## Union of Compacta

1. Let $\{K_n:n\ge1\}$ be a sequence of compact sets in $\mathbb{R}^d$.
If $K=\bigcup_{n\ge1}K_n$ is compact, does there exist a finite collection of indexes $I\subset\mathbb{N}$ such that $K=\bigcup_{n\in I}K_n$?

Example: Let $K_0=\{0\}$ and $K_n=\{1/n\}$ for all $n\ge1$. Then $K=\bigcup_{n\ge0}K_n$ is compact and any finite union of $K_n$‘s is smaller than $K$.

2. Now return to the case that $\Lambda_n$ is a compact uniformly hyperbolic invariant subset of $f:M\to M$ for each $n\ge1$. If $\Lambda=\bigcup_{n\ge1}\Lambda_n$ is compact, could it be uniformly hyperbolic for $f$?

In this case we have a neighborhood $U_n$ containing $\Lambda_n$ for each $n\ge1$ such that $U_n$ admits stable and unstable cones. Now $\{U_n:n\ge1\}$ forms an open covering of $\Lambda$. So a finitely many of $U_n$ will cover $\Lambda$. Then $\Lambda$ also admits a neighborhood $U$ on which we have stable and unstable cones. So $\Lambda$ is uniformly hyperbolic for $f:M\to M$.

3. Does the situation become better for hyperbolic case? That is, if $\Lambda_n$ is hyperbolic and $\Lambda=\bigcup_{n\ge1}\Lambda_n$ is compact, is there some finitely many $\Lambda_n$ that make the whole $\Lambda$?

Anosov rigidity. Let $\Lambda$ be an invariant set on which a diffeomorphism $f$ is nonuniformly hyperbolic. It is compact only if $f$ is uniformly hyperbolic on $\Lambda$. This phenomenon is called Anosov rigidity.

Theorem. A smooth hyperbolic invariant measure for a $C^{1+\alpha}$ diffeomorphism of a compact manifold $M$ decomposes into countably many ergodic components of positive measure.

If the unstable foliation extends to a continuous foliation of $M$ with smooth leaves and if every local unstable leaf (regardless of its size) eventually expands to a certain uniform size when pushed forward, then each ergodic component can be taken to be an open set. This is referred to as local ergodicity.

—————–

Let $X$ be a compact metric space and $\mathcal{K}(X)$ be the collection of nonempty compact subsets of $X$ endowed with Hausdorff metric $d_H$. Then $(\mathcal{K}(X),d_H)$ is a compact metric space.

Let $K_n\in\mathcal{K}(X)$ and consider its limsup and liminf:

$\limsup_{n\to\infty}K_n=\bigcap_{m\ge1}\overline{\bigcup_{n\ge m}K_n}$.

$\liminf_{n\to\infty}K_n=\{x\in X| \forall U\ni x, \exists N\ge1:K_n\cap U\neq\empty\ \forall n\ge N\}$.