Real Analysis - Continuity Problem?
I was given this question in my real analysis course, and I'm totally stuck.
Suppose that f is defined on (0, ∞) to be 0 at irrational points, and to be 1/q at rational points p/q expressed in lowest terms (i.e., x = p/q, and p and q have no common divisors (i.e., gcd(p,q) = 1)). Show that f is continuous precisely on the irrationals.
My thinking was that different rationals can all be the same thing, i.e., 1/2 = 2/4 = 3/6 = ...
So wouldn't they all map to the same x co-ordinate, which implies discontinuity?? And, for the irrationals, because they all map to 0, then they're continuous, on the straight line y = 0??
But hey, any help would be fantastic. I'd assume we'd need to use some sort of epsilon proof, but I'm honestly not too sure.
Comments
Look at the solution to #2 in the following link:
http://faculty.csuci.edu/brian.sittinger/351_hw7so...
I hope this helps!
My thinking could be that we ought to continually use the actuality that the exponent in our function for f '' is even. This seems to me to show that the function f '' is often continually going to be unfavorable by using fact the function x^2f(x) must be universally helpful. this could point out that the function f ' is lowering. We even have that f ' (0)= 0 . subsequently we are able to end that f is lowering for all helpful values of x, and increasing for all unfavorable values of x (x=0 is a worldwide optimal of f). We end that f ought to shrink at an a transforming into form of unfavorable fee, so we glance on the preliminary unfavorable slope call it -m. f will proceed lowering and could go the x axis at a component previously a million/m.