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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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    Try to use the *definition* of the limit — not to only rely on intuition (when the domain is disconnected or discrete, intuition may and often will fail). – 2017-01-16

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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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    What's topology on $\mathbb{Z}$.? – 2017-01-16
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    The open sets of $\mathbb{Z}$. If we use the epsilon and delta definition of continuity, then the open sets of $\mathbb{Z}$are all elements of $P(\mathbb{Z})$, the power set of $\mathbb{Z}$. – 2017-01-16
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    so $f$ is continuous on $\mathbb{Z}$, then what's $f^{-1}([0,1])$.? – 2017-01-16
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    @IgorD. I don't really know how to apply the epsilon delta definition for all the $\mathbb {Z}$... but anyway, doesn't continuity mean that you can draw the graph of the function without lifting the pen up ? – 2017-01-16
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    @MyGlasses, $f^{-1}([0,1]) = \mathbb{N}\cup\{0\}$ in this case, isn't it ? – 2017-01-16
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    Now I see! The idea is the following: Pick a point $n_0 \in \mathbb{Z}$. Then if you take $\delta = 0.5$ (for example) then the set of points satisfying $$|n-n_0|<\delta = 0.5,~n\in \mathbb{Z}$$ should be equal to the singleton $\{n_0\}$. – 2017-01-16
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    So, for every epsilon you take, $\delta = 0.5$ is enough to show continuity. Is this clear for you @Liviu ? – 2017-01-16
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    @IgorD. It is, thank you ! – 2017-01-16