Need help understanding Axiom of Extensionality I'm attempting to learn Set Theory and I'm currently working through Halmos' Naive Set Theory.  I will say that I completely understand the essence of the Axiom of Extensionality.  However, where I'm stumbling is grasping its formalization in set-builder notation.  So, looking at:
$$\forall x\forall y \, \left(x=y \leftrightarrow \forall z \, \left(z\in x \leftrightarrow z\in y\right)\right)$$
The part I do not follow is the $\left(z\in x\leftrightarrow z\in y\right)$
So, in my head, if I imagine that I have a set $x$ that is $\{ “a”, “b”, “c”\}$ and I have a set $y$ that is $\{“a”, “b”, “c”, “d” \}$, I intuitively understand that by the Axiom of Extension, sets $x$ and $y$ are not equal.  However, I don't follow how the formal statement would yield this.  So for example, suppose we have a set $z$ that is $\{ “a”, “b”, “c”\}$.  
So, in my mind - here's what I do to mentally work through the formal statement of the axiom.  First, I'll ask myself: Is $z$ in $y$?  Well, since $y$ is $a, b, c, d$ and $z$ is $a, b, c$ - so it would certainly seem that $z$ is in fact in $y$.  So then I'll ask myself: Is $z$ in $x$?  Well, $x$ is $a, b, c$ and $z$ is $a, b, c$ - so it would certainly seem that $z$ is also in fact in $x$!  Therefore, the expression $\left(z \in x \leftrightarrow z \in y\right)$ should evaluate to True and therefore $x = y$ (which of course I know is wrong).
So, hopefully you can see why I am stuck on this.  I assume the root of my misunderstanding is a flaw in how I'm interpreting the formal statement; specifically the iff biconditional connective; but I'm just not seeing how/why I am.
Any help in providing clarity on where / how I'm stumbling on this would be greatly appreciated.  
Thank you for you time.
 A: You are confusing $z\in y$ ($z$ is an element of $y$) with $z\subseteq y$ ($z$ is a subset of $y$).  There is also possible misunderstanding of the meaning of $\forall z$.
Let's look at $x$ and $y$ as in your example, and let $z=d$. Then, on the assumption that $d$ is not equal to any of $a$, $b$, or $c$, we have $z\in y$ but $z \notin x$. So by Extensionality, we conclude that $\lnot(x=y)$.  Remember that $x=y$ if and only if for all $z$, $z\in x \iff z\in y$.
You don't get to pick the $z$ that you will consider.  Also, if you pick $z$ as you did, namely $z=\{a,b,c\}$, then we do not have $z\in x$, though we do have $z \subseteq x$. That is why I conjecture that you are having some trouble distinguishing between $z \in x$ and $z \subseteq x$.
Informally, Extensionality says that two sets $x$ and $y$ are equal if and only if $x$ and $y$ have the same elements.  
Comment: The book you are going through is a classic work, very well written.  Despite the passage of many years, it remains I think the best introduction to Set theory.  Good choice!
A: Please see the answer I posted here.
The definition of the Axiom of Extensionality you referenced presumes a particular definition of the set $z$ which is different from the "arbitrary" sets $x$ and $y$.
I believe $z$ is presumed to be a "universal" set over the entire Domain of Discourse, in which case this definition of the Axiom of Extensionality suddenly makes perfect sense.
