Variables as Elements of Sets Suppose we have a set $A = \{1, 2, 6\}$. Let's also say we have a variable $x$. If you were asked if $x \in A$ is true, without knowing the value of $x$, how would you respond? Would the answer be false? Is there insufficient information so far?
Further, suppose we have another set $B = \{x, y, 7\}$. Is this notation even valid? Can we say that the variable $y \in B$ without knowing the value of $y$?
One last example: if we have a set $C =\{x, y, z\}$, to determine if $a \notin C$, would we need to know all of the following: $a \neq x$, $a \neq y$, $a \neq z$?
 A: If someone asked you, "Is it true?" you'd need more information, and you'd probably reply with a question: "Is what true?" If you were asked, "Is he going to the party?", you couldn't answer without knowing whom "he" refers to.
Variables are like pronouns. If I were told that $A = \{1,2,6\}$ is and then asked, "Is $x\in A$?", without more information about $x$ no reply is possible. If I'm told that $x=4$ then I can say, No, $x\notin A$; if I'm told that $1\le x\le 2$ and that $x$ is an integer, then I know that, yes, $x\in A$.
If $x$ and $y$ have values, denote entities, then $\{x,y,100\}$ is well-defined; if they don't, it isn't.
Variables are not members of sets; sets (/mathematical entities) are, and variables denote them. Pronouns don't go to parties; people do, and pronouns denote them.

Re your last two examples: 

For all $x,y$, if $B = \{x,y,100\}$ then $y\in B$. This is true. Notice that $x,y$ are bound variables here. In your first example, $x$ is free — it's not bound by a quantifier, and has no value.
If $C = \{x,y,z\}$, then $a\in C \iff (a=x \lor a=y \lor a=z)$, so if you  negate both sides the results are equivalent too. If you know $a\notin C$ then you know that $a$ is not equal to any of $x,y,z$, and if you know the latter, then you know $a\notin C$. This holds for all $x,y,z,a$ (bound variables again). 
