Define f : R → R by f(x) = 5x if x is rational and f(x) = x² + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other point besides 2 at which f is continuous?
at x=1 any open neighborhood about it will contain both rational and irrational numbers, but the irrational part is converging to 7 while the rational part is converging to 5 hear.
however at 2 both parts converge to 10.
to find the other point consider where
$\displaystyle x^2+6=5x \Rightarrow x^2-5x + 6 =0 \Rightarrow (x-3)(x-2)=0$
3 might be another good place to look.
Hi bearej50.
To prove that $\displaystyle f$ is discontinuous at 1, let $\displaystyle \epsilon=1.$ Then for any $\displaystyle \delta>0,$ the interval $\displaystyle (1-\delta,\,1+\delta)$ contains irrational numbers. If $\displaystyle y$ is one of them, then $\displaystyle |f(y)-f(1)|=|y^2+6-5|=|y^2+1|\geq1=\epsilon.$
To prove that $\displaystyle f$ is continuous at 2, let $\displaystyle \epsilon>0$ be given. Then if $\displaystyle \delta=\min\left\{1,\frac\epsilon5\right\},$ consider $\displaystyle y$ in the interval $\displaystyle (2-\delta,\,2+\delta).$ If $\displaystyle y$ is rational, we have $\displaystyle |f(y)-f(2)|=|5y-10|=5|y-2|<5\delta\leq\epsilon.$ If $\displaystyle y$ is irrational, then $\displaystyle |f(y)-f(2)|=|y^2-4|=|y+2||y-2|<3\delta<\epsilon.$
There is one other point at which $\displaystyle f$ is continuous, and that is found by solving the equation $\displaystyle 5x=x^2+6$ for $\displaystyle x.$
Or consider $\displaystyle (a_n) = 1+ \frac{\sqrt{2}}{n} $. Then $\displaystyle a_n \to 1 $. But $\displaystyle f(a_n) \not \to f(1) $. Thus $\displaystyle f $ is discontinuous at $\displaystyle x = 1 $. It seems that if both rational parts and the irrational parts are equal at $\displaystyle x $ then $\displaystyle f $ is continuous at $\displaystyle x $ (e.g. in the case of $\displaystyle x = 2 $).