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Find a subset of the middle-thirds Cantor set which is never the discontinuity set of a function $f:\mathbb{R} \rightarrow \mathbb{R}$. Infer that some zero sets are never discontinuity sets of Riemann integrable functions. [Hint: How many subsets of C are there? How many can be countable unions of closed sets?]

This was one of the problems in Pugh's Real Mathematical Analysis. I am even unable to have any concrete idea to solve this problem. Looking at the hint I can only assume that the number of subsets of Cantor set which are countable unions of closed sets might be equal to card(C). But I have no idea how to prove (or disprove) it.

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  • $\begingroup$ Can you prove that there are $\mathrm{card}(C) = \mathrm{card}(\mathbb R) = \mathfrak c$-many closed subsets of $C$? Can you prove that the set of countable subsets of $C$ (or $\mathbb R$) still has $\mathfrak c$ elements? $\endgroup$ – Mees de Vries Mar 29 '18 at 12:32
  • $\begingroup$ @MeesdeVries Since a countably infinite set can be written in bijection with $\mathbb{N}$, I think that implies the set of countable subsets of C (or $\mathbb{R}$) has card($2^{\mathbb{N}}) = card(\mathbb{R}$) elements. But I am not sure how to prove the first statement. Perhaps I should give it a bit more thought. $\endgroup$ – Prabhat Mar 29 '18 at 13:07
  • $\begingroup$ That's correct, but do you understand how you get to $\mathrm{card}(2^{\mathbb N})$ from $\mathrm{card}(\mathbb{R}^{\mathbb N})$? $\endgroup$ – Mees de Vries Mar 29 '18 at 13:09
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    $\begingroup$ I think if we split an infinite subset $\mathbb{N}$ into countable union of disjoint countable sets (like $A_p = \{p^n\}$ for each primes), write [0,1] in its decimal expansion, take $B_p$ as the set of real numbers in [0,1] whose decimal expansion is all zero except if the position of the decimal is in $A_p$, then we can construct a bijection between a subset of [0,1] and $[0,1]^\omega$. $\endgroup$ – Prabhat Mar 29 '18 at 13:28
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The complement of a closed set (in the real line) is open. An open set is the union of a countable sequence of open intervals with end points at rational numbers. Therefore, there is an injection from closed sets to sequences of rational numbers. The set of sequences of rational numbers has the same cardinality as the real numbers, and this can be injected into the closed sets just by sending each real to the singleton that contains it.

Therefore, the set of countable unions of closed ($F_\sigma$) sets also has the same cardinality as the reals.

The Cantor set has the same cardinality of the reals. Therefore the number of its subsets has a cardinality strictly larger than that of the $F_{\sigma}$ subsets and in particular more than those included in the Cantor set.

Therefore, there is a subset of the Cantor set that is not an $F_\sigma$.

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