So I'm reviewing baby Rudin in studying for an Intro Exam for PhD program. I'm in Chapter 2, specifically theorems 2.38 – 2.40. Having independently done the proofs prior to these theorems, I am wondering why a simpler approach is not warranted.
Consider Theorem 2.38: If $\{I_n\}$ is a sequence of intervals in $R^1$, such that $I_n$ contains $I_{n+1}$ ($n = 1, 2, 3, \dots$), then infinite intersection of the sets $I_n$ is not empty.
By Rudin's definition of interval in definition 2.17, the interval is a closed set in $R^1$. Since $R^1$ is a metric space, the intervals $I_n$ are thus compact sets. So isn't Theorem 2.38 just the corollary to Theorem 2.36 directly above on the page? Why go through the proof Rudin uses to prove Theorem 2.38.
I must be missing something basic?
Best Answer
Just to have an answer here: yes, something was amiss. It is very straightforward to prove that a nested sequence of nonempty compact sets has a common point. One could hope to get the nested interval lemma from here, but ... first it must be proved that a closed bounded interval is compact, which is at least as hard as the nested intervals lemma itself. In fact Rudin uses the nested interval lemma in the proof that a closed bounded interval is compact.