Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
This is (I think) essentially the same confusion as in math.stackexchange.com/questions/125209/…. –  Chris Eagle Apr 13 '12 at 19:25
1  
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
1  
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

3 Answers 3

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.)

share|improve this answer

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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.