive, it has dense periodic points in $\Lambda$, and it has sensitive dependence on initial conditions. A one-dimensional map is Glendinning-chaotic when some iterate has a horseshoe for an interval map: there are an interval $J$ and two disjoint subintervals $K_0,K_1\subseteq J$ such that $F^n(K_0)=F^n(K_1)=J$.

The map $F(x)=ax(\mathrm{mod}1)$ is the beta transformation with slope $a>1$. On every interval of continuity of $F^n$, it is affine with slope $a^n$. The endpoints of these intervals are iterated preimages of the discontinuity set. Choosing $n$ with $a^n>2$, the standard full-branch lemma for the beta transformation supplies two disjoint continuity intervals $K_0,K_1$ on which $F^n$ maps bijectively onto $[0,1)$. Thus $F^n$ has a horseshoe, proving Glendinning chaos.

For Devaney chaos one may either use the standard theorem that a full-branch expanding interval map is Devaney-chaotic, or verify its ingredients directly. Every nonempty open interval expands by the factor $a$ on each uninterrupted iterate; after subdivision at preimages of discontinuities, one of its images contains a full branch and hence eventually covers $[0,1)$. This proves topological transitivity. Applying an inverse branch of a sufficiently high iterate to a full branch gives a contraction whose unique fixed point is periodic; choosing the branch inside any prescribed open interval proves that periodic points are dense. Finally, two sufficiently close points that follow the same branch separate by the factor $a$ on each iterate until their distance exceeds a fixed positive constant, proving sensitive dependence. Hence $F$ is also Devaney-chaotic for every real $a>1$.

The map is chaotic in Glendinning's sense when some positive iterate $F^n$ has a horseshoe.

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ such that

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.

An interval map $F:I\to I$ has a horseshoe if there are two closed subintervals $J_0,J_1\subseteq I$ with disjoint interiors such thatIt is chaotic in Glendinning's sense if some positive iterate $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a 3-cycle and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. Ifthen the intermediate value theorem givesConsequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If insteadthenand again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

For a continuous interval map $F$ with a three-cycle $a<b<c$, after reflecting the coordinate if needed its ordering is $F(a)=b$, $F(b)=c$, $F(c)=a$. Then $F^2(a)=c$, $F^2(b)=a$. Also the intermediate value theorem gives $d\in (b,c)$ with $F(d)=b$, hence $F^2(d)=c$. Each of $[a,b]$ and $[b,d]$ therefore maps across $[a,c]$ under $F^2$. Choose first-passage subintervals between endpoint levels to avoid overshoots; their interiors are disjoint and both map exactly onto $(a,c)$. This is a horseshoe for an interval map, and establishes Glendinning chaos.