A positive Lyapunov exponent at a hyperbolic equilibrium does not produce a chaotic invariant set of the time-one map: the three-dimensional repulsive model has rate $1$ and no compact invariant set with more than one point

The antecedent holds: by [L1] the model has equilibrium $(0,0)$, the linearisation has eigenvector $({\mathbf{e}}_1,{\mathbf{e}}_1)$ with eigenvalue $1>0$, and the Lyapunov exponent along that direction exists and equals $1$. The explicit flow is defined for every real time. This satisfies the antecedent of the refuted statement.

Classification of compact invariant sets of $f$: let $K$ be compact with $f(K)=K$, and let a state of $K$ have transformed coordinate $\mathbf{u}\mathrm{\ne }0$. By [L1], $\parallel f^n(\mathbf{q},\mathbf{v}){\parallel }_2\ge \frac{1}{\sqrt{2}}\parallel \mathbf{u}{\parallel }_2{e}^n$ for $n\ge 0$, which is unbounded; this contradicts boundedness of $K$. Hence $\mathbf{u}=0$ on $K$. Since $f$ is a bijection and $f(K)=K$, applying the same argument to $f^{-n}(K)=K$ shows $\mathbf{w}=0$ on $K$. Therefore every nonempty compact invariant set of $f$ is the singleton $\{(0,0)\}$.

The one-point invariant set fails the chaos list: for $X=\{(0,0)\}$ every admissible $y$ in the sensitivity condition of [L2] coincides with the base point, so $d(f^n(x),f^n(y))=0<\delta$ for every $n$ and every $\delta >0$; the sensitivity clause fails, and with it the four-clause list of Chaos in the stated sense for maps of the model: topological mixing, dense periodic points and sensitive dependence, as one precise list among several.

Conclusion: the antecedent of the refuted statement holds for the declared model (step 1.1), while its time-one map has no compact invariant set with more than one point (step 1.2) and its only compact invariant set is not chaotic in the stated sense (step 2.1); the implication is therefore false. Nonlinear instability of the equilibrium from [L3] does not repair it: instability of a single state and chaos of a compact invariant set are different properties.