Consider family $\{I_j\}_{j=1}^\infty$ of disjoint intervals $I_j=(a_j,b_j]$ such that $\bigcup_{j=1}^\infty I_j=(a,b] \subset \mathbb{R}.$
Question: Can I claim that it always exists a way to reindex the intervals $I_j$ s.t $\bigcup_{j=1}^nI_j$ is an interval contained in $(a,b]$?
I'm trying to use compacticity properties of $\mathbb{R}$ but I can't do it yet...
Such a way does not always exist. Let $$ I_n=\begin{cases} (\,(\frac12)^{k+1},(\frac12)^k] & n=2k \text{ is even}\\ (\,(\frac12)^{k+1}+\frac12,(\frac12)^k+\frac12] & n=2k-1 \text{ is odd}\\ \end{cases} $$
The union of $I_1,I_2,I_3,\dots$ is $(0,1]$, where the even intervals are a partition of $(0,1/2]$ and the odd intervals are a partition of $(1/2,1]$. Two intervals $I_i$ and $I_j$ can only be adjacent if both $i$ and $j$ are both even or both odd. Therefore, in a reordering of $\{I_j\}_{j=1}^\infty$ so that the union of any initial segment is an interval, the parity of each index would have to match the parity of the first index, which means that not every interval appears in the reordering.