I need to prove that $f:[0, 1] \to \Bbb R$ given by $f(x) = \begin{cases}
1, & \text{if $x=\frac{1}{n}$ for any positive integer $n$} \\
0, & \text{otherwise}
\end{cases}$ has an infinite number of discontinuities.
I've identified that the discontinuities exist at $x=\frac{1}{n}$ for positive integers $n \ge 2$.
My first attempt included trying to use the epsilon-delta definition, however, I've figured it'd be easier to use the limit definition (if $f$ is continuous at $x_0$, then $\lim\limits_{x \to x_0} f(x) = f(x_0)$).
I'm just not sure how to satisfy the 'infinite' aspect of the question. I figure I need to prove the existence of one of the discontinuities and show how there exist many more of its kind?
Any hints into the right direction from here would be much appreciated.
Edit: It has been recommended I prove this using the epsilon-delta definition of continuity and do a proof by contradiction.
Thus, I claim $f$ is continuous, and for any $\epsilon >0$, there is a $\delta >0$ such that $$|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon $$
Since $f$ is not actually continuous, I will contradict myself and show that the second part of the above implication fails. So, $|f(x)-f(x_0)|\ge \epsilon$.
I am not sure which such $\delta>0$ will achieve this? Any suggestions on how I can work that out?
Best Answer
Let $n >4$. Then $\frac 1 n +\frac 1 {\sqrt 2},\frac 1 n +\frac 1 {2\sqrt 2},\frac 1 n +\frac 1 {3\sqrt 2},...$ is a sequence in $[0,1]$ converging to $\frac 1 n$. What happens to the values of $f$ at these points? [Note that $\frac 1 n +\frac 1 {m\sqrt 2}$ can never be of the form $\frac 1 k$ for any integer $k$ in view of irrationality of $\sqrt 2$].