Sorry tried to formulate this question the other day, got myself into a bit of a pickle.

(By dense in [0,1] I mean there is an element in our set arbitrarily close to every point in [0,1]. If that isn't what dense in [0,1] means apologies.)

Some examples I have considered;

  • The rationals (Countable, so no)
  • The real line with the rationals removed (Measure 1 so no)
  • Cantor set doesn't have anything within 1/6 of 0.5, right?

If not, does anyone know of a proof of the result?


1 Answer 1


Sure. Take $[0,\frac{1}{2}]\cup ([\frac{1}{2},1]\cap\mathbb{Q}])$. This set is dense, uncountable and has Lebesgue measure $\frac{1}{2}$.

  • $\begingroup$ Nice and simple. $\endgroup$
    – user70147
    Dec 8, 2013 at 16:26
  • $\begingroup$ Is there one that has an uncountable intersection with all intervals in [0,1]? $\endgroup$
    – user70147
    Dec 8, 2013 at 16:27
  • 1
    $\begingroup$ @user70147: You could do something like this. Enumerate the rationals in $[0,1]$ by $\{q_{k}:k\in\mathbb{N}\}$, and take $\frac{1}{2^{k+2}}$ radii ball around each $q_{k}$. Take the union of these balls intersected with $[0,1]$. It has measure at most $\frac{1}{2}$ and bigger than $0$, it is dense, and what can you say about its intersection with every interval on $[0,1]$? (since every interval contains rationals) $\endgroup$
    – T. Eskin
    Dec 8, 2013 at 16:31
  • $\begingroup$ Sure, just replace the appearance of $\mathbb Q$ in the response with a set that’s uncountable and of measure zero, such as (I think) the set of reals with only finitely many $7$’s in the decimal expansion. $\endgroup$
    – Lubin
    Dec 8, 2013 at 16:33
  • $\begingroup$ I really like the second Thomas E. And thank you also Lubin. $\endgroup$
    – user70147
    Dec 8, 2013 at 16:38

You must log in to answer this question.

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