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The image of the full interval under the tent map is $T_s([-1,1])=[1-s,1]$. Since $s>1$, this image contains the critical point zero and the right endpoint one; its next image is again $[1-s,1]$. Moreover, any interval $B$ with $T_s(B)=B$ lies inside the image of $[-1,1]$. Thus the core interval of an expanding tent map isEvery point outside $A$ enters $A$ after one iterate and then stays there. For $s=2$ there are no points of the domain outside $A$.

To prove interval exactness of a tent-map core, first take a nondegenerate closed interval $L\subseteq A$ and put $L_n=T_s^n(L)$. Each $L_n$ is an interval. If $L_n$ does not straddle zero, the corresponding branch is affine with slope of magnitude $s$, so $|L_{n+1}|=s|L_n|$. If it straddles zero, its image has right endpoint one andIf both $L_n$ and $L_{n+1}$ contain zero, then $L_{n+1}$ contains $[0,1]$, whence $L_{n+2}=A$. Otherwise at most one of these two steps folds across the critical point. Consequently, unless the full core has already been reached,Since $s^2/2>1$, indefinite failure to reach $A$ would give intervals of unbounded length inside an interval of length $s$. This is impossible, proving $\boxed{T_s^k(L)=A\text{ for some }k}$. For an interval with open endpoints, apply the argument to any nondegenerate closed subinterval inside it; its image already covers $A$, while the image of the original interval cannot leave $A$.

Now let $U$ be any nonempty relatively open subset of $A$ and choose a closed nondegenerate $J=[u,v]\subset U$. Some $F=T_s^k$ maps $J$ onto $A$, so there are points $p,q\in J$ with $F(p)=1-s$ and $F(q)=1$. Therefore $F(p)-p\le 0$ and $F(q)-q\ge 0$. The intermediate value theorem yields $z\in J$ with $F(z)=z$. This is a periodic point of $T_s$, of period dividing $k$. Every such $U$ contains one, so the periodic points are dense in the whole core $A$.

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.