Free Variables in Set Builder Notation My math textbooks says the following set can not exist:
$$
\{\text{ y : y = 2x for each x }\in \mathbb{N} \}
$$
Because $y = 2 \cdot 1 = 2$ and $4 = 2 \cdot 2 = 4$.
My understanding of this argument is that, for their reading of set builder notation, y is not a free variable. That is to say, the set builder notation claims that there is one y for every x in the natural numbers.
In other words:
$$
(\ \forall x \in \mathbb{N} )\ (\ !\exists y\ )\ (\ y = 2x\ )
$$
where $!\exists$ is read "there exists exactly one". Which seems strange to me, because--isn't $\{\ y : definition\ \}$ meant to both create a set of objects and a free variable that can range over the entire set? ( In other words, y is by definition not fixed. )
Is my understanding of their reading of the set builder notation correct? Is their reading of set builder notation standard across mathematics?
 A: The set builder notation you're using here works like
$$ \{ y \mid \text{some property of }y \} $$
In order to find out what the elements of your set is, you take (in principle) each possible $y$ one by one and ask whether that particular $y$ has the $\text{some property}$.
In this case the property is (presumably; writing "for all" after the formula it applies to is not good logic syntax and invites mistakes):
$$ (\forall x\in\mathbb N)\; y=2x $$
This property has $y$ as a free variable -- as it well should; every nontrivial property of an $y$ must be denoted by a formula where $y$ is free.
The set builder notation binds the $y$, but when you're just looking at the subformula $(\forall x\in \mathbb N)\; y=2x$, then $y$ is free there.
(A variable is not "free" or "bound" in a vacuum; it always depends on which context you're considering it in. Taking a larger context -- such as the entire set builder rather than just the condition itself -- can cause the variable to be bound where it was free).
For the above property, no matter which particular $y$ we try with, the property is not true -- $y$ cannot equal all of $0, 2, -2, 4,$ etc. at once.
Your textbook is wrong, however, when it claims that the set "cannot exist" -- it exists perfectly well; it just happens to be the empty set.

The set you're thinking about would be written
$$ \{ y\mid (\exists x\in \mathbb N)\;y=2x \} $$
or (appealing to the axiom of replacement instead of selection)
$$ \{ 2x \mid x\in \mathbb N \} $$
