Is completeness a necessary assumption for the Birkhoff Transitivity Theorem?

Question

The Birkhoff Transitivity Theorem asserts that any dynamical system $T:X \to X$ on a complete separable metric space without isolated points is topologically transitive if and only if there is a point with dense orbit.

A friend of mine and me were discussing whether completeness is a necessary assumption, but couldn't come up with a counterexample. We tried the restricted doubling map to the rationals mod $1$, but finding the obstruction seems rather hard, as the Baire Category Theorem is non-constructive.

Does anyone know a counterexample?

Thanks a lot!

Answer 1

The tent map $T:x\mapsto1-|2x-1|$ has the property that any nontrivial closed interval $I\subseteq[0,1]$ will eventually evolve into $[0,1]$: $T^n(I)=[0,1]$ all $n$ greater than some $N$. I prove this here.

Let’s now restrict to the separable metric space, without isolated points, $[0,1]\cap\Bbb Q$ and consider the topological dynamical system obtained by applying $T$. $T$ remains continuous and well-defined as it maps rationals to rationals. In particular, it still has the property that it evolves any nontrivial interval into the whole space (which can be seen from $T(x)\in\Bbb Q\iff x\in\Bbb Q$).

By your definition, it is not too hard to see from this property that $([0,1]\cap\Bbb Q;T)$ is then a transitive system. However, it cannot contain any transitive points; following Greg Martin’s comment under this related post, if $x$ is a rational with denominator $d$ then $T^n(x)$ is always a rational with denominator (dividing) $d$, hence the orbit of $x$ will never fall into $(0,1/d)$ and the point is not transitive.

This is why completeness is necessary! There are infinitely many transitive points to the system $([0,1];T)$ but these all fall through the “holes” when we restrict to the incomplete subspace. A close study of the Baire Category Theorem’s proof will reveal why.

Answer 2

Let us also note that (some) separability is also needed for the existence of a point with dense orbit (by definition).

As an explicit example, consider the space $X_0=F(A;[0,1])$ of all functions from a set $A$ to $[0,1]$, where $A$ is with $\operatorname{card}(A)>\operatorname{card}(\mathbb{R})$, the target is endowed with discrete topology and $X_0$ is endowed with product topology (= topology of pointwise convergence). Then $X_0$ is compact Hausdorff non-separable (see https://math.stackexchange.com/a/3360831/169085). Then define the space $X=F(\mathbb{Z};X_0)$ of bi-infinite sequences in $X_0$, again with product topology. Then $X$ too is compact Hausdorff (see The product of Hausdorff spaces is Hausdorff) and non-separable. Define $T:X\to X$ to be the shift homeomorphism, $\omega_\bullet\mapsto \omega_{\bullet+1}$. Then $T$ is topologically strong mixing (as defined in Topological weakly mixing implies totally transitive), hence also topologically transitive.

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In fact, analyzing the proof of the Birkhoff Transitivity Theorem one can obtain an optimal statement. For a continuous self-map $T:X\to X$ of a topological space $X$, for a point $x\in X$ and two subsets $A,B\subseteq X$ put

$$\mathcal{O}_x(T)=\{T^n(x)\,|\, n\in\mathbb{Z}_{\geq0}\},$$

$$\mathcal{D}(T)=\{x\in X\,|\, \overline{\mathcal{O}_x(T)}=X\},$$

$$N^T(A\to B)=\{n\in\mathbb{Z}_{\geq0}\,|\, T^n(A)\cap B\neq\emptyset\},$$

so that $\mathcal{O}_x(T)$ is the orbit of $x$ under $T$, $\mathcal{D}(T)$ is the subset of $X$ that consists of all points whose $T$-orbits are dense, and $N^T(A\to B)$ is the set of return times from $A$ to $B$. Denote by $\mathcal{T}(X)^\ast$ the collection of all nonempty open subsets of $X$. $T$ is topologically transitive if

$$\forall U,V\in\mathcal{T}(X)^\ast: N^T(U\to V)\neq\emptyset.$$

Theorem: Let $X$ be a Hausdorff ($\ast$) topological space, $T:X\to X$ be a continuous self-map. Then

1. If $\mathcal{D}(T)$ contains a non-isolated point ($\dagger$), then $T$ is topologically transitive.
2. If $T$ is topologically transitive, and $X$ is second countable (= its topology has a countable base) ($\dagger\dagger$) and Baire (= intersection of countably many open dense subsets is dense), then $\mathcal{D}(T)$ is a dense $G_\delta$.

Proof: (1) Let $x^\ast\in \mathcal{D}(T)$ be non-isolated. Then any open neighborhood of $x^\ast$ must contain points other than $x^\ast$. Let $U\in\mathcal{T}(X)^\ast$. We claim that $N^T(\{x^\ast\}\to U)$ is infinite. Indeed, by the density of the orbit of $x^\ast$, there is an $n_1\in N^T(\{x^\ast\}\to U)$; whence $T^{-n_1}(U)$ is an open neighborhood of $x^\ast$. Since $x^\ast$ is not isolated (and singletons are closed) $T^{-n_1}(U)\setminus\{x^\ast\}$ is also nonempty open. Using the density again there is an $m_1\in\mathbb{Z}_{\geq0}$ such that $T^{m_1}(x^\ast)\in T^{-n_1}(U)\setminus\{x^\ast\}$; thus by construction $T^{n_2}(x^\ast)\in U$ for $n_2=n_1+m_1>n_1\geq 0$, i.e. $n_2\in N^T(\{x^\ast\}\to U)\cap \mathbb{Z}_{\geq1}$. By induction $N^T(\{x^\ast\}\to U)$ is infinite; that is, the orbit of $x^\ast$ hits $U$ infinitely many times.

Now let $U,V\in\mathcal{T}(X)^\ast$ and pick $a\in N^T(\{x^\ast\}\to U)$ and $b\in N^T(\{x^\ast\}\to V)\cap \mathbb{Z}_{>a}$ (such an $a$ exists by assumption and such a $b$ exists because $N^T(\{x^\ast\}\to V)$ is infinite). Then $b-a\in N^T(U\to V)\cap \mathbb{Z}_{\geq1}$.

(2) Let $U_\bullet:\mathbb{Z}_{\geq1}\to\mathcal{T}(X)^\ast$ be an ordered base for the topology of $X$. Then we have:

$$\mathcal{D}(T)=\bigcap_{m\in\mathbb{Z}_{\geq1}}\bigcup_{n\in\mathbb{Z}_{\geq0}} T^{-n}(U_m).$$

By topological transitivity, $\bigcup_{n\in\mathbb{Z}_{\geq0}} T^{-n}(U_m)$ is dense for any $m\in\mathbb{Z}_{\geq1}$; any such set is also open. Since $X$ is Baire, $\mathcal{D}(T)$ is dense; it's also $G_\delta$ by construction.

($\ast$) Singletons being closed is enough, i.e. $T_1$.

($\dagger$) This is needed, e.g. consider $T:\mathbb{Z}_{\geq0}\to \mathbb{Z}_{\geq0}, x\mapsto x+1$; which is not topologically transitive (one can't go back) but $\mathcal{D}(T)=\{0\}\neq\emptyset$.

($\dagger\dagger$) Note that this is a priori stronger than separability.

Finally, to compare, note that in FShrike's example $X=[0,1]\cap\mathbb{Q}$ is not a Baire space (let alone completeness; also see Example of a Baire metric space which is not completely metrizable).