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?

  • 3
    $\begingroup$ "Since R is a metric space" is not a valid argument: closed subset of a metric space need not be compact. // Welcome to Math.SE $\endgroup$
    – user53153
    Jan 9, 2013 at 14:56
  • $\begingroup$ For those that don't have the book can you add theorem 2.36. $\endgroup$ Jan 9, 2013 at 14:57
  • $\begingroup$ Thanks Pavel M!! That's my misunderstanding - I mashed together Theorems 2.34 and 2.35, thinking a closed subset of a metric space is compact. In R^1 the set would have to be both closed and bounded to be compact. $\endgroup$
    – Chad Hawes
    Jan 9, 2013 at 14:59
  • 3
    $\begingroup$ At this point in the text, it hasn't been established that closed, bounded sets in $\Bbb R$ are compact. $\endgroup$ Jan 9, 2013 at 14:59
  • 2
    $\begingroup$ We never used Rudin's books to study in my university (at least back then), and some years after I graduated I needed to check some stuff for a calculus course I taught and I didn't find it comfortable for me...and upon reading this question and Rudin's book I know now why: his definitions (e.g., "interval") are sometimes quite different from almost any other books', theorem 2.36 is a particular case of a rather important characterization of general compact topological spaces, very handy for proving Tychonov's theorem but pretty weird in this particular spot, etc. $\endgroup$
    – DonAntonio
    Jan 9, 2013 at 15:14

1 Answer 1


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.

  • $\begingroup$ The nested interval lemma looks a lot more basic than Heine-Borel. $\endgroup$
    – Don Larynx
    Nov 8, 2013 at 12:30

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