I've tried using the definition of the continuity of a function to ultimately get
|f(x) - f(c)| < ɛ and |g(x) - g(c)| < ɛ for all ɛ > 0, where x, c ∈ (a,b).
I know that to show the set in question is open, I have to show that there exists an x ∈ (a,b) such that for all ɛ > 0, N (x,ɛ) is a subset of (a,b) such that f(x) < g(x).
But I don't really see how I can reach to this conclusion.
kb
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Let E = {x ∈ (a,b): f(x) < g(x)} = {x ∈ (a,b): f(x) - g(x) < 0}
Fix c ∈ (a, b).
We need to find δ > 0 so that B_δ (c) = {x in (a, b) : |x - c| < δ} is contained in E.
Since f and g are both continuous on (a, b), so is f - g. In particular, f - g is continuous at x = c, and note that (f - g)(c) = d for some d < 0.
Letting ε = -d > 0, by the continuity of (f - g) at x = c, there exists δ > 0 (and less than min{c - a, b - c}, so that the set of x such that |x - c| < δ is contained inside the interval (a, b))
such that |x - c| < δ implies that |(f - g)(x) - (f - g)(c)| < ε.
This last inequality is equivalent to writing
|(f - g)(x) - d| < -d
<==> -(-d) < (f - g)(x) - d < -d
<==> d < (f - g)(x) < 0 for all x satisfying |x - c| < δ, as required.
Therefore, E is an open set.
I hope this helps!