# 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?

• 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 $\mathbb{R}^n$ – Andres Koropecki Jun 10 '20 at 15:12

To close this question, let me present a more explicit answer than the one given in the comments.

First, let $$w:[0,+\infty)\to [0,+\infty)$$ be constructed as follows. On $$[1,2]$$ define $$w$$ to be the tent map, with $$w(1)=w(2)=0$$, and $$w(3/2)=4$$. Then copy-paste scaled versions of this tent infinitely in both sides so that the graph consists of a sequence of congruent triangles.

It is proven in Silverman - On Maps with Dense Orbits and the Definition of Chaos, p 360, that this map has a property that for every open non-empty $$U\subset [0,+\infty)$$ there is
$$k\in\mathbb{Z}$$ and $$m\in\mathbb{N}$$ such that $$[0,2^{k+2r}]\subset w^{2r}(U)$$, for $$r>m$$.

Now consider $$v:[0,+\infty)^n\to [0,+\infty)^n$$ defined 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 property of $$w$$ there are $$k_1,...,k_n\in\mathbb{Z}$$ and $$m\in\mathbb{N}$$ such that $$[0,2^{k_1+2r}]\times...\times[0,2^{k_m+2r}] \subset v^{2r}(U_1\times...\times U_n)\subset v^{2r}(U)$$, for $$r>m$$.

Now observe that $$[0,+\infty)^n$$ is homeomorphic to $$W=[0,+\infty)\times(-\infty,+\infty) ^{n-1}$$ with the boundary corresponding to the boundary. 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 $$v$$, and also if $$U\subset W$$ is open and nonempty we have $$\bigcup v^{r}(U)= W$$.

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

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

Note that when $$x_1=0$$, $$h$$ does nothing, but also since $$v$$ maps boundary into the boundary, the first coordinate of $$v(x_1,...,x_n)$$ is $$0$$, from where the map is well-defined, and therefore 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 TT.

• The last couple steps you use to modify the n-fold tent map is very nice. – BLBA Jun 14 '20 at 6:52
• @AnnieLeKatsu thanks, but for the most part it's just a modification of the Silverman's construction – erz Jun 14 '20 at 8:16