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$(0,1] \cup \left(\frac{1}{2}+\frac{1}{2^2},\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right] \cup \cdots$ 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

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