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(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.

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EthanAlvaree
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    How many elements does $S^n$ have? So how many should $S^0$ have? – Christoph Sep 06 '14 at 19:26
  • 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. – EthanAlvaree Sep 06 '14 at 19:28
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    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, $S^0={\emptyset}=1$. – Hayden Sep 06 '14 at 19:28
  • Did you mean to write $0=\emptyset$? Last I checked, $\emptyset$ is a set and $0$ is not a set. – EthanAlvaree Sep 06 '14 at 19:29
  • Alternatively if we think of $S^n$ as the set of $n$-tuples with entries in $S$, then $S^0$ has the empty tuple as its only element. – Christoph Sep 06 '14 at 19:29
  • @Mathemanic Yes, that is precisely what I meant to write. $0$ as an ordinal is precisely $\emptyset$. Me writing $0=\emptyset$ is in effect me defining $0$ to stand for $\emptyset$, but as stated before this is not some wild assignment. – Hayden Sep 06 '14 at 19:31
  • @Mathemanic: There is a 'canonical' model of the natural numbers in a set-theoretic universe. In that model, each natural number is interpreted as the set of all smaller natural numbers. So $0 = \emptyset$, $3 = { 0, 1, 2 }$, et cetera. –  Sep 06 '14 at 19:31
  • Ah! Okay. That is interesting. I have heard of the "Peano" axioms as one construction of the natural numbers. Is this a different construction, or one consistent with the Peano axioms, etc.? Also, I'd like to know why my question has two downvotes. Is it not well thought-out question? I would like to think others encountering the notion of a "nullary operation" have wondered the same thing. – EthanAlvaree Sep 06 '14 at 19:34
  • @Mathemanic The Peano Axioms are not a "construction" of the natural numbers, they are an axiomatization of them with the model Hurkyl cites being a model of those axioms. Also, use "@" followed by a person's name (without spaces) so that they are informed that you are replying to them. – Hayden Sep 06 '14 at 19:36
  • @Christoph, that's right, however I didn't think of the "empty tuple" as an element. That's why I thought of $\emptyset \times S = \emptyset$. If the empty $n$-tuple is an element, how would one denote it? Still trying to piece everything together. – EthanAlvaree Sep 06 '14 at 19:37
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    If you write a $3$-tuple $(a,b,c)$, feel free to write $()$ for the $0$-tuple. If you think of words $abc$ instead of tuples, the empty word is usually denoted $\varepsilon$. – Christoph Sep 06 '14 at 19:39
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    @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. – Frunobulax Sep 06 '14 at 19:41
  • Thanks, @Christoff. Then $\emptyset \times S = {()}$? And then a unary operation is a function $f:\emptyset \to S$, which is a subset of the Cartesian product of $\emptyset \times S$ (and there is only one element in this cartesian product, denoted $()$, so we write $f={()}$? Would this be correct? – EthanAlvaree Sep 06 '14 at 19:47
  • Good point, @Frunobulax. Based on my understanding, there was a problem, and I thought pointing that out would be motivational for people (as it is for me). – EthanAlvaree Sep 06 '14 at 19:48
  • @Mathemanic No, $\emptyset\times S=\emptyset$ for every set $S$. – Hayden Sep 06 '14 at 19:51
  • @Mathemanic A unary operation is a function $f\colon S^0 \to S$, where $S^0={()}$ is a set with one element. So a function $f\colon S^0\to S$ is uniquely determined by the image of $()$ under $f$. This defines a bijection between the elements of $S$ and unary operations on $S$, i.e. unirary operations are really just elements of $S$. – Christoph Sep 06 '14 at 19:59
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    Awesome comment @Christoph, thanks! So in summary, if $S={1,2,...,k}$, then a nullary operation $f:S^0 \to S$ is a function, i.e. some subset of $S^0 \times S = {()} \times S$. Let's look at ${()} \times S$, which when expanding, 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 again for your help. – EthanAlvaree Sep 06 '14 at 20:13
  • Now that's correct! – Christoph Sep 06 '14 at 20:14
  • @Christoph When you wrote "[...] a function $f\colon S^0 \to S$ [...]", I think you intended to mean "a nullary operation"? – linear_combinatori_probabi May 11 '25 at 04:11
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    @linear_combinatori_probabi Yes, all the "unary operation" in my comment from Sep 6 2014 should have been "nullary operation". A nullary operation is a map $S^0={()}\to S$ and a unary operation is a map $S^1=S\to S$, sorry. – Christoph May 12 '25 at 06:56

2 Answers2

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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.)

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

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

(where I mean everything up to bijection)