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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 $z \in S$?

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

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up vote 6 down vote accepted

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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