Wisniewool

Answered

2022-07-03

All examples of a dense and co-dense set I have seen are either of full Lebesgue measure or of measure zero. For instance, in restriction to the unit interval $\mathbb{I}=[0\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1px}{0ex}},\phantom{\rule{thinmathspace}{0ex}}1]$, we could have respectively $\mathbb{I}\cap \mathbb{Q}$ or $\mathbb{I}\setminus \mathbb{Q}$. What I am looking for is a dense and co-dense subset $A\subset \mathbb{I}$ such that

$\mathrm{m}(A)=\mathrm{m}(\mathbb{I}\setminus A)={\textstyle \frac{1}{2}}.$

I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\mathbb{I}$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

$\mathrm{m}(A)=\mathrm{m}(\mathbb{I}\setminus A)={\textstyle \frac{1}{2}}.$

I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\mathbb{I}$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

Answer & Explanation

conveneau71

Expert

2022-07-04Added 17 answers

You can take $A=[0,\frac{1}{2}]\cup ([\frac{1}{2},1]\cap \mathbb{Q})$

Ciara Mcdaniel

Expert

2022-07-05Added 4 answers

Let $C$ be a fat Cantor set with measure 1/2. Set $A=C\cup \mathbb{Q}\cap [0,1]$ and you're done.

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