Is completeness a necessary assumption for the Birkhoff Transitivity Theorem? 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!
 A: 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.
A: 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.

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

*

*If $\mathcal{D}(T)$ contains a non-isolated point ($\dagger$), then $T$ is topologically transitive.

*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).
