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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.

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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 $x_0$

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