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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

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    $\begingroup$ The open intervals $(a_i, b_i)$ are fixed. They cover $[a,b]$ and they overlap. You must show that you can divide the interval into closed intervals that only meet at their endpoints and, moreover, each individual one is contained completely in one of the open sets. $\endgroup$
    – John Douma
    Commented Oct 27, 2015 at 1:15
  • $\begingroup$ Could induction help? $\endgroup$
    – Matthew Leingang
    Commented Oct 27, 2015 at 1:20
  • $\begingroup$ Note: although each of the closed intervals $[x_{j-1}.x_j]$ is contained completely in one of the open sets, it may intersect one or more of the others. In fact if you draw the case with just two open sets, you'll see what I mean. $\endgroup$
    – John Douma
    Commented Oct 27, 2015 at 1:31
  • $\begingroup$ "Intuitively seeing it's true" Well there can't be any gaps so the ($a_i, b_i$) must overlap. So for ($a_i, b_i$) there must be some $b_j$ that overlaps the $a_i$ and an $a_m$ that overlaps the $b_i$. (in other words $a_i < b_j < a_m < b_i$). Choose $a_i < x_k < b_j$ and $a_m < x_{k+1} < a_i$. [$x_k, x_{k+1}$] $\subset$ ($a_i, b_i$) and $\{[x_k,x_{k+1}]\}$ cover all the [a, b]. That's intuitive. We have to consider all cases. $\endgroup$
    – fleablood
    Commented Oct 27, 2015 at 1:59
  • $\begingroup$ Thank you, these comments are helping my understanding. I'm going to try it by induction $\endgroup$
    – Burgundy
    Commented Oct 27, 2015 at 2:05

1 Answer 1

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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.

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