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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.