Let f:D-->R be continuous at c belonging to D and suppose that f(c)>0. Prove that there exists an $>0 and a

Question

Let f:D-->R be continuous at c belonging to D and suppose that f(c)>0. Prove that there exists an $>0 and a neighborhood U of c such that f(x)>$ for all x belonging to U intersection D.

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Answer 1

Let $ be any number such that 0 < $ < f(c). Then, f(c) - $ >0. According to the definition of continuity, for every eps >0 there exists a neighborhood U of c such that |f(x) - f(c)| < eps for every x in U untersection D.

Applying this definition with eps = f(c) - $, we get a neighborhood U of C such that |f(x) - f(c)| < f(c) - $ for every x in U intersection D. This implies that, for such elements x, we have

$ - f(c) < f(x) - f(c) < f(c) - $.

From the left hand side inequality it follows that f(x) > $ for every every x in U intersection D, as desired.

Answer 2

Looks to me like Steiner nailed it, I'm wondering why someone gave him a thumbs down.