Equivalence between two definitions of closure of $A\subseteq B$ under a function $f\colon B\to B$ Synopsis
In one of the exercises in Enderton's Elements of Set Theory, we are introduced to the concept of the closure of a set to another set under a function $f$ with two constructions. 
Let $f$ be a function from $B$ into $B$ and assume that $A \subseteq B$. We have two possible methods for constructing the "closure" $C$ of $A$ under $f$.
First, we have the construction: $$C^* = \bigcap\{X \mid A \subseteq X \subseteq B \wedge f [\![X]\!] \subseteq X \}.$$
Alternatively, we could apply the recursion theorem to obtain the function $h$ for which $$h(0) = A$$ $$h(n^+) = h(n) \cup f[\![h(n)]\!].$$
Define $C_*$ to be $\bigcup \text{ran } h$; in other words: $$C_* = \bigcup_{i \in \omega} h(i).$$
In the exercise, we were asked to prove that $C^* = C_*$, which I was able to do without too much effort. The next exercise, though, which asks us to find the closure $C$ if we take $B$ to be the set of real numbers, $f(x) = x^2$, and $A$ the closed interval $[\frac{1}{2}, 1]$, confused me a great deal.
This is because the sets $C^*$ and $C_*$ described above make little sense to me intuitively. I have a general idea that the closure of $A$ under $f$ describe how the function $f$ always maps to values within $A$, but I'm not sure how this closure is a set, or how the constructions above elucidate this concept. For example, when people say that the natural numbers are closed under multiplication, how is the closure under multiplication a set? Isn't that just a property? 
Anyways, all I've been able to make out from the above exercise with my incomplete understanding is that $1/2$ must be in $C$ and as such, $1/4$, $1/8$, $1/16$ ... all the way to $0$. But I'm not sure if this means that the closure $C$ is the set $[0,1]$. It seems like it might be right, but I can't understand how it fits into the definitions above!
In conclusion, can someone please explain to me why the sets $C^*$ and $C_*$ are defined the way they are, and how I can understand the intuition behind them better? I feel like this is all just an issue with me not understanding the formal definitions. I'm self studying this course before I start my first year at university, so you guys are the closest thing I have to a professor! Thank you!
 A: The closure of $A$ under $f$ should be the 'smallest' containing $A$ and closed under $f$.


*

*$C^*$ characterises what this means from the top down: consider all sets that contain $A$ and are closed under $f$. Any such set will all contain the smallest one as a subset, and so the value of the intersection will be the smallest one.

*$C_*$ characterises what this means from the bottom up: take the elements of $A$, and keep adding what you get by applying $f$ to what you already have—at the end, you'll have a set containing $A$ and closed under $f$. This is what your function $h$ does: $h(i)$ is the set of things that can be obtained from elements of $A$ by iterating $f$ at most $i$ times.


The fact that $C^*=C_*$ then tells you that the top-down and bottom-up characterisations of 'smallest' coincide.
A: Clive Newstead has addressed the general idea; I’ll take you through the specific example that you mentioned.
If we build $A_*$, forming the closure of $A$ from the bottom up, we start with $h(0)=\left[\frac12,1\right]$. Then
$$\begin{align*}
h(1)&=h(0)\cup\{x^2:x\in h(0)\}\\
&=\left[\frac12,1\right]\cup\left\{x^2:x\in\left[\frac12,1\right]\right\}\\
&=\left[\frac12,1\right]\cup\left[\frac14,1\right]\\
&=\left[\frac14,1\right]\;,
\end{align*}$$
$$\begin{align*}
h(2)&=h(1)\cup\{x^2:x\in h(1)\}\\
&=\left[\frac14,1\right]\cup\left\{x^2:x\in\left[\frac14,1\right]\right\}\\
&=\left[\frac14,1\right]\cup\left[\frac1{16},1\right]\\
&=\left[\frac1{16},1\right]\;,
\end{align*}$$
and so on. You should be able to convince yourself fairly easily that
$$h(n)=\left[\frac1{2^{2^n}},1\right]$$
and hence that
$$A_*=\bigcup_{n\in\omega}\left[\frac1{2^{2n}},1\right]=(0,1]\;.$$
Suppose that instead we work from the top down and compute $A^*$, the intersection of all sets of real numbers that contain $A$ and are closed under the squaring function. A set $X$ of real numbers is closed under the squaring function if $\{x^2:x\in X\}\subseteq X$: $X$ contains the squares of all of its members. Thus,
$$A^*=\bigcap\big\{X\subseteq\Bbb R:A\subseteq X\text{ and }\{x^2:x\in X\}\subseteq X\big\}\;.\tag{1}$$
It’s easy enough to see that $\{x^2:x\in[0,1]\}=[0,1]$, so $[0,1]$ is closed under the squaring function, and certainly $A\subseteq[0,1]$. Thus, $[0,1]$ is one of the sets $X$ that are intersected in $(1)$ to get $A^*$, and therefore we know that $A^*\subseteq[0,1]$. The question is whether $A^*$ is any smaller, i.e., whether there is some set $X$ that is closed under the squaring function — it contains the squares of all of its members — and satisfies the condition $A\subseteq X\subsetneqq[0,1]$. And in fact there is: $(0,1]$ contains $A$ and is closed under the squaring function. We now know that $A^*\subseteq(0,1]$.
It takes a bit of work to show that we can’t cut away any more and still have a set that both contains $A$ and is closed under the squaring function, and I’ll just sketch the argument. Let $g:[0,1]\to[0,1]:x\mapsto\sqrt{x}$, and suppose that $X$ is closed under the squaring function, $A\subseteq X$, and $x_0\in(0,1)$. Given $x_n$ for some $n\in\omega$, let $x_{n+1}=g(x_n)$. Then $\langle x_n:n\in\omega\rangle$ is an increasing sequence with limit $1$, so there is an $n\in\omega$ such that $x_n\in A\subseteq X$. But then $x_{n-1}=x_n^2\in X$ (since $X$ is closed under squaring), and then $x_{n-2}=x_{n-1}^2\in X$, and so on all the way down to $x_0\in X$. (This should really be formalized as a downward induction, but the idea should be clear.) This shows that $(0,1)\subseteq X$, and of course we already knew that $1\in A\subseteq X$, so $(0,1]\subseteq X$. That is, on the one hand $(0,1]$ is a set that contains $A$ and is closed under the squaring function, and on the other hand we’ve just seen that any set $X$ that contains $A$ and is closed under the squaring function must contain $(0,1]$, so the intersection of all such sets must be $(0,1]$: $A^*=(0,1]$.
