# sequences of pairwise disjoint intervals — sums of lengths

Let $(I_n)$ and $(J_n)$ be sequences of bounded intervals (in $\mathbb{R}$) such that $\cup_n I_n = \cup_n J_n$. Then page 268 of Carothers' Real Analysis states that if the $I_n$'s are pairwise disjoint then $\sum_{n=1}^\infty \text{length}(I_n) \leq \sum_{n=1}^\infty \text{length}(J_n)$.

I don't understand the proof he gives. He begins by shrinking the $I_n$'s slightly to make them all compact and expanding the $J_n$'s slightly to make them all open. I'm OK with that. Then he supposes that $\sum_{n=1}^\infty \text{length}(I_n) > \sum_{n=1}^\infty \text{length}(J_n)$, so there is an $N$ such that $\sum_{n=1}^N \text{length}(I_n) > \sum_{n=1}^M \text{length}(J_n)$ for each $M$. That's also OK. But then he says that since that since $\cup_1^N I_n$ is compact, there is a subcover of the $J_n$s which somehow gives us a contradiction. How does it give us a contradiction?

edit: I mean, if $(J_1,\ldots,J_s)$ is a subcover of $(J_n)$ then in order to get a contradiction we would need to show that $\sum_{n=1}^N \text{length}(I_n) \leq \sum_{n=1}^s \text{length}(J_n)$, which he doesn't do.

Thanks.

• The collection of all the $J_n$'s cover the compact set $\cup_{n=1}^N I_n$. But this cover has no finite subcover, since $\sum_{n=1}^N \text{length}(I_n) > \sum_{n=1}^M \text{length}(J_n)$ for each $M$. He spells this out in the text ... – David Mitra Mar 15 '14 at 17:29
• But why does that imply that $(J_n)$ admits no finite subcover... – nigel Mar 15 '14 at 17:34
• The $I_n$'s are pairwise disjoint. From this, it follows that if $J_1\cup\cdots\cup J_M$ covered $\cup_{n=1}^N I_n$, then $\sum_{i=1}^M |J_M| \ge \sum_{i=1}^N |I_n|$. – David Mitra Mar 15 '14 at 17:44
• See this post for an idea of how to prove the above. – David Mitra Mar 15 '14 at 18:05

We have that $\sum\limits_{n=1}^N \text{length}(I_n) > \sum\limits_{n=1}^M \text{length}(J_n)$ for each $M \in \mathbb{N}$ as above. Then as $I = \bigcup\limits_{n=1}^N I_n$ is compact, and $\{J_n\}$ is an open cover of $I$, we can take a finite subcover of $I$, say $\{J_{n_1},\ldots,J_{n_k}\}$. Let $K = \max\{n_1,\ldots,n_k\}$. Then we have that $\{J_1,\ldots,J_K\}$ is an open cover of $I$. So for each $1 \leq i \leq N$ we have $I_i \subseteq \cup_{i=1}^M J_i$. Since each $J_i$ is open and each $I_i$ is connected, we have that $I_i \subseteq J_j$ for some $1 \leq j \leq M$.
Now to get really pedantic and technical we do the following. For each $1 \leq i \leq M$ let $A_i = \{ 1 \leq j \leq N : I_j \subseteq J_i\}$. Then we have
$$$$\begin{split} \sum\limits_{i=1}^N \text{length}(I_n) \\ &= \sum\limits_{i=1}^M \sum\limits_{j \in A_i} \text{length}(I_j) \\ & \leq \sum\limits_{i=1}^M \text{length}(J_i) \end{split}$$$$