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Math Help - [SOLVED] Prove: if f(A) open subset of R the R is continuous

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    [SOLVED] Prove: if f(A) open subset of R the R is continuous

    Let A be an open subset of \Re and let f: A \rightarrow \Re be an increasing function. Prove that if f(A) is an open subset of \Re then f is continuous.
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    Quote Originally Posted by Amanda1990 View Post
    Let A be an open subset of \Re and let f: A \rightarrow \Re be an increasing function. Prove that if f(A) is an open subset of \Re then f is continuous.
    Set B=f(A), we are told that A\text{ and }B are open subsets of \mathbb{R}.

    Let a\in A, so f(a) \in B but B is open so there exists r>0 so that (f(a)-r,f(a)+r) \subseteq B. From here and on we will prove that the \delta,\epsilon definition of continuity is satisfied for all \epsilon < r. Since f(a) - \epsilon,f(a)+\epsilon \in B it means there exists a_1\in A \text{ and }a_2\in A so that f(a_1) = f(a) - \epsilon \text{ and }f(a_2) = f(a) + \epsilon. Now if a_1 < x < a_2 \implies f(a_1) < f(x) < f(a_2) and so f(a) - \epsilon < f(x) < f(a) + \epsilon \implies |f(x) - f(a)| < \epsilon. Certainly, a_1 < a < a_2, so the neighborhood N = \{ x\in \mathbb{R} | a_1 < x < a_2 \} of a gets mapped within \epsilon of f(a). Thus, f is continous at any point a\in A.
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