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Please verify my proof for the following claim.

Define $f$ on $[0,2]$ by $\displaystyle{f(x) = \begin{cases} x & \textrm{ if } x \in \mathbb{Q} \\ 2-x & \textrm{ if } x \in [0,2]\backslash \mathbb{Q}. \end{cases}}$.

Claim. $f$ is continuous at $c=1$.

Proof. Let $\epsilon >0$. Choose $\delta = \epsilon$. If $x\in [0,2]$, then $x\in [0,2]\cap \mathbb{Q}$ or $x\in [0,2]\setminus \mathbb{Q}$. If $x\in [0,2]\cap \mathbb{Q}$ and $|x-1|<\delta$, then $|f(x) - f(1)| = |x - 1| < \delta = \epsilon.$ Otherwise, if $x \in [0,2]\setminus\mathbb{Q}$ and $|x-1|<\delta$, then $|f(x) - f(1)| = |2-x - 1| = |1-x| = |x-1| < \delta = \epsilon.$ This proves that $f$ is continuous at $c=1$.

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    $\begingroup$ You can also try by sequential criteria of continuous function. $\endgroup$ – neelkanth Nov 3 '18 at 3:25
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Yes, your proof is correct. You have explained the two possible cases very clearly and correctly.

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