If f:R--->R is continuous on a dense set of points in R, then f is continuous on

Question

If f:R--->R is continuous on a dense set of points in R, then f is continuous on R.

Explanation / Answer

Are you sure that the question is correct? Cause it has some very famous counter examples ex. f(x) = 1/q if x is rational and x = p/q (where gcd(p,q) = 1) = 0 if x is irrational Now lets prove that x is continuous on irrational points Consider any irrational number l; Consider epsilon > 0, we must prove that there exists delta st f(x) epsilon are the candidates for violating the inequality) There are at most finite number of such n, Consider the set S = {a/n | 1/n > epsilon and a/n < 2l} It has atmost finite elements and hence there is an element which is nearest to l let k = u/v be that element then let delta = |k-l| Hence by the construction of delta none of the elements of S are delta close to l hence for all |x-l| inf f(an) = f(l) (if f is continuous at l) so lim n->inf 0 = 0 = f(l) but f(l) is non zero Hence f is discontinuous at l Hence we have our counter example an f which is continuous on a dense set of reals but not continuous on R If you have a doubt message me before rating the problem

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