Initial conditions that converge to an unstable equilibrium

Question

Consider the discrete-time dynamical system \begin{align} x_{k+1}=T(x_k), \end{align} where $k\in\mathbb{N}$, $x_k\in\mathbb{R}^n$, $x_0$ is the initial condition, and $T:\mathbb{R}^n\to\mathbb{R}^n$ is a diffeomorphism on $\mathbb{R}^n$. Assume $x_{\rm u}$ is an unstable equilibrium. I think that it's well-known that the set of initial conditions which converge to $x_{\rm u}$ is a set of Lebesgue measure zero. I can't find the proof, but I believe that I can prove it using the center manifold theorem.

1) Is that right?

2) How about the case of uncountable number of unstable equilibria? I mean assuming that the set $\mathcal{U}\triangleq\{x_{\rm u}\in\mathbb{R}^n:x_{\rm u} {\rm ~is~ an~ unstable~ equilibrium~ of~ the~ system}\}$ is uncountable with Lebesgue measure zero, is that true that the set of initial conditions which converge to $\mathcal{U}$ is a set of Lebesegue measure zero?

Any hint, counterexample, proof, or reference? Thanks

Answer 1

I don't have any proof but let me try to give intuition for linear systems, or linearization around the equilibrium if you wish. Let the linearized model is $x_{k+1} = A x_k$. If $Au_i = \lambda_i u_i$, then $x_k = \lambda_i^k u_i$ is a solution. If $|\lambda_i| < 1$, then the system converges to the equilibrium for $x_0 = u_i$. Since this system is unstable $A$ has at least one eigenvalue such that $|\lambda| > 1$. This means the set of initial conditions which converge to the equilibrium are at most $(n-1)$-dimensional subspace spanned by the stable eigenvectors of $A$ and I think this subspace has Lebesgue measure of $0$ inside $\mathbb{R}^n$.

Answer 2

Take ODE $\dot x=x^2$. Then $\hat{x}=0$ is unstable, and at the same time it attracts any point from $(-\infty,0)$.

If you are unhappy with this example, take a system in polar coordinates $$ \dot r=r(1-r),\\ \dot \theta=\sin^2(\theta/2)+(r-1)^2 $$ The equilibrium $(1,0)$ is unstable, however it attracts orbits starting at any point in $\mathbb R^2$.