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

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

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

fixed syntax in title; removed superfluous digression
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In a non-compact metric space, topologicallytopological 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 converse of the above is not true and the compactness hypothesis cannot be removed.

To show that converse is not true, I let $X=\{0,1\}$ with discrete topology and $f$ be the identity map on $X.$ Then $f$ is onto but not topologically transitive.

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

In a non-compact metric space, topologically 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 converse of the above is not true and the compactness hypothesis cannot be removed.

To show that converse is not true, I let $X=\{0,1\}$ with discrete topology and $f$ be the identity map on $X.$ Then $f$ is onto but not topologically transitive.

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

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

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In a non-compact metric space, topologically 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 converse of the above is not true and the compactness hypothesis cannot be removed.

To show that converse is not true, I let $X=\{0,1\}$ with discrete topology and $f$ be the identity map on $X.$ Then $f$ is onto but not topologically transitive.

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