Intensional Set Definitions like $\{ x | A(y) \}$ Let $x = 1$. Is it valid to define sets like $Y = \{ x | 1 = 1 \} = \{ 1 \}$ and $Z = \{ x | 1 \neq 1 \} = \emptyset$?
What I want to know: Are we allowed to define sets like $\{ y | A(z) \}$ where $A(z)$ is a condition that is independent from $y$ and $y$ is defined somewhere outside the set (intensional definition)? Normally we define sets like $\{ y | A(y) \}$ where $A(y)$ is a condition depending on $y$.
 A: In the framework of axiomatic set theory, you cannot do that as you'll need some axiom to justify the existence of a set and in cases like these you'd use the axiom of separation for which you'd need another set.  In other words, if $Y$ is a set, then $\{x \in Y \mid A(x)\}$ is a set, no matter what $A(x)$ is.
But there's no general axiom that makes something like $\{x \mid A(x)\}$ a set.
$\{x \mid 1 = 1\}$ would be the class (note: class, not set) of all sets.  And even something like $\{x \mid x \neq 42\}$ would result in a proper class.  See Russell's paradox.
A: It's fine to use "$x$" outside of "$\{x:\phi\}$" in the way you do. The reason is that the "$x$" in "$\{x:\phi\}$" is bound. So, for instance, when you say "$Y = \{x: 1= 1\}$", what you mean is "there is a set $y$ such that for all $x$, $x$ is in $y$ iff $1=1$; and $Y=y$". Any use of "$x$" outside this sentence doesn't fall within the scope of "for all $x$" and is thus independent of the "$x$" in "$Y = \{x:1=1\}$". In general, terms of the form "$\{x:\phi\}$" involve "$x$" bound in this way, and uses of "$x$" outside of them will be independent of uses inside them.  
