Continuity points of a function

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Let $f\colon\mathbb {Z}\to\mathbb{R}$ with $f(n) = \begin{cases} \frac{1}{n} & \text{ for } n\in \mathbb {Z}\setminus\{0\}\\ 0 & \text{ for } n = 0 .\end{cases}$

I have to find the continuity points of $f$ (the answer is $\mathbb {Z}$). I would have said the answer is $\emptyset$ because the left and right hand limits do not exist in any of the points. Can someone explain to me why the answer is $\mathbb {Z}$ ?

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1 Answer

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The answer depends on the topology you are using! I'm assuming that you're using the epsilon and delta definition of continuity. In this case, since $\mathbb{Z} \subset \mathbb{R}$ is a discrete subset, then every function from $f: \mathbb{Z} \mapsto \mathbb{R}$ is continuous (in all points of $\mathbb{Z}$).

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