Set Theory Elementhood Notation From How to Prove it:
Given $A=\{n^2|n \in N\}$ where $N$ is the set of all natural numbers.
I want to express A in terms of elementhood test notation.
Velleman says $A=\{x| \exists n \in N (x=n^2)\}=\{x| \exists n(x=n^2 \land n\in N)\}=\{n^2|n \in N\}$ 
Question is:
Why is $A=\{x|(x=n^2 \land n\in N)\}$ incorrect(why is the $\exists$ necessary) and what is the difference between the one with the existential quantifier and the one without ? 
 A: If you say $$\exists x. \text{$x$ is white}$$ (“there is some $x$ that is white” or “there exists an $x$, such that $x$ is white”)  you are asserting the existence of some white thing.  This is a  completely self-contained statement; it is true, because for example snow is white.
But if you say just $$\text{$x$ is white}$$ this is not a self-contained statement, because it asserts the whiteness of a particular object which we may have agreed to refer to as $x$ in some larger discussion; without such a prior agreement, you cannot say whether the statement is true of false, and if someone asks you if it is true, you have ask “I don't know, what is $x$?”
If you say that 

$A$ is the set of all $x$ such that there is some $n$ with $x=n^2$,

then one can decide, for any proposed element of $A$, whether it is in $A$ or not:


*

*Is $2$ in $A$?  Yes, if there is some $n$ for which $2 = n^2$; there is no such (whole number) $n$, so $2$ is not in $A$.  

*Is $16$ in $A$?  Yes, if there is some $n$ for which $16 = n^2$; there is such an $n$, namely $4$, so $16$ is in $A$.


So this is a self-contained definition of $A$.
But if you say that

$A$ is the set of all $x$ such that $x=n^2$

then nobody can decide anything about $A$ without first knowing what $n$ is supposed to represent.  Is $16$ in $A$?  Yes, if $16=n^2$.  Wait, what is $n$? If $n$ was previously agreed to be 4, then $16$ is in $A$, but if $n$ was previously agreed to be 57, or a potato, then $16$ is not in $A$.  And in the absence of such an understanding of $n$, one cannot say anything at all about whether 16, or anything else, is in $A$.  This statement is not self-contained; it depends on some previous agreement about the meaning of $n$ in this context.
The technical description of the difference between these two definitions of $A$ is that in “the set of all $x$ such that there is some $n$ with $x=n^2$”, the variable $n$ is bound by the quantifier “there is some”, but in “the set of all $x$ such that $x=n^2$” the variable $n$ is free.   Expressions with free variables are open, which means they are not self-contained.
