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In the language of first order logic we sometimes have certain symbols which are $n$-ary functions.

We are able to extend an axiomatic theory by introducing new functions and corresponding axioms. But the name "function" here is overloaded. In the context of LFOL an $n$-ary function is a syntactic construction which will appear as the function symbol followed by $n$ terms: $ft_1\ldots t_n$. However, in set theory, a function from set $X$ to set $Y$ is a subset of the cartesian product $X\times Y$ satisfying certain criteria. In this latter case the function $f$ is itself a set, and, in the context of LFOL, would be symbolized by a $0$-ary function (a constant). However, once the function is defined we will be interested in writing down something like $f(x)$ which would require, in the context of LFOL, a 1-ary function.

So the surprise to me, is that to define the function $f$ we require the introduction of TWO new function symbols to the language. One $0$-ary function for the set theoretic function $f$, and one $n$-ary function to facilitate the $f(x_1, \ldots, x_n)$ construction.

My question: Am I correct in my analysis here or am I somehow overcomplicating things? and if so, where am I going wrong?

J.-E. Pin
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Jagerber48
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  • A function in set theory is a set. When we add to the language of first-order set theory a new symbol for e.g. the "function" $\cap$ we add a "name" for the set $((a,b),c)$ such that $c=a \cap b$ – Mauro ALLEGRANZA Jun 14 '22 at 16:45
  • See this post for details. – Mauro ALLEGRANZA Jun 14 '22 at 16:54
  • @MauroALLEGRANZA I guess I'm asking about mathematical, rather than logical, functions like $f: \mathbb{R} \to \mathbb{R}$, $f(x) \mapsto x^2$ which is formalized as $f = {z \in \mathbb{R} \times \mathbb{R} : \exists x \exists y (z = (x, y) \land y = x^2)}$ but $f(x)$ is defined as the unique element of $\mathbb{R}$ satisfying $f(x) = x^2$. In the first case $f$ is a 0-ary function symbol (a constant) represent the subset of $\mathbb{R}\times \mathbb{R}$ and in the second case $f$ is a 1-ary predicate that takes meaning when we adjoing a single term, like $f(3)$. – Jagerber48 Jun 15 '22 at 03:23
  • I make this distinction because $\cap$ can't be thought of as a function in the mathematical sense because it's domain would be $\mathbb{S} \times \mathbb{S}$ and its range would be $\mathbb{S}$ where $\mathbb{S}$ is the set of all sets, but this of course can't exist. So $\cap$ can exist as a function in the logical sense (in that it has a unique output for each pair of inputs), but it can't be defined as a mathematical function (in that it is a subset of a cartesian product satisfying certain rules). – Jagerber48 Jun 15 '22 at 03:25
  • A function in set theory is a set: thus it is an "object" whose symbol is a "name". The "name" refers to the intersection of two sets. – Mauro ALLEGRANZA Jun 15 '22 at 06:09
  • @MauroALLEGRANZA sorry but I feel like you’re not following my question. It sounds like the set you’re referring to is what I’m calling the 0-ary function $f$ (the subset of the Cartesian product of the domain and range). But my point is that we conventionally ALSO introduce what must be ANOTHER syntactical function to the theory that conventionally has the same name $f$, but this latter case is a 1-ary function (meaning it must be paired with a term syntactically). I.e. $f(x)$ where $x$ is a variable or $f(3)$ where 3 is a constant (a named set). – Jagerber48 Jun 15 '22 at 11:01
  • My question is if it is correct that these two definitions of $f$ (both sets in their own right when the 1-ary version is paired with a constant…) are indeed separate function symbols that must be added to the theory. And therefore if it’s true that we actually add 2 function symbols when we define a new mathematical function (such as the real squaring function, not f logical functions like set intersection). – Jagerber48 Jun 15 '22 at 11:05
  • This is not my point of view... We prove that $\forall a,b \exists z \forall x \ (x \in z \leftrightarrow (x \in a \land x \in b))$ i.e. we prove that for every pair of sets there exists the intersection (and we prove that the resulting set is unique). Having done that, we add the following definition: $z= a \cap b \leftrightarrow \forall x (x \in z \leftrightarrow (x \in a \land x \in b))$ – Mauro ALLEGRANZA Jun 15 '22 at 12:53
  • Syntactically, $\cap$ is a binary function symbol that "names" the intersection of two sets. The "named" set $z$ is not a function in the sense of set theory (i.e. a set of pairs) but obviously is a set. – Mauro ALLEGRANZA Jun 15 '22 at 12:56
  • Let us continue this discussion in chat. – Jagerber48 Jun 15 '22 at 15:38
  • @MauroALLEGRANZA Above you said that "a function in set theory is a set" but in your last comment you said that "$\cap$ is a binary binary function symbol". This is my point. Sometimes functions are sets (i.e. special subsets of Cartesians products of their domains and ranges). In this case, syntactically, a function is a 0-ary function. but sometimes functions are n-ary functions that map elements of their domain(s) into their range. I'm looking for either confirmation that I'm interpreting this correctly or pointing out where I'm going wrong. – Jagerber48 Dec 22 '22 at 22:14

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