Absoluteness and Extensionality In the set theory text that I am reading, the author writes:

Relative to the set $A = \{ 0, \{\{0\}\} \}$, the sets $0$ and $\{\{0\}\}$ are indistinguishable in the sense that $[$for all $x$ in $A$, $x$ is a member of $0$ just in case it is a member of $\{\{0\}\}]$, so $[0 = \{\{0\}\}]$ restricted to $A$ is true but $[0 = \{\{0\}\}]$ "restricted" to $V$ is not.

I don't understand the first statement.  Why is $0 = \{\{0\}\}$ not false in $A$ even though "$x$ is an element of $0$ just in case it is an element of $\{\{0\}\}$" is true there?  (Isn't the truth-condition for "$x = y$" that the thing assigned to $x$ is the numerically same thing as that assigned to $y$?)  Thanks in advance.
 A: This is a tricky explanation. Let's say that two sets $a, b$ are $\in$-equivalent if they satisfy $\forall x(x\in a\leftrightarrow x\in b) $.
The axiom of extensionality states that two $\in$-equivalent sets are actually equal. 
But what happens if we only look at $A$ as in the question? Now the quantifier on $x$ is only allowed to go over elements of $A$. So now the two sets $0, \{\{0\}\}$ are $\in$-equivalent as far as $A$ is concerned. Namely $A$ "knows" these are two different objects, but it doesn't "know" how to separate them using only the $\in$ relation. 
So in the structure $(A, \in)$ the axiom of extensionality fails. 
A: Extensionality ($E$) is $\forall a\;\forall b \; (a=b\iff (\forall x\; (x\in a \iff x \in b))).$ Now relativized to $A$ we have $E^A\iff \forall a\in A\;\forall b\in A\; (b=a\iff (\forall x\in A\; (x\in a\iff x\in b))).$ Now no member of $A$ belongs to $0$ and no member of $A$ belongs to $\{\{0\}\}.$ So if $E^A$ were true we would have $0=\{\{0\}\},$ which is absurd by the def'n of $0$. It is not correct to say that "$0=\{\{0\}\}$ is true in A". It is correct to say that $A$ does not satisfy $E.$ 
