Show that there is a strictly increasing function on [0,1] that is continuous on

Question

Show that there is a strictly increasing function on [0,1] that is continuous only at the irrational numbers in [0,1].

Explanation / Answer

Define f(x) = Eq (1/2q) q(x) Then. f(x) has all the properties described. 2) f(x) is a convergent series since 0 x + 1/s. Now, each (q(y) - q(x)) >= 0, but s(y) - s(x) > 0. So, by construction, f(y) - f(x) > 0 and f is strictly increasing. 3) f is discontinuous for each rational number. Let y = p/n > 0 (handle the case of y = 0 separately.) Choose epsilon = (1/2n) (1/n) . Then for all delta, we can find an x such that y-x 1/2n (p/n - (p-1)/n) = 1/2n (1/n) = epsilon. Thus, f is discontinuous for y rational. 4) Say y is irrational. Let epsilon be given. We can find an integer n such that 1/n

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