Prove that a nonempty bounded closed set can be obtained by removing countable disjoint collection of open intervals from closed interval in R1

Question

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?

Answer 1

• 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

Answer 2

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