Continuity of a function
Q: f is function from the reals to the reals. f(x) = x for x rational, f(x) = 1 - x otherwise. Find the set of all a such that f is continous at a.
A: I think f is continous at x = 1/2. Is this correct? Are there any other points? I thought not because if we choose any a not equal to one half, then we can always find some rational number between x and a, as we proceed to the limit. Is this correct reasoning?
Your reasoning isn't very clear. Yes, given a not equal to 1/2 we choose some rational number between x and a- but that is true with any number replacing 1/2! What's special about 1/2?
Originally Posted by Rudipoo
And why do you mention only rational numbers? In order for a function to be continous at a point, its limit must exist and be equal to the value of the function. But, given any x, there will be rational numbers as close as we please to x and so values close to x itself. There will also be IRRATIONAL numbers arbitrarily close to x so there will be values close to 1- x. In order that the limit exist, we must have x= 1- x. That's what is special about 1/2! I'm sure that's what you were thinking but you need to state clearly why this would work with 1/2 only and include the irrational numbers.
Thanks HallsofIvy, that makes it clearer. I had noticed 1/2 is the only number which satisfied x = 1 - x, which is why i went for that one!
I have another attempted answer to a question, maybe you could help we with this?
Q: Prove that f(x) tends to infinity as x tends to infinity iff f(x_n) tends to infinity for every sequence such that x_n tends to infinity:
IF: if f(x_n) tends to infinity for all (x_n) such that x_n tends to infinity, we have (x_n) increasing, and unbounded, so if we take x = x_i, y = x_j, then, for j >= i, y>=x.
Now, for j >= i, f(x_j) >= f(x_i), i.e. for y>=x, f(y)>=f(x), so we have f(x) increasing and unbounded, therefore f(x) tends to infinity as x tends to infinity.
ONLY IF: if f(x) tends to infinity as x tends to infinity, and we take a sequence (x_n) which tends to infinity, we have x_(n+1) >= x_n, so f(x_(n+1)) >= f(x_n), i.e. f(x_n) increasing, and unbounded. So f(x_n) tends to infinity.
Is this clear/correct reasoning?