For the function $f: [0,2 \pi] \rightarrow \mathbb{R}$ ,state at which points $c \in [0, \pi]$ is $f$ continuous or discontinuous.
$$f(x)=\begin{array}{cc} ( & \begin{array}{cc} 0 & x \in \mathbb{Q} \\ sin(x) & x \notin \mathbb{Q} \end{array} \end{array}) $$

My Attempt
If $c \in \mathbb{Q}$ then there is a sequence $c_n \notin \mathbb{Q}$ with $ cn \rightarrow c$ and $f$ being continuous at $c$ requires $$\lim_{n \to \infty}f(c_n)=\lim_{n \to \infty}sin(c_n)=sin(c)=0=f(c)$$ Now this condition is only necessary not sufficient.
On the otherhand, if $c \notin \mathbb{Q}$ then there's a sequence $c_n \in \mathbb{Q}$ with $c_n \rightarrow c$ and just as above the necessary condition is; $$\lim_{n \to \infty}f(c_n)=\lim_{n \to \infty}0=0=sin(c)=f(c)$$ for $f$ to be continuous at $c$.

So we know that $f$ is discontinuous at all points $[0,2 \pi]$ except possibly at points where $0=sin(c)$ i.e $c=0$ $c= \pi$ and $c=2\pi$ Since the conditions above are only necessary not sufficient we cannot immediately conclude that $f$ is continuous at these points we need to verify;
i.e at $c=\pi$ $$\lim_{x \to \pi}f(x)=f(\pi)=sin(\pi)=0$$ Let $|x-\pi|<\epsilon$ then if $x \in \mathbb{Q}$ $$|f(x)-f(\pi)|=|f(x)-f(\pi)|=|0|<\epsilon$$ if $x \notin \mathbb{Q}$ then $$|f(x)-f(\pi)|=|sin(x)-0|<\epsilon$$

I am not sure about the last part of the proof i.e the verification part, is this correct any feedback would be much appreciated.


In order to make your argument complete, note that if $x_0 \in \{0,\pi,2\pi\}$, and $x \in [0,2\pi]$, then one has $|\sin x| = |\sin(x-x_0)| \leq |x-x_0|$ to justify that $|x-x_0| < \varepsilon$ implies $|\sin x| < \varepsilon$.

Otherwise, a 'softer argument' to show that $f$ is continuous at $x_0$, can be given as follows: Let $\varepsilon > 0$ be arbitrary, and choose $\delta > 0$ such that $|\sin x| <\varepsilon$ whenever $|x-x_0|<\delta$ and $x \in [0,2\pi]$. This is possible since $\sin x$ is continuous and $\sin x_0 = 0$. Then observe that $|f(x)-f(x_0)| = |f(x)|$ equals either $0$ or $|\sin x|$ and thus is less than $\varepsilon$ for all $x \in [0,2\pi]$ with $|x-x_0| < \delta$. Therefore $f(x)$ is continuous at $x_0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.