Thread: Can You find real values where f is continous?

1. Can You find real values where f is continous?

define f(x)= 1 if x is rational
cos x if x is irrational

and f(x)= x^2 if x is rational
1/x^2 if x is irrational

i understand its by using sequences of rationals that converge to irrationals and vice versa...but can someone explain to me how to find the points at which they are continous??

2. Originally Posted by mtlchris
define f(x)= 1 if x is rational
cos x if x is irrational

and f(x)= x^2 if x is rational
1/x^2 if x is irrational

i understand its by using sequences of rationals that converge to irrationals and vice versa...but can someone explain to me how to find the points at which they are continous??

There is this nice and rather easy lemma:

Lemma: if we can put $\displaystyle \{x_n\}_{1\le n\le\infty}=\{y_{n_1}\}\cup\{z_{n_2}\}\,,\,\,with\ ,\,\,\{y_{n_1}\}\cap\{z_{n_2}\}=\emptyset$ , then $\displaystyle \{x_n\}$ converges iff each of the subsequences $\displaystyle \{y_{n_1}\}\,,\,\{z_{n_2}\}$ converge and to the same exact limit.

From this you can see your first f(x) is continuous at $\displaystyle 2k\pi\,,\,k\in\mathbb{Z}$ and the second one at $\displaystyle \pm 1$

Tonio