# Subset of Cantor set which is never the discontinuity set of a real function

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.

• 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? – Mees de Vries Mar 29 '18 at 12:32
• @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. – Prabhat Mar 29 '18 at 13:07
• That's correct, but do you understand how you get to $\mathrm{card}(2^{\mathbb N})$ from $\mathrm{card}(\mathbb{R}^{\mathbb N})$? – Mees de Vries Mar 29 '18 at 13:09
• 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 $[0,1]^\omega$. – Prabhat Mar 29 '18 at 13:28

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 $F_\sigma$.