Existence of topologically transitive map on Euclidean space 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?
 A: 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.
