I had asked [this](https://math.stackexchange.com/questions/2601034/in-a-non-compact-metric-space-topologically-transitivity-need-not-imply-onto) question on [Mathematics Stack Exchange](https://math.stackexchange.com/) yesterday but it got no response so I'm asking here.




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



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> I'm trying to show that the compactness hypothesis cannot be removed.
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> I couldn't find any example of a non-compact metric space and a continuous function which is topologically transitive but not onto.
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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 $f^n(U) \cap V\neq \emptyset.$



In a non-compact metric space, topological transitivity need not imply onto
By [Birkhoff's theorem](http://home.iitk.ac.in/~chavan/linear_dynamics.pdf), a bounded linear operator on a Banach space is topologically transitive if and only if it is [hypercyclic](https://en.wikipedia.org/wiki/Hypercyclic_operator). Charles Read has developed a whole machinery for constructing non-surjective, hypercyclic operators on spaces of the form $\ell\_1(X)$:



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> C.J. Read, [The invariant subspace problem for a class of Banach spaces, II. Hypercyclic operators](https://link.springer.com/article/10.1007/BF02765019). *Israel J. Math.*, **63** (1) (1988), 1-40.
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Of course, possibly these are not the easiest counter-examples for your purposes, although I find them quite instructive.


