Equivalence between two definitions of closure of $A\subseteq B$ under a function $f\colon B\to B$

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.

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 $(0,1]$: $A^*=(0,1]$.