12
$\begingroup$

(I'm going to link to this Stackexchange post concerning Nullary Operations: why is a nullary operation a special element, usually 0 or 1?)

In general an operation is a function $f:S^n \to S$, where $n$ denotes the arity of the operation, and a nullary operation is a function $f:S^0 \to S$.

It is clear that $S^n=S \times S \times \cdots \times S$ denotes the Cartesian product of a set with itself ($n$ times), however $S^0$ makes little sense to me.

One might define $S^0=\emptyset$. At first glance, that seems like a good choice. However then a nullary operation makes no sense, since it must be a function $f: A \to B$ where $A$ and $B$ are two sets representing the domain and codomain of the function. And note a function is defined in terms of relation, where a relation is a subset of the Cartesian product $A \times B$. So when we think of $A=S^0=\emptyset$ and $B=S$, so that we have a nullary operation $f:\emptyset \to S$, what is a a subset of $\emptyset \times S$? What even is $\emptyset \times S$? By definition of the Cartesian product, \begin{equation} \emptyset \times S = \{(a,b):a \in \emptyset\textrm{ and }b \in S\}. \end{equation} But there exists no $a \in \emptyset$! Therefore there exist no ordered pairs in our Cartesian product! Therefore $\emptyset \times S = \emptyset !$. Remember a function is a relation which is a subset of the Cartesian product. Therefore our relation is the empty set, and therefore also $f=\emptyset$.


Edit: We got the definition sorted out in the comments. Thanks especially to Hayden and Christoff for their great help. In summary, we figured out that $S^0=\{()\}$, the set containing only one element, the "empty $n$-tuple". Then if taking $S=\{1,2,...,k\}$, then a nullary operation $f:S^0 \to S$ is a function, i.e. some subset of $S^0 \times S = \{()\} \times S$. Looking at $\{()\} \times S$, this equals $\{((),1),((),2),...,((),k)\}$. Then a function is a special subset of this, so that each element of the domain (and there is only one element, namely $()$) is paired with exactly one element of the codomain. Therefore a function would have the form $f=\{((),a)\}$ where $a$ is one of $1,2,...,k$. Thank you for your help.

$\endgroup$
  • 2
    $\begingroup$ How many elements does $S^n$ have? So how many should $S^0$ have? $\endgroup$ – Christoph Sep 6 '14 at 19:26
  • $\begingroup$ To answer your comment. Suppose $S$ has $k$ elements. Then $S^n$ has $k^n$ elements. Then $S^0$ theoretically has $k^0=1$ element. Still doesn't answer how we define $S^0$. Clearly $S^0=\emptyset$ is a bad definition. $\endgroup$ – Mathemanic Sep 6 '14 at 19:28
  • 2
    $\begingroup$ If we think of $S^0$ as the set of all functions from $0=\emptyset$ to $S$, then there is exactly one such function, the empty-function $\emptyset$. Thus, $S^0=\{\emptyset\}=1$. $\endgroup$ – Hayden Sep 6 '14 at 19:28
  • $\begingroup$ Did you mean to write $0=\emptyset$? Last I checked, $\emptyset$ is a set and $0$ is not a set. $\endgroup$ – Mathemanic Sep 6 '14 at 19:29
  • 3
    $\begingroup$ @Mathemanic: I guess the downvotes are mainly because of the tone of your question. Had you just written that you don't understand something and want clarification, that would have been fine. However, writing that something is a "huge problem" just because you misunderstood it, probably piqued some people. $\endgroup$ – Frunobulax Sep 6 '14 at 19:41
18
$\begingroup$

In set theory, $A^B$ usually means the set of all functions from $B$ to $A$. In that sense, you can see $A^n$ as the set of all functions from $n$ to $A$ where $n$ is the von Neumann ordinal $\{0,1,2,\dots,n-1\}$ (which is a set with $n$ elements). In that sense, an $n$-tuple of elements of $A$ and a function from $n$ to $A$ are just two different ways of interpreting the same thing.

So, $A^0$ would then just be the set of all functions from the von Neumann ordinal $0$ (which is the empty set $\emptyset$) to $A$. And there's only one such function which is $\emptyset$, so $A^0$ must be $\{\emptyset\}$ - a set with one element.

This all fits perfectly. It seems you have no issues with, say, $2^0$ being defined as $1$. This is a similar construction. Actually, $A^0=A^\emptyset=\{\emptyset\}$ not only has cardinality $1$ but is $1$ in the von Neumann sense.

(And this interpretation makes a $0$-ary function a constant, BTW.)

$\endgroup$
  • 1
    $\begingroup$ Thanks for the great answer, Frunobulax. I wanted to let you know that your answer inspired me to study axiomatic set theory, just so I could understand this. I went and purchased Herbert Enderton's 1977 "Elements of Set Theory" and have been learning about how we define $0=\emptyset$, $1=\{\emptyset\}$, $2=\{\emptyset,\{\emptyset\}\}$, etc. This actually led me to post a follow-up question regarding the properties of the member of $S^0$: math.stackexchange.com/questions/964092/… $\endgroup$ – Mathemanic Oct 8 '14 at 22:22
  • $\begingroup$ The problem is, we need $(\emptyset,s_1,\ldots,s_n)=(s_1,\ldots,s_n)$, however we haven't figured out (in the other question I linked to) how to make that happen. But it's the condition we need if we want $S^0 \times S^n = S^n$. $\endgroup$ – Mathemanic Oct 12 '14 at 1:03
0
$\begingroup$

$S^0$ is the set with the property that

$$ S^0 \times S = S^1 $$

(where I mean everything up to bijection)

$\endgroup$
  • 4
    $\begingroup$ It is worth mentioning that there are multiple sets that fulfill that property. $\endgroup$ – Hayden Sep 6 '14 at 19:31
  • $\begingroup$ Hi @Hurkyl, this led me to post a follow-up question regarding this property of the single member of $S^0$: math.stackexchange.com/questions/964092/… $\endgroup$ – Mathemanic Oct 8 '14 at 22:28
  • 1
    $\begingroup$ The definition fails for $S=\varnothing$, though. Even if you only consider this "up to bijection". $\endgroup$ – Asaf Karagila Jan 11 '17 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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