The question was taken from Apostol's Mathematical Analysis, question 3.7.

I have taken a look at the solution for this from this website: https://www.csie.ntu.edu.tw/~b89089/book/Apostol/ch3-all.pdf

Also posted here:

3.7 Proof Snippet

However, the notation makes absolutely no sense, and the argument is not at all satisfying to me. That, or perhaps I just don't understand it. For example, I do not accept the argument that $[inf S, sup S] - S$, for a closed and bounded set, is anything but $\phi$ (This explanation helped a little bit: Representation Theorem for Open Sets on The Real Line (Proof-explanation).).

I would like to prove this using only the machinery developed in prior chapters in the book, but I have absolutely no idea where to start. I will also post my solution when it is done. Thanks guys!

Edit: I have been spending time trying to work from Munkres's Topology in hopes of understanding the topic from a different perspective--do you guys think that this is the right way to go about solving this problem?

  • $\begingroup$ The complement is an open set, which can always be written as a countable union of disjoint open intervals. Two of these must be unbounded. You’re done. $\endgroup$
    – MPW
    Jan 23 '19 at 22:37
  • $\begingroup$ @MPW I'm not sure I follow. The complement can always be written as a countable union of disjoint open intervals, and two of them must be unbounded because the complement is C = R1-A, which has infC = $-\inf$ and supC = $+\inf$. I agree with you up to here. How this proves the statement however, I'm not quite making the jump. $\endgroup$ Jan 25 '19 at 18:24
  • $\begingroup$ @RohanCasukhela Supppose $S=\{1,2\}$. What is $[\inf \, S, \sup \, S]\setminus S$? Surely not empty right? $\endgroup$ Jan 30 '19 at 6:10
  • $\begingroup$ @KaviRamaMurthy sure, that works for $S$ in that sense, but is $S$ closed and bounded? Bounded, I believe so, but it doesn't contain any of its accumulation points, hence it is not closed (indeed, I do not believe that the set is open either!). Therefore it doesn't work for this example. $\endgroup$ Jan 30 '19 at 15:01
  • S is closed and bounded
  • $R^1$ - S is open
  • ($R^1$ - S) intersection (inf S, sup S) is open
  • ($R^1$ - S) intersection (inf S, sup S) = (inf S, sup S) - S
  • (inf S, sup S) - S = [inf S, sup S] - S
  • [inf S, sup S] - S is open and is therefore a countable union of open intervals
  • S is therefore [inf S, sup S] - countable union of open intervals

I think maybe I can help by explicitly giving you an example where A=[infS,supS]-S is not empty. Take S = {0,1,2}, then A={1} which, in particular is not empty.

Hopefully this helps a bit; I can try to help out more if you'd like.

All the best, John

  • $\begingroup$ I'm sorry, I still don't quite understand--the notation [infS, supS] stands for the closed interval in R1 where the lower bound is infS and the upper bound is supS, right? That would mean for your example, the interval is [0, 2], and A would be the set of elements from [0, 2] whose elements don't belong to the set {0,1,2}. So A = (0,1) U (1, 2) instead. $\endgroup$ Jan 24 '19 at 17:08

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