Q:

Domain of Predicate

Exercise from the book A Transition to Advanced Mathematics:
For every positive real number $x$, there is a positive real number $y$ with the property that if $y<x$, then for all positive real numbers $z$, $yz\geq z$.
I'm totally fine with the above exercise. However, i have a confusion over the Domain of the predicate variable in the exercise below.
Edited Exercise: 
For every positive real number $x$, there is a positive real number $y<x$ with the property that for every positive real number $z$, $yz\geq z$.
Is a) or b) the domain of $y$? 
a) $y\in \{y\in \mathbb{R^+} : y<x\}$
b) $y\in \mathbb{R^+}$ and $P(x,y): y<x$ is a property.
Are they different and how are they different?
What is the difference between the original exercise and the edited exercise in the case that $y$ satisfies the condition $y<x$?
I have tried using both a) and b) in forming the negation of the original statement to prove that the original statement is false. a) turns out well but b) gives me a very weird statement.

A:

In logic you can't make your domain dependent on some variable, as you seem to be trying to do in b): all variables are quantified over the same domain.
Now, if you specify that the doamin is $\mathbb{R}^+$, then your statement can be symbolized as such:
$\forall x \ \exists y \ (y < x \rightarrow \forall z \ yz \ge z)$
If your domain is larger than that, say all real numbers  $\mathbb{R}$, then you can write:
$\forall x \in \mathbb{R}^+ \exists y \in \mathbb{R}^+ \ (y < x \rightarrow \forall z \in \mathbb{R}^+ \ yz \ge z)$
though this is really just shorthand for:
$\forall x ( x \in \mathbb{R}^+ \rightarrow \exists y (y \in \mathbb{R}^+ \land (y < x \rightarrow \forall z (z \in \mathbb{R}^+ \rightarrow yz \ge z)))$
This statement is true, since for any $x$, you can simply pick $y=x$, making $y<x$ false, and thus the whole conditional true.
For your edited exercise, you get:
$\forall x \in \mathbb{R}^+ \exists y \in \mathbb{R}^+ \ (y < x \land \forall z \in \mathbb{R}^+ \ yz \ge z)$
This statement is false, since if $x < 1$, then that means $y < 1$, and then $yz < z$.
The negation of this is:
$\neg \forall x \in \mathbb{R}^+ \exists y \in \mathbb{R}^+ \ (y < x \land \forall z \in \mathbb{R}^+ \ yz \ge z)$
which is equivalent to:
$\exists x \in \mathbb{R}^+ \forall y \in \mathbb{R}^+ \ (y \ge x \lor \exists z \in \mathbb{R}^+ \ yz < z)$
This statement is true, since you can just pick $x =1 $, meaning that for any $y \ge 1$ you have $y \ge x$, and for $y<1$ you have $yz < z$, so for any $y$, one of the disjuncts will hold.

