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

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 $$f^n(U) \cap V\neq \emptyset.$$

• It's now answered on MathSE. – YCor Jan 12 '18 at 11:27

## 1 Answer

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.

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