# Reversal of open cover with topologically transitive dynamical system

Let $$X$$ be a separable metric space, $$\phi\in C(X,X)$$ be a topologically transitive dynamical system, and $$V$$ be a non-empty open subset of $$X$$, and $$\nu$$ be a locally-positive and atomless Borel probability measure on $$X$$.

Then, for every $$\delta \in (0,1)$$, does there exist:

• $$\{V_i\}_{i \in \mathbb{N}}$$ are open subsets of $$V$$ satisfying $$\nu\left( V-\cup_{n \in \mathbb{N}} V_n \right)=0$$

• A sequence $$\{N_i\}_{i \in \mathbb{N}}$$, such that the following holds: $$\nu\left(X - \cup_{i \in \mathbb{N}} \phi^{-N_i}[V_i] \right)<\delta .$$

In intuitive words: there exists an open cover of a non-empty open set be reversed into an almost-everywhere cover of the entire space by appropriately reversing the dynamical sytem?

Auxiliary Definitions:

• A Borel measure $$\nu$$ on $$X$$ is said to be locally-positive iff for every non-empty open subset $$U\subseteq X$$, $$\nu(U)>0$$. For example, if $$X$$ has more than two points then the Dirac is not such a measure.

• $$\phi$$ is said to be topologically transitive iff for every two non-empty open subsets $$U,V\subseteq X$$ there exists some $$N\in \mathbb{N}$$ such that $$\phi^N(U)\cap V \neq \emptyset.$$

• If a solution $(V_i, N_i)_i$ exists, then $(V, N_i)_i$ is also a solution, so $\delta_1$ is unimportant. Is this intentional? Oct 30, 2019 at 15:25
• Do you require $V_i$ to be distinct? If not, since the existence of a solution implies $(V, N_i)_i$ is a solution, and if $(V, N_i)_i$ is allowed to be a solution, why not just ignore $V_i$ and work only with $V$? Oct 30, 2019 at 15:47
• Are you assuming that $\nu$ is finite? at least locally?
– YCor
Oct 30, 2019 at 16:23
• (Deleted a bunch of comments since they have been clarified and now there are no obvious counterexamples.) Oct 30, 2019 at 16:37
• Thanks you Willie; I'll do the same.
– ABIM
Oct 30, 2019 at 16:38

No, even if we assume $$\nu$$ to be invariant under $$\phi$$.

Let $$X = \{0,1\}^\mathbb{Z}$$ be the set of two-way infinite binary sequences with the prodiscrete topology, and let $$\phi$$ be the left shift on $$X$$. Let $$\nu = (\mu_1 + \mu_2)/2$$ where $$\mu_1$$ is the uniform Bernoulli measure on $$X$$ and $$\mu_2$$ is an atomless $$\phi$$-invariant probability measure on some proper subshift of $$X$$. For simplicity, let's choose $$\mu_2$$ as the Parry measure on the shift of finite type $$Y \subsetneq X$$ where $$0 0$$ is forbidden. Let $$V = \{ x \in X : x_0 = x_1 = 0 \}$$ be the set of sequences with an occurrence of the forbidden word $$0 0$$ at the origin. These definitions satisfy your requirements: $$\phi$$ is well-known to be transitive, $$\nu$$ gives positive measure to each nonempty clopen set (which form a basis of the topology) and has no atoms, and $$V$$ is a nonempty open set.

Consider an open cover $$(V_i)_{i \in \mathbb{N}}$$ of $$V$$ and a sequence $$(N_i)_{i \in \mathbb{N}}$$ of integers. For each $$i$$ the translate $$\phi^{-N_i} V_i$$ is disjoint from $$Y$$, so $$\nu(X - \bigcup_{i \in \mathbb{N}} \phi^{-N_i} V_i) \geq \nu(Y) = 1/2$$.

But yes, if we strengthen the assumptions further.

In my counterexample the ergodic decomposition of $$\nu$$ features a positive-weight measure $$\mu_2$$ which is not locally positive. Let's thus assume that $$\nu$$ has an ergodic decomposition as an integral $$\nu = \int_{E(M_\phi)} x \, d\mu(x)$$ over the $$\phi$$-ergodic probability measures on $$X$$ and $$\mu$$-a.e. $$x \in E(M_\phi)$$ is locally positive. Then $$x(V) > 0$$ holds for those measures $$x$$. Since they are ergodic, this implies $$x(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) = 1$$, so that $$\nu(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) = \int x(\bigcup_{i \in \mathbb{N}} \phi^{-i} V) \, d\mu(x) = 1$$. Then $$V_i = V$$ and $$N_i = i$$ give the sequence you're looking for, for every $$\delta > 0$$. Note that even if I didn't use transitivity in this proof, it's implied by the existence of a locally positive ergodic measure.

Depending on your application, the assumptions of $$\phi$$-invariance and local positivity of the ergodic decomposition may be too strong. In the context of dynamical systems invariance seems natural, but by itself it's not enough.

• If instead, we strengthen the assumptions on $X$ and $\phi$; namely that it is a separable Banach space and we assume that $\phi$ is linear. Then the counterexample seems to vanish. In this case, it's not clear that the added assumptions are still necessary. Am I right?
– ABIM
Oct 31, 2019 at 9:03
• @MrMMS We can do something similar. Let $X = \ell^1$ and $\phi(x)_n = 2 x_{n+1}$. Define $\mu_1$ by first choosing $c$ from a distribution that is locally positive on $[0, \infty)$, and then choosing each $x_n$ independently and uniformly from $[-c/2^n, c/2^n]$. Define $\mu_2$ similarly but with $[0, c/2^n]$. These are invariant, atomless, and $\mu_1$ is locally positive: For an open ball $B_r(y)$, take $N$ with $\sum_{n \geq N} |y| < r/2$. For a $\mu_1$-random $x$, $\sum_{n \geq N} |x| < r/2$ and $\sum_{n < N} |x-y| < r/2$ with positive probability. Take $V = \{ x \in X : x_0 \in (-1,0) \}$. Oct 31, 2019 at 10:42
• By the way, it feels like I'm chasing some moving goalposts here. Do you have a specific dynamical system in mind? Or is the case of linear continuous transitive functions on separable Banach spaces the exact setting that you need? Oct 31, 2019 at 10:46
• It's indeed the exact setting (I was hoping to get something more general) if possible.
– ABIM
Oct 31, 2019 at 10:46
• In both examples, $\nu(X - \bigcup_i \phi^{-i} V)$ can be any number $a \in [0,1]$, since we can choose $\nu = (1-a) \mu_1 + a \mu_2$. Oct 31, 2019 at 10:53