Closure Question in Enderton's 'Elements of Set Theory'

Question

I am currently working on a follow-up question to the one I did here: Closure question from Enderton's 'Elements of Set-theory'

I am unsure though whether I am on the right track with the mechanics of this question. In particular, I am unsure whether I am meeting the intersection ($\bigcap$) condition on $C^{*}$ with my set $S = \mathbb{Z^{-}} \cup \{0\}$. Any commentary would be appreciated on this point.

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Conditions of the question: Let $f$ be a function from $B$ into $B$, and assume that $A \subseteq B$. Define $C^{*} = \bigcap \{X \ | \ A \subseteq X \subseteq B \ \& \ f[X] \subseteq X\}$. Further, define the function $h$ where $h(0) = A, h(n^{+}) = h(n) \cup f[h(n)]$. We also define $C_{*} = \cup_{i \in \omega} h(i)$. We assume that $B = \mathbb{R}$, $f(x) = x - 1$, and $A=\{0\}$. What is $C^{*}$ and $C_{*}$?

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We claim that $C^{*} = \mathbb{Z^{-}} \cup \{0\}$. Considering $f(0) = -1$, we know $f(A) \not \subseteq A$ since $f(A) = \{-1\} \not \subseteq A = \{0\}$. Continuing on, we have $f(1) = -2$, and $\{-2\} \not \subseteq \{-1\}$. So beginning with 0, and applying $f(0)$, each iteration produces a larger negative number with each $f(x) \not \subseteq x$. The only set S that has $A \subseteq X \subseteq \mathbb{R}$ and $f(X) \subseteq X$ is $S = \mathbb{Z^{-}} \cup \{0\} = C^{*}$.

For $C_{*}$, we have $h(0) = A = \{0\}$, and $h(n^{+}) = h(n) \cup f[h(n)]$. By continuous recursion, we get $C_{*} = \cup_{i \in \omega} h(i) = \mathbb{Z^{-}} \cup \{0\}$.

Answer 1

It is clear that $C^* \subseteq C_*$ since $A \subseteq C_* \subseteq B$ and $f(C_*) \subseteq C_*$.

Now suppose $X$ is such that $A \subseteq X \subseteq B$ with the property that $f(X) \subseteq X$. By induction, we will show that $h(n) \subseteq X$ for all $n$. $A \subseteq X$ implies $h(0) \subseteq X$. Suppose that $h(n) \subseteq X$. Since $f(X) \subseteq X$, one has that $f(h(n)) \subseteq f(X) \subseteq X$. Hence $h(n + 1) \subseteq X$. By induction it has been shown that $h(k) \subseteq X$ for all $k$. Thus $C_* \subseteq X$. Thus $C_* \subseteq C^*$.

Equality has been shown.