Exercise 2.5.14 from 'The Real Numbers and Real Analysis' by Ethan D.Bloch.
Hello, today I tried to prove this statement, but I'm stuck so I would like some help. Below I will put a drawing that more or less expresses the idea I had when making the proof for the second case, in this way I hope to give more clarity to some definitions.
Solution. We prove the statement by induction. Let $n=1$. Then$$\operatorname*{\bigcup}_{i=1}^{1}[a_i,b_i]=[a_1,b_1].$$
Hence $\operatorname*{\bigcup}_{i=1}^{1}[a_i,b_i]$ equals the union of finitely many disjoint closed bounded intervals. For the inductive step, suppose that $\operatorname*{\bigcup}_{i=1}^{n}[a_i,b_i]$ equals the union of finitely many disjoint closed bounded intervals for some $n=k$. Then $\operatorname*{\bigcup}_{i=1}^{k}[a_i,b_i]=\operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i]$ where for all $i \in \{1,\cdots,m\}$ the intervals $[\alpha_i,\beta_i]$ are disjoint. Consequently\begin{equation}\begin{aligned}\operatorname*{\bigcup}_{i=1}^{k+1}[a_i,b_i] &= \left ( \operatorname*{\bigcup}_{i=1}^{k}[a_i,b_i] \right )\bigcup [a_{k+1},b_{k+1}] \\&= \left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right ) \bigcup [a_{k+1},b_{k+1}].\end{aligned}\end{equation}
Now we proceed by cases.
Case 1. Suppose that $[a_{k+1},b_{k+1}]\cap[\alpha_i,\beta_i]=\varnothing$ for all $i \in \{1,\cdots,m\}$. Let $[\alpha_{m+1},\beta_{m+1}]:=[a_{k+1},b_{k+1}]$. Then$$\operatorname*{\bigcup}_{i=1}^{k+1}[a_i,b_i] = \operatorname*{\bigcup}_{i=1}^{m+1}[\alpha_i,\beta_i]$$
where for all $i \in \{1,\cdots,m+1\}$ the intervals $[\alpha_i,\beta_i]$ are disjoint.
Case 2. Suppose that $[a_{k+1},b_{k+1}]\cap[\alpha_i,\beta_i]\neq\varnothing$ for some $j \in \{1,\cdots,m\}$. Let$$I:=\{i\in \{1,\cdots,m\}:[a_{k+1},b_{k+1}]\cap[\alpha_i,\beta_i]\neq\varnothing\}.$$We define $\alpha:=min\{a_{k+1},min\{\alpha_i\}_{i\in I}\},\beta:=max\{b_{k+1},max\{\beta_i\}_{i\in I}\},$ and $J:=\{1,\cdots,m\}-I.$Now, we prove that
$$\left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right ) \bigcup [a_{k+1},b_{k+1}]=\left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )\bigcup[\alpha,\beta].$$
($\subseteq)$. Let $x \in \left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right ) \bigcup [a_{k+1},b_{k+1}]$. If $x\in \left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right )$, then $x\in [\alpha_j,\beta_j]$ for some $j\in\{1,\cdots,m\}$. Suppose that $j \in J$. Then $x \in \left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )$. Now let $j\notin J$. Then $j \in I$, and so$$\alpha \leq min\{\alpha_i\}_{i\in I}\leq \alpha_j \leq x \leq \beta_j \leq max\{\beta_i\}_{i\in I}\leq \beta.$$Consequently $x \in [\alpha,\beta]$. On the other hand, If $x \in [a_{k+1},b_{k+1}]$, then $\alpha\leq a_{k+1}\leq x\leq b_{k+1}\leq \beta,$and hence $x \in [\alpha,\beta]$. Therefore, based on the previous two cases we have showed that
$$\left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right ) \bigcup [a_{k+1},b_{k+1}]\subseteq\left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )\bigcup[\alpha,\beta].$$
$(\supseteq)$. Let $x \in \left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )\bigcup [\alpha,\beta]$. If $x\in \left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )$, then $x\in [\alpha_j,\beta_j]$ for some $j \in J$. Hence $j \in \{1,\cdots,m\}$, and so $x \in \left ({\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right )$. If $x\in [\alpha,\beta]$, then ...
The statements I need to prove in order to finish the proof are
$$(1) x \in [\alpha,\beta] \implies x \in \left ( \operatorname*{\bigcup}_{i=1}^{m}[\alpha_i,\beta_i] \right ) \bigcup [a_{k+1},b_{k+1}]$$
and
$$(2) \left ( \operatorname*{\bigcup}_{i\in J}[\alpha_i,\beta_i] \right )\bigcap[\alpha,\beta]=\varnothing.$$
For (1), I observed in the drawing that there are two things that can happen to $x$, either $x$ is exclusively in $[a_{k+1},b_{k+1}]$ or not, we can express that mathematically as $x \notin [\alpha_i,\beta_i]$ for all $i\in I$ or $x \in [\alpha_j,\beta_j]$ for some $j \in I$. I tried to proceed from here, but couldn't find something useful. For (2), I just don't see anything I can do yet. Any ideas or hints are welcomed. Thank you for reading.