The fat Cantor set is a nowhere dense subset of $\mathbb{R}$ with positive Lebesgue measure. My question is, does there exist a $G_\delta$ set dense in the fat Cantor set with Lebesgue measure $0$?

If such a set does exist, is it possible to produce an actual example of it?

  • $\begingroup$ Dense in what? The original fat Cantor set? $\endgroup$ – Ian Nov 11 '18 at 0:12
  • $\begingroup$ @Ian Yeah, dense in the fat Cantor set. I edited to clarify. $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 0:16
  • $\begingroup$ And it is also $G_\delta$ in the fat Cantor set, or is it $G_\delta$ in $\mathbb{R}$? (The two aren't homeomorphic so it matters.) $\endgroup$ – Ian Nov 11 '18 at 0:28
  • $\begingroup$ @Ian Yes, $G_\delta$ in the fat Cantor set. $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 0:42
  • $\begingroup$ @Ian Why does it matter? The fat Cantor set is a closed set, so it';s a $G_\delta$ set, so any relative $G_\delta$ set in the fat Cantor set is a real $G_\delta$ set, right? $\endgroup$ – bof Nov 11 '18 at 1:40

Your fat Cantor set is closed, so it's a $G_\delta$ set. If $C$ is any $G_\delta$ subset of $\mathbb R$, there is a $G_\delta$ subset of $C$ which is dense in $C$ and has measure zero. Namely, let $D$ be a countable dense subset of $C$, and let $A$ be a $G_\delta$ set of measure zero containing $D$. Then $A\cap C$ is a $G_\delta$ set of measure zero and is dense in $C$.

To give an explicit example, you would start by defining an explicit fat Cantor set. Next, you need a countable dense subset $D$; you can do that by taking, for each interval $[a,b]$ with rational endpoints such that $[a,b]\cap C\ne\emptyset$, the least element of $[a,b]\cap C$. Next, you need to specify an enumeration of $D$; that is easily obtained an enumeration of the rational intervals $[a,b]$. So now we have $D=\{d_n:n\in\mathbb N\}$, a countable dense subset of $C$. Finally, define $$A=\bigcap_{k=1}^\infty\bigcup_{n=1}^\infty\left(d_n-2^{-n-k},d_n+2^{-n-k}\right).$$

  • $\begingroup$ Why do $D$ and $A$ exist? $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 2:03
  • $\begingroup$ Every subset of $\mathbb R$ (or any separable metric space) has a countable dense subset. Every countable set has Lebesgue measure zero. Every set of Lebesgue measure zero is contained in a $G_\delta$ set of measure zero. $\endgroup$ – bof Nov 11 '18 at 2:07
  • $\begingroup$ OK thanks for your answer. I used it to post this answer: math.stackexchange.com/a/2993377/71829 $\endgroup$ – Keshav Srinivasan Nov 11 '18 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.