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\}.

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