I was reading this [post](https://mathoverflow.net/questions/336489/existence-of-topologically-mixing-discrete-dynamical-system-on-manifold) 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?



Existence of topologically transitive map on Euclidean space
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.


