Can the empty set be the image of a function on $\mathbb{N}$? I cannot find any example of function $f:\mathbb{N}\rightarrow\mathbb{N}$ of which we can say that 
$$
f(\mathbb{N})=\emptyset
$$
Does there exists any?
 A: There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(\mathbb{N})$ cannot be the empty set, because contains $f(1)$. 
A: Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.
A: Since for every $a\in \mathbb{N}$ the $f(a)$ is also in $\mathbb{N}$ we see that $f(\mathbb{N})\ne \emptyset $ so there is no such function. 
A: No.  By definition of $f:A \to B$, then for every $a\in A$ then $f(a)$ must exist and $f(a) \in B$.  So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)
However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.
$f: \emptyset \to B$ is the empty function in this case.
=====
Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.
A: This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.
