How is the function $f: \mathbb{Z} \to \mathbb{R}$ continuous? Where $\mathbb{Z}$ is the set of integers and $\mathbb{R}$ the set of real numbers. 
In a question in a problem sheet, it said this statement was correct, however I do not understand how.
You clearly cannot even begin to draw this function without a lot of gaps.
I suppose when the $\lim_{x\to Z_1} f(x) = f(Z_1)$. So is that the reason why the function is continuous?
Edit: This question came up in a first year university analysis module so I'm not too sure what topology means. Also, I use the standard epsilon, delta definition of continuity. 

$f:\Bbb Z \to \Bbb R$ is continuous since it cannot be noncontinuous.
Let me remind you, a function, $f$, is noncontinuous if there is an $x_0$ in its domain in which $f$ is noncontinuous in $x_0$.
By definition${}^\dagger$:

*

*If a function $f$ is noncontinuous in $x_0$, then (there exists a small distance $\epsilon$, where for all neighborhoods of $x_0$ there are points, $p$, in which $|f(y)-f(x_0)|>\epsilon$).

But this couldn't happen since there are neighborhoods in which no or just a single point, $x_0$, exist. so $|f(p)-f(x_0)|=0.$
This denies the consequent of the of the previous proposition, so by modus tollens, $f$ is continuous.
${}^\dagger$ You may have not seen a definition for a function to be noncontinuous in a point, so you may define it as a negation to the definition of being continuous in a point. Since the definition of being continuous in a point is in the form of $\forall \epsilon \exists \delta P$, the definition of being noncontinuous in a point should be of the form of $\exists \epsilon \forall \delta \neg P$, with same $\epsilon,\delta,P$ as the previous definition. 
