In a non-compact metric space, topological transitivity need not imply onto

Question

I had asked this question on Mathematics Stack Exchange yesterday but it got no response so I'm asking here.

Let $X$ be a compact metric space and $f:X \to X$ be continuous. If $f$ is topologically transitive. Then $f$ is onto.

I'm trying to show that the compactness hypothesis cannot be removed.

I couldn't find any example of a non-compact metric space and a continuous function which is topologically transitive but not onto.

Any hints will be appreciated.

Note: If $(X,f)$ is a dynamical system. Then $f$ is said to be topologically transitive if for every pair of non-empty open sets $U$ and $V$ in $X$ there exists $n \geq 1$ such that $f^n(U) \cap V\neq \emptyset.$

score 11 · Accepted Answer · answered Jan 12 '18 at 11:25

11

By Birkhoff's theorem, a bounded linear operator on a Banach space is topologically transitive if and only if it is hypercyclic. Charles Read has developed a whole machinery for constructing non-surjective, hypercyclic operators on spaces of the form $\ell_1(X)$:

C.J. Read, The invariant subspace problem for a class of Banach spaces, II. Hypercyclic operators. Israel J. Math., 63 (1) (1988), 1-40.

Of course, possibly these are not the easiest counter-examples for your purposes, although I find them quite instructive.

answered Jan 12 '18 at 11:25

Tomasz Kania

11,291

Your counter-example, while is interesrting, lacks compactness. – Leandro Jan 12 '18 at 13:02
2

@Leandro this was precisely the question, since there's no compact counterexample – YCor Jan 12 '18 at 13:24
You right @Ycor. – Leandro Jan 12 '18 at 16:57
@TomekKania That's a very interesting example. Thank you. – Mark Jan 12 '18 at 17:57

In a non-compact metric space, topological transitivity need not imply onto

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