Monday, November 25, 2019

real analysis - Continuous Function and a Dense Set



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 zS?


Answer



No, it does not. Let S=R{0}, and let f(x)=x; there is no αS such that f(α)=0.


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