Can $f:\Bbb{N}\rightarrow\Bbb{N}$ return an empty set?

I have a function $f:\Bbb{N}\rightarrow\Bbb{N}$. An empty set is not a member of $\Bbb{N}$. Can $f$ still return an empty set for some arguments $x\in\Bbb{N}$?

• $f$ cannot return a set at all; it can only 'return' a number. If you had $f:\mathbb{N}\mapsto\mathcal{P}(\mathbb{N})$, then the empty set would be a legitimate value of $f$. – Steven Stadnicki Oct 15 '14 at 19:36

It depends. If you are working with "Pure set-theory", where all objects are sets, then the natural numbers are defined by $0 = \emptyset$ and $n = \{0, \dotso, n-1\}$, so $$\begin{array}{rcl} 0 & = & \emptyset; \\ 1 & = & \{0 \} = \{\emptyset\}; \\ 2 & = & \{0, 1\} = \{\emptyset, \{\emptyset\}\}; \\ 3 & = & \{0, 1, 2\} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}; \\ 4 & = & \{0, 1, 2, 3\} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}\}; \\ 5 & = & \{0, 1, 2, 3, 4\}; \text{ etc.} \end{array}$$
Except ... there is a curious ambiguity involved when applying functions to more than one point at a time. A function $$f:A\rightarrow B$$ induces a function (almost always referred to by the same name $f$, though this is really an abuse of notation) $$\hat f : \mathscr P(A)\rightarrow \mathscr P(B)$$ defined by $$\hat f(A) = \hat f \left(\bigcup_{a\in A}\{a\}\right) = \bigcup_{a\in A}\{f(a)\}.$$ Through the aforementioned abuse of notation, one can indeed have that $$f(\varnothing) = \varnothing$$ though what is really meant is that $$\hat f(\varnothing) = \varnothing.$$