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Let's take a set builder notation, if I define a set {$x$|$x$∈N} Is the scope of the variable supposed to be limited to the set-builder notation?

Would using $x$ again imply that the second use of $x$ is new variable, and different to the first, because the scope of $x$ in set builder notation is limited to it?

Is it more formally correct to define new variables for use in set-builder notation? Would this imply that re-using symbols is a sort of abuse of notation?

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The variables inside the set definition are dummy ones.

For instance if you define the set $S=\{a\mid \exists x\in\mathbb N, a=x^2\}$

Then $x\in S$ means $x$ is a square integer, i.e. it is actually the '$a$' from the inner statement, and not the '$x$' of the same statement.

Of course it is always preferable to keep consistent notations for understanding, same as you would probably rather not work with the complex number $y+ix$ (unless you are masochistic). It's not forbidden, but so error prone...

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    $\begingroup$ So if I were to previously define $x∈N$ and then use $x$ in a set-building notation then it would be a slightly confusing as $x$ is a sort of quantity in one case, and a 'dummy' variable in the second case, so it's better to define a new variable for either one of them. $\endgroup$
    – user1096856
    Commented Sep 20, 2022 at 19:39
  • $\begingroup$ I think so yes, now it also depends of the context, if the problem already uses a lot of different letters and you are coming short of new names, probably you'll be forced to do some 'reuse' at some point. $\endgroup$
    – zwim
    Commented Sep 21, 2022 at 20:04
  • $\begingroup$ I think I will, because it's a case that in definitions we see $x$ and $y$ as existing only in he definition, so for convenience we re-use them but we probably shouldn't, if we explicitly define a variable $a$ which we want to serve some purpose. $\endgroup$
    – user1096856
    Commented Sep 21, 2022 at 20:47
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In logic, set builder (or, set abstraction) operator is syntactically a variable-binding term operator.

Variable-binding term operators constitute a class by themselves: They form terms from terms like functions, but differ from them in that functions do not bind variables. Like quantifiers, variable-binding term operators bind variables, but differ from them in that quantifiers operate on formulas, not on terms. Russell's definite description operator $\iota$ (originally, inverted iota) and Hilbert's $\epsilon$ operator are examples from logic. See also the Wikipedia article for mathematical examples.

The usual notation for set builder/abstraction operator is $\{\;\mid\;\}$ or $\{\;:\;\}$. In order to display its logical form, it may be helpful to represent it as $\sigma x\phi(x)$ in which $\phi$ is a formula defining the comprehension of the set. Thus, for example,

$$\{x\mid x\in\mathbb{N}\}$$

can be represented as $$\sigma x(x\in\mathbb{N})$$

The occurrences of the variable $x$ have distinct syntactic roles. In the first occurrence, $x$ is a constituent of the operator prefix $\sigma x$, and in the second occurrence, it is a constituent of the matrix $(x\in\mathbb{N})$, which is an open formula.

Hence, $\sigma y(y\in\mathbb{N})$, $\sigma z(z\in\mathbb{N})$ (and so on with other variable symbols in the language so long as a clash of variables is avoided), are merely alphabetic variations of the same closed term.

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    $\begingroup$ Another important example of a variable-binding term operator would be the $\lambda$ operator from lambda calculus - though there's a difference in this case in that it takes a term formula as its body, whereas the set builder, $\iota$, Hilbert $\epsilon$ take a propositional formula as their respective bodies. $\endgroup$
    – Daniel Schepler
    Commented Sep 20, 2022 at 21:38

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