For any interval $I = (a,b] \subseteq \mathbb{R}$, how can we find disjoint partitions (which maybe countably infinite)?
Specifically, for any $I$, how can we construct
$$ I:= \cup_{i \in \mathbb{N}}I_i $$
such that
$I_i \cap I_j = \emptyset$ for all $i\neq j$,
$a = \inf\{I_1\}$,
$b = \sup \left\{ \cup_{i \in \mathbb{N}}I_i \right\}$
$| I_i | \leq \epsilon$ for all $\epsilon > 0$ and for all $i \in \mathbb{N}$?
My idea for this problem is that for any given $\epsilon>0$ (and $\epsilon \leq b-a$), set $L \triangleq \frac{b-a}{\epsilon}$, and construct $I_i$ as follow:
$$ I_i = \begin{cases} (a + (i-1)\epsilon, a+i\epsilon ] , \text{ if } i \leq \lceil L\rceil -1 \\ (a + (i-1)\epsilon, a+L\epsilon ] , \text{ if } i =\lceil L \rceil. \end{cases} $$
Then this interval can be countably infinite because for any $N \in \mathbb{N}$, we can always find $L$ such that $L > N$.
Could you please help me elaborating such construction? Also, could this logic be extended to all real-line not just $I$?
Quick answer: Impossible.
From the last condition $|I_{i}|\leq\epsilon$ for all $\epsilon>0$ and all $i\in\mathbb{N}$, it implies that $|I_{i}|=0$ for each $i$.
Being a countable union of sets of measure zero, $|I|=0$.