There is a sequential listing of the rational numbers: .
Define a function
What happens if
what does it mean when a function from real no to real no is continuous at infinitely many pts but between every two points of continuity, there is a point of discontinuity and vice versa?
i dont get how the function will look..
also, in the proof, i do not understand the purpose behind saying that q> ε and then q>1/ε...isit contradicting?
im not really sure ..
but i guess that if t is a quotient then f(t+)-f(t-) <0 and
im not sure what happens to f(t+)-f(t-) if t is not a quotient..i assume that it will still give <0 becos for a^j which are less than t, there will still be values that are quotient.and that f(t+) <1 while f(t-)>1
ya i think i should do that. its just that im just starting this module and so i thought my foundations for it aint that good as of yet..
btw, are you giving an example for the case where: a function from real no to real no is continuous at infinitely many pts but between every two points of continuity, there is a point of discontinuity and vice versa or trying to explain the photo that i have attached?
This function is discontinuous at every rational point x = p/q. The reason is that as you approach p/q along a sequence of irrational numbers, the value of the function jumps from 0 to 1/q when you reach the point p/q.
Less obviously, the function is continuous at every irrational point c. The reason for that is that given , you can choose a neighbourhood of c that is small enough to avoid all the (finitely many) rational points at which the function takes values greater than . This means that the function gets closer and closer to 0 as you approach the point c, so that , which is the definition of f being continuous at c.