0
$\begingroup$

Let $S$ be an open set in $\mathbb{R}$. Let $f$ be a continuous function such that $f(S)$ is dense in $\mathbb{R}$. Let $\alpha$ be an arbitrary element of $\mathbb{R}$. Then as $f(S)$ is dense in $\mathbb{R}$, there exists $\{z_{i}\} \subset S$ such that $\lim f(z_{i}) = \alpha$. Since $f$ is continuous does this imply that $\alpha = f(z)$ for some $z \in S$?

  • 0
    What is $S$ ? And did you mean $\lim f(z_i) = \alpha$ (because otherwise, there would be no question).2011-10-02
  • 0
    Edited. Thanks for pointing it out.2011-10-02
  • 0
    Take $S = \mathbb{R} \smallsetminus \{\alpha\}$ and the inclusion $f: S \to \mathbb{R}, f(s) = s$ then the image is dense but no $z \in S$ is mapped to $\alpha$.2011-10-02
  • 0
    What is the domain of $f$?2011-10-02
  • 0
    I think you need the image f(S) to be a closed subset of $\mathbb R$2011-10-02
  • 0
    @MAK: Closed+Dense = $\mathbb R$. In which case it is trivial that there is such $z\in S$.2011-10-03
  • 0
    @MAK : Wouldn't the condition be that $S$ is closed instead ?2011-10-03
  • 0
    @Asaf: True, but I was coming from the perspective that a subset is closed iff it contains all its limit points. Then, if there was a sequence converging to z, z would be a limit point on the range, and then the limit points are in the range iff the range is closed.2011-10-04

1 Answers 1

7

No, it does not. Let $S = \mathbb{R}\setminus \{0\}$, and let $f(x)=x$; there is no $\alpha\in S$ such that $f(\alpha)=0$.