Proof of continuity of all functions on N My task is to basically proof that any function defined on $\mathbb N$ is a continuous function.
I wanted to use the definition that states that if $f$ is continuos at every point a in the domain then $f$ is included in the set of all continuous functions on D, and basically trying to prove that the set of all continuous function on $\mathbb N$ (which is the domain) is $\mathbb N$ itself. 
But i am completely stuck!!!
Please help me.
 A: This is really simple to show from basic definitions.
If $X$ and $Y$ are topological spaces and $X$ has the discrete topology (every set is open), then every function $f:X\rightarrow Y$ is continuous. That's because $f^{-1}(U)$ is open in $X$ for every $U$ open in $Y$.
Since $\mathbb N$ has the discrete topology, the result is immediate.
A: This is a case where the literal definition makes something trivial but the intuitive concept is impossible.
The definition of f being continuous at x is that for ever e> 0 there exists a d > 0 so that for all y; |x - y | < 0 it follows that |f(x) - f(y)| < e.
Well, simply always choose d = 1.  Then for all y; |x - y| < 1 would mean that y = x (!!!!)  then for any e > 0.  |f(x) - f(y)| = |f(x) - f(x)| = 0 < e.
It's counter-intuitive as continuous seems to mean "doesn't jump about" but being restricted to domain N, means it has to "jump about".  But restricting to N means you can find neighborhoods with only one point in them and those will have images with only one point in them. Therefore at that point there is no "jumping about" because there are no other points to "jump" to! (Conter-intuitive but trivial once you wrap your mind around it.)
