Vrftjkry

I cannot find any example of function $f:mathbb{N}rightarrowmathbb{N}$ of which we can say that
$$
f(mathbb{N})=emptyset
$$
Does there exists any?

No. By definition of $f:A to B$, then for every $ain A$ 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.

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Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.

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 $f(1)$.

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.

Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.

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.