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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 $x\in\Bbb{N}$?

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    $\begingroup$ $f$ 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 $f$. $\endgroup$ – Steven Stadnicki Oct 15 '14 at 19:36
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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 $$ \begin{array}{rcl} 0 & = & \emptyset; \\ 1 & = & \{0 \} = \{\emptyset\}; \\ 2 & = & \{0, 1\} = \{\emptyset, \{\emptyset\}\}; \\ 3 & = & \{0, 1, 2\} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}; \\ 4 & = & \{0, 1, 2, 3\} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}\}; \\ 5 & = & \{0, 1, 2, 3, 4\}; \text{ etc.} \end{array} $$

If you are working with a set-theory with sets and another "objects", also, your naturales numbers are part of these objects, then the answer is "no".

NOTE. The two types of theories are more or less equivalent for the purposes of doing mathematics, and so you can take an agnostic position as to whether all objects are sets or not.

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No.

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 $$\hat f(\varnothing) = \varnothing.$$

So, normally, I would say "no" to your question, and add "don't abuse the notation".

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