$(0,1] \cup \left(\frac{1}{2}+\frac{1}{2^2},\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right] \cup \cdots$ is not in $\mathbf{A}$ I have a little problem with understanding following example. If I want to explain this to someone so I should be sure of my understanding. (sorry for not native English) 
 
I know (i) 
since we can choose a partition (with finite member) on $(0,1]$ like $$\left(0,\frac{1}{2}\right]\cup \left(\frac{1}{2},1\right]$$.  Am i right? (1)
For part (ii) if $A\in \mathbf{A}$ so $A=\cup_{i=1}^{m} (a_i,a_i^{\prime}]$.
I can make a partition on $(0,1]$ like
 $$(0,1]=A \cup_{i=1}^{m} (b_i,b_i^{\prime}]$$ where $B=\cup_{i=1}^{m} (b_i,b_i^{\prime}]$ 
look like Figure (1) and (2)(the parts of $(0,1]$ that not selected by $A=\cup_{i=1}^{m} (a_i,a_i^{\prime}]$) . so $A^{c}=B\in \mathbf{A}$.   Am i right? (2)
part (iii) if $A=\cup_{i=1}^{m} (a_i,a_i^{\prime}]$ and $B=\cup_{i=1}^{m} (b_i,b_i^{\prime}]$.
I think $A\cap B$ is look like 
$$\cap_{some \, i \, j} \left(a_i \vee b_j , a_i^{\prime} \wedge b_j^{\prime}   \right]$$
so $A\cap B\in \mathbf{A}$. Am i right? (3)
And  last ,I am stuck in following step ,(4)
$$D=\left(0,\frac{1}{2}\right] \cup \left(\frac{1}{2}+\frac{1}{2^2},\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right] \cup \left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4},\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\right] \cdots  $$.
I know  $D=\cup_{n=1}^{\infty} C_n$ is constructed from countable unions  sets of $\mathbf{A}$. But Why $D$ is not in  $\mathbf{A}$. I see  $1\notin D$ but I don't know how can I use it? Am I wrong something?   
Source: A Probability Path, Sidney I. Resnick  page 15
 A: (i) is easier than that.  $a = 0$, $a' = 1$ is allowed, so $(0,1]$ has been explicitly permitted as an element of $\mathcal{A}$.
(ii)  This might be easier.  Your $A = \bigcup_{i=1}^m (a_i, a_i']$ can be rewritten as a finite union of disjoint elements $\bigcup_{j=1}^n (b_j,b_j']$, where $n \leq m$  and 
$$  0 \leq b_1 < b_1' < b_2 < b_2' < \cdots < b_n < b_n' \leq 1  \text{.}  $$
(This is such a useful fact, you should probably prove it as a lemma and then reuse it : We can always make $A$ a union of disjoint intervals where the interval endpoints are sorted to be monotonically strictly increasing.)
Then the complement, 
$$  \Omega \smallsetminus A = (0,b_1] \cup \left( \bigcup_{j = 1}^{n-1} (b_i',b_{i+1}] \right) \cup  (b_n',1]  \text{,}  $$
where we acknowledge that the first and/or last interval listed may be empty if both of its endpoints are the same.
(iii)(3): Yes, taking the Cartesian product of intersections works and such a product is finite here.
(4):  $D$ has already been written as a (-n infinite) disjoint union with ordered endpoints.  $A$ is given to (at least) contain finite disjoint ordered unions, "FDOU"s.  The number of disjoint components in $\Omega \smallsetminus A$ is at most one more than the number in $A$ (see the formula above), so complementation turns FDOUs into FDOUs.  Intersection and union each take two FDOUs and produce a new FDOU.  In all cases, only FDOUs are produced by the operations, so any finite sequence of operations performed on a collection of FDOUs produces only an FDOU.  Therefore, $A$ only contains FDOUs.  The given $D$ is not an FDOU, so $D \not\in \mathcal{A}$.
