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Let $(X,d)$ be a separable metric space with no isolated points. Recall that a continuous function $T:(X,d)\rightarrow (X,d)$ is called topologically transitive if

Given $U,V$ non-empty open subsets of $(X,d)$ there is some positive integer $n$ such that $$ T^n(U)\cap V \neq \emptyset . $$

My questions are: does there necessarily exist a topologically transitive function on $(X,d)$? Is there a simple test (by simple I mean something easy to verify like Stone-Weierstrass for linear spaces) to verify if $T$ is topologically transitive of not?

Note: If it makes it easier we can also assume that $X$ is a topological group. (I know this helps when $X$ is the unit circle).

Alp Uzman
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  • It's often the case that transitivity is equivalent to the existence of a dense orbit. These are normally easier to identify. Be careful though, they are not precisely the same - see this page. – Dan Rust Jun 21 '19 at 17:37
  • True, thanks for pointing that out. In that case, is there an easy way to find dense orbits?

    Also, I believe my assumptions are enough to apply Birkhoff's theorem (so in this case they are the same).

    –  Jun 21 '19 at 19:19

1 Answers1

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No. As an example take $X$ which is the disjoint union of the Knaster–Kuratowski fan $X_1$ and interval $X_2=[0,1]$. Since $X_1$ contains a dispersion point, every continuous map $X_2\to X_1$ is constant. Hence, for every continuous map $T: X\to X$ either $T(X_1)\subset X_1, T(X_2)\subset X_2$ (in which case $T$ of course cannot be topologically transitive) or $T(X)\subset X_1$ (ditto), or $T(X)\subset X_2$ (ditto), or $T(X_1)\subset X_2$, $T(X_2)=\{x_1\}\subset X_1$. Consider the last case and set $x_2=T(x_1)\in X_2$. Then for every $n\ge 2$, $T^n(X)\subset \{x_1, x_2\}$. It is clear that such $T$ cannot be topologically transitive either.

Edit 1: Actually, one can even take $X=(-1,0) \cup [1,2]$. A similar argument will work since (apart form three trivial cases) the image of $[1,2]$ in $(-1,0)$ will be a compact subinterval $I$ and $T^n(X)\subset I \cup [1,2]$ for all $n\ge 2$, hence, $T$ cannot be topologically transitive.

Edit 2: It follows from the results of

S. Alpern and V. S. Prasad: Typical dynamics of volume preserving homeomorphisms. Cambridge Tracts in Mathematics, 139, Cambridge University Press, Cambridge, 2000.

that for every connected compact or open and tame manifold $M$ of dimension $\ge 2$, there exists a topologically transitive homeomorphism $T: M\to M$.

Here a manifold $M$ is called tame if it is homeomorphic to the interior of a compact manifold with boundary. In particular, each ${\mathbb R}^n$, $n\ge 2$, admits a topologically transitive self-homeomorphisms.

Remark. It appears that the first example of such self-map of the plane is due to L.Shnirelman, 1930 (but published in a barely accessible place). The most common reference for topologically transitive homeomorphisms of the plane is

A. S. Besicovitch, A problem on topological transformation of the plane, Fund. Math. 28(1937), 61-65.

As for connected manifolds $M$ dimension 1, while (apart from circle) there will be no topologically transitive homeomorphisms. However, there is always a topologically transitive continuous self-map. For instance, if $M={\mathbb R}$, then an example is given by
$$ T(x)= \begin{cases} −3x+8n+2, & \hbox{if} ~~2n\le x\le 2n+1\\ 5x−8n+2, & \hbox{if} ~~2n−1\le x\le 2n \end{cases} $$ with $n\in {\mathbb Z}$. See

A. Nagar and S. P. Sesha Sai, Some classes of transitive maps on ${\mathbb R}$, Jour. of Anal., 8 (2000), 103–111.

In particular, if $M$ is homeomorphic to a connected Lie group then $M$ always admits a topologically transitive continuous self-map.

In particular, each finite-dimensional Banach space admits a topologically transitive continuous self-map. This leaves open the case of infinite-dimensional separable Banach spaces. (As Dan Rust noted in a comment, there is only one such space up to a homeomorphism, so it suffices to understand the case of $\ell_2$.)

As for compact spaces, the situation is unclear. It is known that if $M$ is a compact metrizable finite-dimensional Peano continuum (i.e. a connected and locally connected space) then $M$ admits a topologically transitive continuous self-map:

S. Agronsky and J. G. Ceder: Each Peano subspace of $E^k$ is an $\omega$-limit set, Real Anal. Exchange 17 (1991/92), no. 1, 371-378.

This gives yet another proof that each compact connected manifold (possibly with boundary) admits a topologically transitive continuous self-map.

Moishe Kohan
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  • If in addition, I assume that the space is homeomorphic to a Banach space would the claim then hold? –  Jun 21 '19 at 20:47
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    @MrMMS Not sure. – Moishe Kohan Jun 21 '19 at 20:58
  • Thanks for the answer it was really insightful (plus I never hear of the Knaster-Kuratowski space before, very interesting). If you do think of the answer to this additional bit, let me know :) –  Jun 21 '19 at 23:01
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    All infinite dimensional separable Banach spaces are homeomorphic, so you only need to show that a transitive map exists for one. – Dan Rust Jun 21 '19 at 23:06
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    It looks like it's not too simple to find transitive maps on non-compact manifolds. Here's there first example I could find for one on the plane: – Dan Rust Jun 21 '19 at 23:11
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    @DanRust: Actually, every tame connected manifold of dimension $\ge 2$ admits a topologically transitive homeomorphism. What I am not sure is the infinite-dimensional Banach space case. – Moishe Kohan Jun 21 '19 at 23:54
  • The Lie group example is very interesting actually. I'll have to lookinto it in more detail. Thanks for all the help :) –  Jun 23 '19 at 14:31