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
Best Answer
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.