Proving a partition is in a union of open intervals If $[a,b] \subseteq (a_1, b_1)\cup\cdots\cup(a_n, b_n)$, show that there is a partition $P = \{x_0, . . . , x_m\}$
of $[a, b]$ such that each interval $[x_{j−1}, x_j ]$ is contained in $(a_i, b_i)$ for some index $i = i(j)$.
So I am trying to prove this, but I'm having a hard time just intuitively seeing that it is true. Is there a certain rule that the $(a_i,b_i)$'s must follow? As all of $[a,b]$ could be a subset of $(a_i,b_i)$ for some $i$ or $(a_i,b_i)\cap[a,b]$ could be the empty set for some $i$... So I guess I don't truly understand what I am trying to prove. Obviously it is not that $[x_0,x_1]\subseteq(a_1,b_1)\dots [x_{m-1},x_m]\subseteq(a_n,b_n)$ because that is true for all $(a_1, b_1)\cup\cdots\cup(a_n, b_n)$. Can someone help me understand what it is I am trying to prove? Thank you in advance
 A: Well, you ask how to intuitively understand what you are expected to prove:
So [a, b] = $\cup_{i}(a_i, b_i)$ a bunch of open intervals that do not include the endpoints.  So the endpoints of an interval must be included in some other open interval.
If I may abuse notation and make assumptions that the intervals are indexed sequentially and no index are completely nested in another we have:
$(a_1 .......(a_2 .... b_1)......(a_3 .... b_2)....(a_4 ....b_3) .....(a_n ...b_{n-1}) .... b_n)$.  [Dang.  This would be clearer is I could do color or animate the writing of the integers one by one.  The point is the intervals overlap]
For instance:  take a, the endpoint of [a,b], a must be in some interval ($a_i, b_i$).  So $a_i < a$.  Call $a = x_0$.  Look at $bi$ if we assume $b_i < b$ (and we pretty much have too, otherwise [a,b] $\subset$ ($a_i, b_i$) which is a bit ... light...) then as $b_i$ $\notin$ ($a_i, b_i$) we must have $b_i$ in some ($a_j, b_j$).  So pick some $x_1$ that $a_j < x_1 < b_i$ so [$x_0, x_1$] $\subset$ ($a_i, b_i$).
Thus we have, if I may abuse notation, overlapping intervals:
($a_i ...[a=x_0 ..... (a_m .... x_1][x_1 ... b_i) .... b_m)...... b]$
We want to keep going to get
$(a_i ...[a ....(a_m .....x_1][x_1 ...b_i) ..(a_j.. x_2][x_2.... b_m)...   (a_n...x_v=b]..a_n)$  So [a,b] = $\cup_{k}[x_k, x_{k+1}$] and each $[x_k, x_{k_1}]$ is a subset of some $(a_j, b_j)$.
Of course, an actual proof needs more precision and rigor.
