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I don't understand continuity of $f:\mathbb{N}\to A$ where $A\not= \emptyset$ because the usual layman's definition of "you are able to draw it without lifting the pencil" fails here, at least if you use a normal coordinate system.

For the sake of having something to talk about, let's take $f(x)=x$ as our example. We know it's continuous when the domain is $\mathbb{R}$.

The problem I have is that I don't understand well enough the definition of neighborhood.

So, can someone illuminate me?

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Continuous means that if you are close to $x$, the values are close to $f(x)$.

So you have your function $f:\mathbb N\to A$. For continuity, points close to $n$ should take values close to $f(n)$. But $\mathbb N$ is discrete, so there is no point (other than $n$), really close to $n$. So the definition of continuity is satisfied: points close to $n$ (only $n$) have values close to $f(n)$.

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    Is there a good memnonic to memorize this? And to give it to others, such that they understand quickly. – 2017-01-18
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    Probably not. What I did is just write "in words" the usual definition: $$\forall\ \varepsilon>0,\ \exists\ \delta>0:\ |x-y|<\delta\implies |f(x)-f(y)|<\varepsilon.$$ After working enough with this definition, it becomes obvious. And then one can switch to the more abstract (and general) version: a function is continuous if $f^{-1}(V)$ is open for all open sets $V$. – 2017-01-18
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    Is (2,3,4) open in N? If yes, case closed^^ – 2017-01-18
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    Yes, because individual points are open (because of the discreteness), and arbitrary unions of open sets are open. – 2017-01-18
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    Are individual points open, because you exclude all other elements of the set? (and the definition of open is, that you exclude something). I.e. (1) would not be open if the set would be {1} – 2017-01-18
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    Of course, deciding what's open and what isn't is arbitrary (in the sense that one can assign different topologies to a set). The most natural point of view here is that one uses the restriction to $\mathbb N$ of the usual topology on $\mathbb R$. In a metric space, a set $X$ is open if every point is interior: that means that there is a ball (an interval, in the case of $\mathbb R$ such that the ball is in $X$. Here, when $X=\{3\}$, consider the ball $B$ of radius $1$ around $3$: that would be the set of $n\in \mathbb N$ such that $|3-n|<1$, which contains precisely $3$ and nothing else. – 2017-01-18