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I was reading this post and wondered. Does there exist a topologically transitive (TT) map $f:\mathbb{R}^n\to\mathbb{R}^n$ when $n\geq 2$? I know that post asks for compactness and topological mixing but if we relax the requirement to only TT is it possible?

Note: If $\mathbb{R}^n$ is replaced by an infinite-dimensional Hilbert space, then the Ansari-Bernal theorem guarantees such a map exists; moreover it can be linear... So maybe it can exist in the finite-dimensional case?

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    $\begingroup$ It's overkill, but the answer to the post you cited in your question says that there is a Baire-generic subset of topologically mixing diffeomorphisms in the space of volume-preserving diffeomorphisms of any closed manifold of dimension greater than $1$. You can use this in the $n+1$-dimensional sphere. If you pick any diffeomorphism of this sphere with a hyperbolic fixed point, then arbitrarily close to it there is a topologically mixing diffeomorphism which still has a fixed point. Removing this point you get a topologically mixing (in particular transitive) diffeomorphism of $\mathbb{R}^n$ $\endgroup$ Jun 10, 2020 at 15:12

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Here is an explicit answer.

First, define a continuous tent map $w$ on $[1,2]$ define $w$ with $w(1)=w(2)=0$, and $w(3/2)=4$. Then define $w:[0,+\infty)\to [0,+\infty)$ by pasting scaled versions of this tent infinitely on both sides so that the graph consists of a sequence of congruent triangles.

Then for every open non-empty $U\subset [0,+\infty)$, there are
$k\in\mathbb{Z}$ and $m\in\mathbb{N}$ such that $$r>m \ \to \ [0,2^{k+2r}]\subset w^{2r}(U).$$ (This is proved In Silverman - On Maps with Dense Orbits and the Definition of Chaos, p 360.)

Now define $v:[0,+\infty)^n\to [0,+\infty)^n$ by $v(x_1,...,x_n)=(w(x_1),...,w(x_n))$. Let $U\subset [0,+\infty)^n$ be open and non-empty. There are $U_i\subset [0,+\infty)$, such that $U_1\times...\times U_n\subset U$. From the above property of $w$ there are $k_1,...,k_n\in\mathbb{Z}$ and $m\in\mathbb{N}$ such that $$r>m \ \to \ [0,2^{k_1+2r}]\times...\times[0,2^{k_m+2r}] \subset v^{2r}(U_1\times...\times U_n)\subset v^{2r}(U)$$

Now $[0,+\infty)^n$ is homeomorphic to $W=[0,+\infty)\times(-\infty,+\infty) ^{n-1}$ and the boundaries are also homeomorhpic. Below I will view $v$ as a map on $W$. Then the set $\{(x_1,...,x_n), x_1=0\}$ is invariant with respect to $v$. Also, if $U\subset W$ is open and nonempty, then $\bigcup v^{r}(U)= W$.

Let $h:\mathbb{R}^n\to \mathbb{R}^n$ be the reflection $h(x_1,...,x_n)=(-x_1,x_2,...,x_n)$. Define $f:\mathbb{R}^n\to \mathbb{R}^n$ by

$$f\left(x_1,...,x_n\right)=\left\{\begin{array}{ll} h(v(x_1,...,x_n)) & x_1\ge 0 \\ v(h(x_1,...,x_n)) & x_1\le 0 \end{array}\right.$$

When $x_1=0$, $h$ does nothing, but also since $v$ maps boundary points into boundary points, the first coordinate of $v(x_1,...,x_n)$ is $0$. So the map is well-defined and continuous. Observe that $f^2|_{W}=v^2$, and so if $U\subset \mathbb{R}^n$ is open and nonempty, one can show that $\bigcup f^{r}(U)= \mathbb{R}^n$, which implies Topological Transitivity.

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    $\begingroup$ The last couple steps you use to modify the n-fold tent map is very nice. $\endgroup$
    – ABIM
    Jun 14, 2020 at 6:52
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    $\begingroup$ @AnnieLeKatsu thanks, but for the most part it's just a modification of the Silverman's construction $\endgroup$
    – erz
    Jun 14, 2020 at 8:16
  • $\begingroup$ I was confused by the definition of $w$ with "scaled" copies and "congruent" triangles -- can you specify it in more detail? $\endgroup$
    – user44143
    Sep 22, 2021 at 20:40

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