# Continuity

• Sep 20th 2010, 06:44 PM
penguinpwn
Continuity
For what value(s) of x is f continuous?

f(x)= { 0 if x is rational
2 if x is irrational
• Sep 20th 2010, 09:10 PM
Pinkk
I am going to use the sequential definition of continuity and show the definition fails at rational and irrational points. It is not continuous anywhere. Let $a$ be rational. Let $x_{n} = \frac{1}{n\sqrt{2}} + a$, which is a sequence of irrational numbers which converges to $a$. Since each $x_{n}$ is irrational, $f(x_{n}) = 2$ for all positive integers $n$, so it trivially converges to $2$, and since $f(a) = 0 \ne 2$, we have that $\lim x_{n} = a$ yet $\lim f(x_{n}) \ne f(a)$, so the function is discontinuous on the rationals. Now let $a$ be irrational. By the denseness of $\mathbb{Q}$, we can construct a sequence of rational numbers $x_{n}$ that converges to $a$. Since each $x_{n}$ is rational, we have that $f(x_{n}) = 0$ for all positive integers $n$. So $f(x_{n})$ trivially converges to $0$, and since $f(a) = 2 \ne 0$, we have that $\lim x_{n} = a$ yet $\lim f(x_{n}) \ne f(a)$, so the function is discontinuous on the irrationals.

If you need to find the values using the epsilon-delta definition of continuity, the basic idea behind the proof is that any interval contains infinitely many rationals and irrationals, so if the function were continuous then $f(x)$ must get arbitrarily close to 0 and 2 simultaneously, which is clearly impossible.