# A mapping of an infinite set into itself has an invariant subset

Let $E$ be an infinite set and $f:E\rightarrow E$. Then there exists a subset $S$ of $E$ such that $\emptyset\neq S\neq E$ and $f(S)\subset S$.

This seems easy, but I don't have an idea. A hint would be nice.

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Hint: Let $a$ be in the set. Look at the set that has the objects $a$, $f(a)$, $f(f(a))$, and so on. If things close up you have a finite (and so proper) subset. If not, then everything is OK too. – André Nicolas May 19 '11 at 22:57
@Stefan: I didn't intend to be scolding; it's less my issue that the issue many seem to have here: posting questions in the "imperative" or simply stating a theorem with as though copied from a text...Some get pretty brutal with their responses. At any rate, I'll delete my comment(s)...this one after you've had a chance to read it. No offense intended. – amWhy May 19 '11 at 23:42
@Amy: In that case, I am sorry for getting your post wrong. I'm sometimes a little touchy when I feel I'm being lectured about something I already know. So hopefully no bad feelings. – Stefan May 20 '11 at 0:03
@Stefan Walker: Has the Hint been enough for you to push things through? If so, you should write the solution up very carefully, and post it. If the Hint is not sufficient, please so indicate, and I can post a solution. – André Nicolas May 20 '11 at 0:06
@Stefan - I understand...no bad feelings whatsoever ;-) – amWhy May 20 '11 at 0:25

Let $a$ be an element of $E$. If there exists $n\geq 1$ such that $f^n(a)=a$, then define $S=\{a,f(a),f^2(a),\ldots,f^{n-1}(a)\}$. Since $S$ is finite, $S\neq E$. Obviously, $f(S)\subset S$.
If $f^n(a)\neq a$ for all $n\geq 1$, then define $S=\{f^n(a)|n\geq 1\}$. Obviously, $f(S)\subset S$, and since $a\notin S$, $S\neq E$.