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 $z \in S$?

  • $\begingroup$ What is $S$ ? And did you mean $\lim f(z_i) = \alpha$ (because otherwise, there would be no question). $\endgroup$ – Joel Cohen Oct 2 '11 at 23:30
  • $\begingroup$ Edited. Thanks for pointing it out. $\endgroup$ – user1205 Oct 2 '11 at 23:32
  • $\begingroup$ 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 $\alpha$. $\endgroup$ – t.b. Oct 2 '11 at 23:35
  • $\begingroup$ What is the domain of $f$? $\endgroup$ – Nate Eldredge Oct 2 '11 at 23:47
  • $\begingroup$ I think you need the image f(S) to be a closed subset of $\mathbb R$ $\endgroup$ – MAK Oct 2 '11 at 23:54

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


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