Wednesday, 26 July 2017

calculus - Is there an everywhere discontinuous increasing function?



Does there exist a function f:RR that is strictly increasing and discontinuous everywhere?




My line of thought (possibly incorrect): I know there are increasing functions such as f(x)=x, and there are everywhere-discontinuous functions such as the Dirichlet function. I also know that when there is a discontinuity at a point c, there is a finite gap ϵ such that there are points d arbitrarily close to c such that |f(d)f(c)|>ϵ. This is where my thinking gets unclear - does it make sense to have a "gap" at every real number?


Answer



There is no such function. Suppose that f:RR is strictly increasing. For each aR let f(a)= lim and f^+(a) = \lim\limits_{x\to a^+}f(x). Then f is discontinuous at a if and only if f^-(a) < f^+(a). Let D = \{a\in\mathbb{R}:f\text{ is not continuous at }a\}, and for each a\in D let q_a be a rational number in the non-empty open interval I_a = (f^-(a),f^+(a)).



It’s not hard to check that if a,b \in D with $a

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