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Prove that the Cantor set cannot be expressed as the union of a countable collection of closed intervals whereas it's complement can be expressed as the union of a countable collection of open intervals!

I have had a chance to look at answer for this somewhat similar question, I guess: Is $[0,1]$ a countable disjoint union of closed sets?

However, the answer uses concepts of compactness and in the book which I am reading ( Bartle's Elements of Real Analysis) , the question has been posed before introducing those concepts.

I seem to have very little idea on how to proceed further. I know that the cantor set is uncountable. But, I don't seem to have an intuition on how to relate any concept to prove/disprove the existence of any such open intervals.

Until now, in the chapter, I am well versed with connectedness, open and closed sets and other elementary topics but haven't started Heine-Borel or Baire's Theorems.

EDIT : One direction in which I thought of was that since the rational numbers are countable in $\mathbb R$, they will be countable in the cantor set as well. Hence, if we isolate all the rational numbers in the cantor set and have a one-one correspondence with the intervals covering complement of the cantor set, we may achieve something. But, no progress after that.

Hence, It will greatly appreciated if you can please give me a guide on how to move forward.

Thank you for your help

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    $\begingroup$ Are you asking about the unit interval or the cantor set? There are no closed intervals contained in the Cantor set except the singletons. So ot just boils down to saying that the cantor set is uncountable. $\endgroup$ – Rene Schipperus Sep 3 '14 at 19:38
  • $\begingroup$ I am asking for the Cantor Set.. So, we should prove that there are no closed intervals contained in the cantor set except the singletons to prove one part of the question right? $\endgroup$ – MathMan Sep 3 '14 at 19:41
  • $\begingroup$ An interval has an interior, it contains the open interval, this cannot be in the cantor set. $\endgroup$ – Rene Schipperus Sep 3 '14 at 19:42
  • $\begingroup$ But by removing the middle $1/3 rd$ from $[0,1]$ we are obtaining closed intervals at each step.? $\endgroup$ – MathMan Sep 3 '14 at 19:44
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By the Cantor set, I will assume you mean the usual $\frac{1}{3}$ Cantor set inside of $[0,1]$.

First observe that any closed interval $[a,b]$ which is not a singleton (i.e. $a = b$) contains an open set, in particular the open interval $(a,b)$.

It follows from the construction of the Cantor set that the Cantor set has no interior (contains no nonempty open subsets). Therefore, if the cantor set contains a closed interval it must be a singleton. Since the Cantor set is uncountable, one can not write the cantor set as countable union of singletons. This shows that the cantor set is not the union of countably many closed intervals.

However, the Cantor set is a closed set. Hence the complement of the Cantor set is an open set. Open subsets of $[0,1]$ are countable union of open intervals. This proves your second statement.

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  • $\begingroup$ Thank you. I get this :-) . Just one query though please. The reason, the cantor set has no interior is because the total length of $F_n$ is $(2/3)^n$. Hence, as $n$ tends to $\infty$ the length of the cantor set tends to $0$. Hence, the cantor set must consist of only singletons. Am I correct? $\endgroup$ – MathMan Sep 3 '14 at 19:55
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    $\begingroup$ @VHP Unless you know measure theory and can actually formalize what "length" means for sets like the Cantor set, I would rather use the following argument: for any $a < b$, there exists a $n$ sufficiently large such that a removed $\frac{1}{3^n}$ interval lies inside of $(a,b)$. Hence $(a,b)$ is not a subset of $F_n$. So $(a,b)$ is not a subset of $C$. $\endgroup$ – William Sep 3 '14 at 20:05
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    $\begingroup$ @DavidMitra There is a difference. I am suggesting to prove $C$ has no interior by showing every open interval is not a subset of some $F_n$ (disjoint union of intervals). The OP line of argument (from first comment) is to say $(a,b)$ is not a subset of $C$ since $C$ has "length" $0$. I am just cautioning the OP in the case that he or she does not understand what "length" for the Cantor set should mean, but it is true these two proofs are essentially the same. $\endgroup$ – William Sep 3 '14 at 20:23
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    $\begingroup$ Yes, that's good advice. I didn't read the OP's first comment carefully... I meant to point out though, one does not have to regard the measure of the Cantor set, just the sum of the lengths of the sets comprising an $F_n$. $\endgroup$ – David Mitra Sep 3 '14 at 20:28
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    $\begingroup$ @majmun yes everything you said is correct. If the $F$ notation for those sets is common, it would only be because they are (unions of) closed sets (countable unions of which are called $F_\sigma$ sets). $\endgroup$ – mathematrucker Jul 13 '17 at 17:16

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