3
$\begingroup$

Talking about an example of an open dense subset of $[0,1]$ with measure $1/2$ I heard talking about 'fat rationals' but I don't find anything on the internt.. Does someone know about this?

$\endgroup$
2
  • 1
    $\begingroup$ Hmmm... Never heard "fat rationals", though I've heard "fat Cantor set". You can take a fat Cantor set of measure $1/2$; it's complement will be open. The fat Cantor set has empty interior, so its complement is dense. $\endgroup$ Commented Apr 17, 2012 at 20:12
  • 3
    $\begingroup$ 'fat rationals' probably means a set of the form $$A = \bigcup_{n\in \mathbb N} B_{\epsilon_n}(q_n)$$ where $\{q_n\}_{n\in \mathbb N}$ is an enumeration of the rationals and the $\epsilon_n>0$ are chosen in such a way that the measure of $A$ is $1/2$. $\endgroup$
    – Sam
    Commented Apr 17, 2012 at 20:21

2 Answers 2

4
$\begingroup$

For a fixed $\varepsilon$, consider the set $S_{\varepsilon}:=\bigcup_{n\in\mathbb N}(q_n-2^{-n}\varepsilon,q_n+\varepsilon 2^{-n})$, where $\{q_n\}$ is an enumeration of the rationals. Then $S_{\varepsilon}$ is open and dense in $[0,1]$, since it contains all the rationals of this interval. The maps $f\colon\varepsilon\mapsto \lambda(S_{\varepsilon})$ is Lipschitz-continuous. Indeed, if $\varepsilon_1\leq\varepsilon_2$, we have \begin{align*}f(\varepsilon_2)-f(\varepsilon_1)&=\lambda(S_{\varepsilon_2}\setminus S_{\varepsilon_1})\\\ &\leq \lambda\left(\bigcup_{n=0}^{+\infty}(q_n-2^{-n}\varepsilon_2,q_n+\varepsilon_2 2^{-n})\setminus (q_n-2^{-n}\varepsilon_1,q_n+\varepsilon_1 2^{-n})\right)\\\ &\leq \sum_{n=0}^{+\infty}\lambda((q_n-2^{-n}\varepsilon_2,q_n+\varepsilon_2 2^{-n})\setminus (q_n-2^{-n}\varepsilon_1,q_n+\varepsilon_1 2^{-n}))\\\ &=2(\varepsilon_2-\varepsilon_1)\sum_{n=0}^{+\infty}2^{-n} \end{align*}

$\endgroup$
1
  • 1
    $\begingroup$ The continuity consideration of $\varepsilon \mapsto \lambda(S_\varepsilon)$ implies of course that by suitably choosing $\varepsilon$ you can obtain any desired real value $\gt 0$ as measure of $\lambda(S_\varepsilon)$. $\endgroup$
    – t.b.
    Commented Apr 17, 2012 at 21:09
2
$\begingroup$

As an alternative to Davide Giraudo's proposal which avoids having to order the rationals, try

$$S_k = [0,1] \cap \bigcup_{a\in\mathbb Z, b\in \mathbb Z^+}\left(\frac{a}{b}-\frac{k}{b 2^{b}},\frac{a}{b}+\frac{k}{b 2^{b}}\right).$$

For $k\approx 0.32184058\ldots$ this will have measure $\frac12$.

This has the same continuity considerations as Davide's and I suspect that as a function of $k$ the measure of $S_k$ has a zero derivative for all rational $k$, though I have not checked.

$\endgroup$
3
  • $\begingroup$ How did you think of this one?! And what about the value of k? $\endgroup$
    – balestrav
    Commented Apr 18, 2012 at 23:08
  • $\begingroup$ @balestrav: It is similar to something I did earlier with dyadic fractions. Working out the measure is simply a question of looking at some overlapping intervals (I took $b$ up to $50$), and then iterating until the desired measure target is achieved. It is nice that the intersection of any interval $[x,x+1]$ with the union produces the same measure. $\endgroup$
    – Henry
    Commented Apr 18, 2012 at 23:53
  • $\begingroup$ That page has now moved to se16.info/hgb/nowhere.htm $\endgroup$
    – Henry
    Commented Mar 2, 2016 at 19:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .