Prove that a nonempty, bounded closed set $S$ in $R^1$ can be obtained from a closed interval by removing a countable disjoint collection of open intervals whose endpoints belong to $S$

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Here is the exercise, where the Representation Theorem for Open Sets on The Real Line is used.

Here i have two questions:

1)Why $R^1 - S = [\inf S,\sup S] - S$ or $R^1 = [\inf S,\sup S]$?

2)I understand proof of the Representation Theorem for Open Sets but it's a bit hard to visualize it. Can you provided some picture of this theorem, if it is possible?

  • $\begingroup$ What is $R^1$? I need more context. $\endgroup$
    – Masacroso
    Nov 24, 2016 at 9:50
  • 1
    $\begingroup$ @Masacroso $R^1$ is real line(the first Euclidean space) $\endgroup$ Nov 24, 2016 at 9:53
  • $\begingroup$ If $R^1=\Bbb R$ with the euclidean topology, then the statement $$R^1 - S = [\inf S,\sup S] - S$$ is senseless. $\endgroup$
    – Masacroso
    Nov 24, 2016 at 9:57

1 Answer 1


I think it means here for the part $(R^1-S)\cap (\inf S,\sup S)=[\inf S,\sup S]-S$ which is open(as both $(R^1-S)$ and $(\inf S,\sup S)$ here are open) and hence equal to the union of some countable disjoint collection of open sets by representation theorem.

Consider the example $S=[1,2] \cup [3,4] \cup [5,6]$ if you want some intuition, now $$(R^1-S)\cap (\inf S,\sup S)=(2,3) \cup (4,5)=[\inf S,\sup S]-S\\\implies S=[1,6]-(2,3) \cup (4,5)$$


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