Where is the function
f(x) = (q^2 - 1)/q^2 if x = p/q meaning x is a rational in reduced form and f(x) = 1 when x is not a rational
continuous in the interval (0,1)? Please also explain how you came up with the answer.
(1) Let x0=p/q be a point in the interval (0,1) and x0 is an rational. By the definition of f(x), we know that
f(x0)=(q^2-1)/q^2. Obviously, f(x0)<1.
Since the irrational is dense in (0,1), we can choose a sequence y1,y2,...,yn,... such that yn->x0 and yi are irrational numbers for all i=1,2,....
By the definition of f(x), we have f(yi)=1 for i=1,2,...,
So, f(yn)->1 as n tends to infinity, which implies that the limit of f(x) (as x tends to x0=p/q) is not equal to f(x0)=(q^2-1)/q^2 if the limit of f(x) (as x->x0) exists.
So, by the ...
The continuity of a given function is investigated across an interval.