Does there exist a real-valued function such that for any proper subset of real numbers the function maps outsidr of the set?

Question

Does there exist a bijective function $f$ from the real numbers to the real numbers such that for any non-empty proper subset of real numbers $R$ there exist $x$ in $R$ such that $f(x)$ is not an element of $R$?

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The negation would be that for every real-valued function $f$ there exists a non-empty proper subset of real numbers $R$ such that for every $x$ in $R$, $f(x)$ is also an element of set $R$?

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A Remark

I think that it helps to consider iterating the function many times. Consider the set of all real numbers $y$ such that there exists a natural number $k$ such that $y = f^{(k)}(0)$ where, $f^{(0)}(0) = 0$ and for any integer $k$, $f^{(k)}(0) = f(f^{(k-1)}(0))$

If we begin at $x = 0$ and we iterate $f(x) = x + 1$ many many times, then we simply get the set of all integers. The function $f$ described by $f(x) = x + 1$ never escapes the integers. Equivilantly, there is no non-integer $y$ such that there exists integer $x$ such that $f(x) = y$.

But what about other functions?

Answer 1

Such a function does not exist.

For any bijection $f$ of $\mathbb{R}$, let $R_f = \{f^{k}(0),\ k \in \mathbb{Z}\}$ (where $f^0$ is the identity function, and for any $n > 0$, $f^n$ is the $n$-th iterate of $f$, $f^{-n}$ is the $n$-th iterate of the inverse of $f$).

$R_f$ is nonempty and countable, so it is a proper subset of $\mathbb{R}$, and for any $x \in R_f$, we also have $f(x) \in R_f$.