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let $f$ be a surjective mapping on the set of positive integers $\mathbb{Z}^{+}$. My question is that does there always exist a proper subset $A$ of $\mathbb{Z}^{+}$, such that $f(A)\subset A$?

Did
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Wei Xia
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  • Say $f : \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ is a function such that every proper subset $A \subset \mathbb{Z}^+$, $f(A) \not\subset A$. I think that this implies that $f$ cannot be surjective, but I am not sure how to show this. – A.E Feb 26 '14 at 04:09
  • the surjective condition is used to exclude the trivial case.Note that $f(\mathbb{Z}^{+})$ is a proper invariant subset if $f$ is not surjective. – Wei Xia Feb 26 '14 at 14:42

1 Answers1

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The answer is YES. To see this, pick an arbitrary $k\in \Bbb Z^+$ and let $A:=\{f^{\circ n}(k): n\ge 1\}$. Then $f(A)\subset A$ and either $k\notin A$ or $A$ is finite.

user104254
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