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Let $F$ a closed subset of the real line without isolated points.

Can we deduce that $F$ is a union of closed intervals of $\mathbb{R}$?

EDIT : As @Arthur said, of course closed intervals with positive length. Thank you for your comment, it is not really easy otherwise ^^.

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    Does a single point count as a closed interval (like ${0} = [0,0]$)? Because if so, then any subset of the real line is a union of closed intervals, and this problem becomes somewhat boring. So if I were to guess, I would say that you want to ask about unions of closed intervals with positive length. – Arthur Nov 18 '19 at 13:50
  • Either "Yes, because of what @Arthur said" or "No because Cantor set". –  Nov 18 '19 at 13:54
  • @JoséCarlosSantos Does it matter? Non-disjoint intervals can be merged into single intervals, and you can reduce a non-disjoint union to a disjoint one, can you not? Are there infinity-related subtleties that I'm missing? Or are you just saying "disjoint" to get a unique representation? – Arthur Nov 18 '19 at 13:56

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Define "closed interval". Typically the definition goes like this: for $a,b\in\mathbb{R}$

$$[a,b]:=\{x\in\mathbb{R}\ |\ a\leq x\leq b\}$$

and therefore singletons $\{c\}$ are closed intervals as well, i.e. $[c,c]=\{c\}$. Thus any subset of $\mathbb{R}$ is a union of closed intervals. Even disjoint union.

However if by "closed interval" you mean "closed interval of non-zero length" (equivalently "closed intervals which are not singletons") then the answer is "no" because the Cantor set has no isolated points. And the Cantor set does not contain any open interval (and thus closed of non-zero length as well) because it is totally disconnected.

freakish
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