Let S be an open set in R. Let f be a continuous function such that f(S) is dense in R. Let α be an arbitrary element of R. Then as f(S) is dense in R, there exists {zi}⊂S such that limf(zi)=α. Since f is continuous does this imply that α=f(z) for some z∈S?
Answer
No, it does not. Let S=R∖{0}, and let f(x)=x; there is no α∈S such that f(α)=0.
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