I have the solutions book and see the answers, but they make no sense to me.
Any/All help is appreciated.
a) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... but continuous at all other points.
b) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... and at 0 but continuous at all other points.
Problem 7 doesn't seem to have enough explanation in the solution either. How is f being continuous 0 used?
7) Suppose f satisfies f(x+y) = f(x) + f(y) and f is continuous at 0. Prove it is continuous at A for all values of A.
In your example in this post, f(x) is really discontinuous at x = 0? It seems I don't even understand continuity properly. At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right?
Wouldn't that mean in your first example, f(x) was also discontinuous at x 0?
Yes. Would you agree that a person's spacial position as a function of time is a continuous function? If f(t) = 0 for t <= 0 and f(1) = 1 for t > 0 described my spacial position, then I would be at position 0 before time 0 and then suddenly move to position 1 (say, 1 mile from position 0). If you know how to construct such transportation, I would certainly like to hear!
Formally, f(t) is not continuous at 0 because any interval containing t = 0 is mapped by f to a set that contains some and such that . In other words, however small you make an interval around 0, the image of that interval contains numbers that are sufficiently far apart, i.e., not arbitrarily close. It does not matter how precisely you approximate t = 0: if you change t just a little bit, f(t) can change by 1.
Haha, you are right. I didn't realise I typed 1/4 twice. Thank you so much for everything. I understand this all now.
For 6b, you can just change it to "otherwise, x = -1" and now x = 0 will be discontinuous as from the left, f(x) approaches -1 but from the right, f(x) still approaches 0.