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In the standard construction of natural numbers in axiomatic set theory (ZFC), zero is defined as being the empty set.

However, if we consider, for instance, the function $f:\mathbb N\rightarrow \mathbb N$ defined by $f(n)=n+1$, we have $f(0)=1$, but $f(\emptyset)=\emptyset$, because the image of the empty set is always empty.

Is this contradictory? What am I missing here?

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This is (I think) essentially the same confusion as in…. – Chris Eagle Apr 13 '12 at 19:25
You confusing the empty set as an argument of a function with the image of the empty set under the function. Unfortunately in the usual notation they are both denoted $f(\emptyset)$. – guy Apr 13 '12 at 19:26
Remember that in set theory everything is a set. So the function is defined on sets. This is why in set theory we often use $f[A]$ or $f''A$ for the set $\{f(a)\mid a\in A\}$. – Asaf Karagila Apr 13 '12 at 21:26

You are confusing the function $f$ with the direct image function that $f$ induces.

Recall that if $X$ and $Y$ are sets, and $f\colon X\to Y$ is a function, then $f$ induces a function, often denoted also by $f$ but which I will call $\underline{f}$, $$\underline{f}\colon\mathcal{P}(X)\to\mathcal{P}(Y),$$ given by $$\underline{f}(A) = \{f(a)\mid a\in A\}$$ for all $A\subseteq X$.

The function $f$ defined by $f(n)=n+1$ will map the element $\emptyset$ to the element $\{\emptyset\}=1$. The function $\underline{f}$ will map the subset $\emptyset$ to the subset $\emptyset$.

(Note that this is not the only problem with the notation: under the usual definition of $\mathbb{N}$, $\mathbb{N}$ is a transitive set: if $a\in \mathbb{N}$, then $a\subseteq \mathbb{N}$. Thus, $1 = 0\cup\{0\} = \{\emptyset\}$, and $\emptyset\subseteq\mathbb{N}$. So while $f(1)=2=\{0,1\} = \{\emptyset,\{\emptyset\}\}$, we also have $\underline{f}(1) = \underline{f}(\{\emptyset\}) = \{f(\emptyset)\} = \{f(0)\} = \{1\}\neq 2$. So here it is important to keep the distinction between $f$ and $\underline{f}$ clear, if it is not obvious from context.)

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You're mistaken. If "the image of the empty set" is taken to mean $\{f(x) : x\in\varnothing\}$, then that must indeed be empty. But when the empty set is itself taken to be a member of the domain, then it's just a member of the domain, and treated accordingly.

In set theory, but not as much in other areas of mathematics, one often has a set $A$ that is both a member and a subset of the domain. Taking $A$ to be a member of the domain, $f(A)$ is a member of the image. But one then uses the notation $f[A]$ to refer to $\{f(x) : x\in A\}$. These are two different things. Then one would say $f[\varnothing]=\varnothing$. But $f(\varnothing)$ can be something else---it depends on what function $f$ is.

The practice of defining $0$ that way is merely a convention. It is used for the purpose of encoding arithmetic within set theory.

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Thank you, that makes sense. – Filipe G. Apr 13 '12 at 19:39

If $\emptyset$ is the domain of the function then you are right, but here $\emptyset$ is not the domain, but an element in the domain. Therefore $f(\emptyset)$ can be for example $\{\emptyset\}$, and it is fine.

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