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Let $X$ be a set and $n \in \mathbb{N}$ a natural number. Then one can define the cartesian $n$-product of $X$ in the following way: $$X^n:=\{f \ | \ f:\{1,...,n\} \to X\}.$$ Now we can extend our language in the following way. For every $x \in X^n$ and all $i \in \{1,...,n\}$ there exists a unique $y \in X$ such that $(i,y) \in x$. We define this to be $x_i$ which gives an extension to our language. How can I now define $(x_1,...,x_n)$? I would like to define it to be the function $x \in X^n$ which has $x_1,...,x_n$ as its values, however this feels somewhat circular to me right now. How can I define this symbol in the sense of extension by definitions?

In the following suppose I have somehow extended the language with the symbols $(x_1,...,x_n)$. I would like to write something like $X^n:=\{y \ | \ \exists x_1,...,x_n \in X: y=(x_1,...,x_n)\}$, however, I am quantifying over function symbols here, namely the $x_1,...,x_n$, which is not allowed in first order language as far as I know. Is there a way to fix this/work around this? Can I quantify over these symbols? Once I established that for every $x:\{1,...,n\} \to X$ we have defined $x_1,...,x_n$ as its values and extended my language in this way, can I define $X^n:=\{y \ | \ \exists x:\{1,...,n\} \to X: y=(x_1,...,x_n)\}$. In this way I would not quantify over function symbols, which should be fine, shouldn't it?

So my questions are: $(1)$ How can I define the symbol $(x_1,...,x_n)$ as an extension of the language?

$(2)$ Can I quantify over function symbols and if not, how do I fix "$X^n=\{y \ | \ \exists x_1,...,x_n \in X: y=(x_1,...,x_n)\}$"?

user1578232
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  • In a definition of the "symbol" $x_i$, $x$ and $i$ would be variables representing a function and element of its domain. In a definition of the $n$-tuple notation using $(x_1,\ldots,x_n)$, the $x_i$ would be variables representing arbitrary sets. For $n=3$ you could write this as $(x,y,z)$ for example. The confusion and perception of circularity is arising from using the same notation in different ways. – blargoner Sep 05 '22 at 15:21

1 Answers1

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I would like to sketch an approach to the question, as I understand it; a full answer may demand a burdensome work.

In a language of first-order logic, terms denote the first-order entities, that is, the elements of the domain of quantification. One can quantify over the terms formed by functions, but not over the functions themselves without resorting to some kind of reduction.

The general definition of an $n$-ary Cartesian product may exhibit the case more clearly:

$$X_{1}\times\cdots\times X_{n} = \{(x_1, \ldots, x_n) \mid x_{i}\in X_{i},\; i=1,\ldots, n\}$$

Drawing on the notion of function, we get an equivalent definition:

$$\{f:I\to\bigcup_{i\in I}X_{i}\mid f(i)\in X_{i},\;i=1,\ldots, n\}$$

To make the case more concrete, consider an example:

$Y =\{1,2,3\},\; Z =\{a, b\},\; n = 2$

$Y\times Z = \{(1,a), (1,b),(2,a),(2,b),(3,a),(3,b)\}$

Using the functional definition $\{f\;|\; f:\{1, 2\} \to Y\cup Z\}$, we get the identical sets if we use index (sequence) notation. For other notations, we get isomorphic definitions.

In order to add a new $n$-ary function symbol $f$, we need a formula $\varphi$ of $\mathcal{L}$ to introduce a non-logical axiom to serve as the defining axiom for $f$ such that

$$y = f(x_{1},\ldots, x_{n})\leftrightarrow\varphi$$

where $x_{1},\ldots, x_{n}, y$ are distinct variables.

$\varphi$ can be written in ZFC, the ordinarily used set theory. ZFC is a first-order theory, thus it allows quantification over sets if needed.

Two constraints on $\varphi$ are that $\varphi$ is to be satisfiable for all $x_{1},\ldots, x_{n}$

$$\mathbf{ZFC}\vdash(\exists y)\varphi$$

and, that $y$ satisfying $\varphi$ is to be unique ($z$ is a distinct variable)

$$\mathbf{ZFC}\vdash\varphi \wedge\varphi(z/y)\rightarrow y = z$$

Tankut Beygu
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