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In an open-interval convention, a continuous interval map $F:I\to I$ has a horseshoe if there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ with $F(K_0)=F(K_1)=J$. Each of the two branches can therefore be followed by either branch on the next step.

For the full tent map on $[0,1]$, choose $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$. Either branch maps onto $J$. Successive preimages specify a binary itinerary; when branches are monotone they have ordinary inverse branches, while more general continuous covering arguments use the intermediate value theorem to select subintervals.

Open endpoints require care: for example, the all-left limiting itinerary of the tent map selects the boundary point zero. Compact symbolic realizations include suitable endpoints and account for boundary coding ambiguities. The branch-covering geometry nevertheless supplies arbitrarily long binary itineraries and is the mechanism used by interval exactness produces a horseshoe and three-cycle forces a two-iterate interval horseshoe.

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.