P(x) is x-special???

P(x) is x-special???

Beforehand, I apologize for my complete inability to write any sort of TeX
commands; fortunately, my question shouldn't require this. (If anyone
knows a good reference for me to start learning, I would greatly
appreciate it.) Also, any occurrence of "0" below refers to the empty set.
I'm reading Smullyan and Fitting's "Set Theory and the Continuum Problem,"
and the authors are trying to show that for a fixed progressing function
g, (i.e., a function g on the universal class V such that x is a subset of
g(x) for all x in V) the existence of the class of all g-sets (x is a
g-set if x belongs to every class superinductive under g) is derivable
without the axiom of substitution. In particular, when g is the successor
function x |--> x U {x} , this shows that the class of all ordinals
exists.
Now, the authors approach this through a series of lemmas. First, they
define:
For sets S and x, "S is closed (under g) relative to x" iff for all z in
S, if z is a subset of x, then g(z) is in S.
A set S is "x-special" iff 0 is in S, S is closed relative to x, and S is
closed under chain unions.
Then they write "Lemma 7.4: P(x) is x-special. Proof: Obvious."
But I don't believe that's even true! For example, if we let x = 0 and g
be the successor function, then P(x) = {0} = 1, so if we take z = 0, we
see that z is in S, z is a subset of x, and clearly g(z) = 0 U {0} = {0} =
1 is not in P(x), so that P(x) is NOT closed under g relative to x, hence,
P(x) is NOT x-special.
Am I misunderstanding something? Please help clarify this for me...it's
been driving me crazy. It seems so simple, yet I can't understand why the
authors would write something like this if it were so "obviously" false.
Thank you! (Sorry for the long read)