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.


  • $\begingroup$ 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 ... $\endgroup$ – David Mitra Mar 15 '14 at 17:29
  • $\begingroup$ But why does that imply that $(J_n)$ admits no finite subcover... $\endgroup$ – nigel Mar 15 '14 at 17:34
  • $\begingroup$ 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 $\sum_{i=1}^M |J_M| \ge \sum_{i=1}^N |I_n|$. $\endgroup$ – David Mitra Mar 15 '14 at 17:44
  • $\begingroup$ See this post for an idea of how to prove the above. $\endgroup$ – David Mitra Mar 15 '14 at 18:05

I just realized this is about connectedness.

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

$$ \begin{equation} \begin{split} \sum\limits_{i=1}^N \text{length}(I_n) \\ &= \sum\limits_{i=1}^M \sum\limits_{j \in A_i} \text{length}(I_j) \\ & \leq \sum\limits_{i=1}^M \text{length}(J_i) \end{split} \end{equation} $$

and we're done.

Please note that is in Carothers' Real Analysis before measures or outer measures are introduced, so any solution based on those are undesirable for my purposes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.