Can I define $C_i$ this way?

Question

Let $A_i=(0,1-\frac{1}{2^i}]$ for $i\in \mathbb N$. Let $A=\bigcup_iA_i=(0, 1]$. Let $B=A\setminus\{1\}=(0,1)$. Can I define $C_i=B\setminus A_i$? I was thinking that since $A_i$ approaches $(0,1]$ as $i \rightarrow \infty$, I am not able to define $C_i$ this way, because $\{1\} \notin B$ but $\{1\} \in A_i$. Am I just wrong?

Answer 1

First off, note that since $1\notin A_i$ for any $i\in \Bbb N$, then$$\bigcup_i A_i=(0,1)$$and therefore $$B=A-\{1\}=A=(0,1)$$therefore $$C_i=B-A_i=\left(1-{1\over 2^i},1\right)$$