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Let $\{x_1,x_2,x_3,...\}$ be an enumeration of the countable set $\mathbb{Q}\cap [0,1]$.

For $\varepsilon>0$ let

$U_{\varepsilon}=\bigcup_{n=1}^{\infty}\big( x_n-\frac{\varepsilon}{2^{n+1}} ,x_n+\frac{\varepsilon}{2^{n+1}} \big), \quad V_{\varepsilon}=U_{\varepsilon}\cap (0,1).$

Show that $V_{\varepsilon}$ is open and dense in $[0,1]$ and that $0<\lambda(V_{\varepsilon})\leq \varepsilon$.

I don't know how to start. Thanks for any help

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  • $\begingroup$ I have a feel that the set you have described is somehow related to the Cantor set $\endgroup$
    – user210387
    Oct 13, 2015 at 14:25
  • $\begingroup$ The firs two parts come more or less for free. Note first that $U_{\epsilon}$ is a union of open sets, so is open. For dense, given any $w$ in $[0,1]$, there is a sequence $r_1, r_2,\dots$ of rationals in $(0,1)$ with limit $w$. These rationals are in $V_{\epsilon}$. $\endgroup$ Oct 13, 2015 at 14:33
  • $\begingroup$ Thanks André, is that the same as saying that $V_{\varepsilon}$ contains all the rationals that exist in [0,1]? $\endgroup$
    – user279562
    Oct 13, 2015 at 14:38

1 Answer 1

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$V_\varepsilon$ is open, since it is a union of open sets. You can also see that $\Bbb{Q}\cap [0,1]$ is a subset of $V_\varepsilon$ and it implies that $V_\varepsilon$ is dense in $[0,1]$.


I didn't see you asking about the measure of $V_\varepsilon$; From $\sigma$-subadditivity, $$\lambda(V_\varepsilon) \le \sum_{n=1}^\infty\lambda\left(\left(x_n - \frac{\varepsilon}{2^{n+1}},x_n + \frac{\varepsilon}{2^{n+1}}\right)\right) = \sum_{n=1}^\infty\frac{\varepsilon}{2^n}.$$

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    $\begingroup$ To be more precise it's $U_\varepsilon$ which is a union of open sets. $V_\varepsilon$ is a finite intersection of open sets (hence open). $\endgroup$
    – M.G
    Oct 13, 2015 at 14:28
  • $\begingroup$ Thanks. So because $\mathbb{Q}$ is dense in [0,1], and $\mathbb{Q}$ is contained in $V_{\varepsilon}$, then $V_{\varepsilon}$ must be dense in [0,1] as well? $\endgroup$
    – user279562
    Oct 13, 2015 at 14:37
  • $\begingroup$ @user279562 Yes. To be precise, the closure of $V_\varepsilon$ is $[0,1]$ since the closure of the set of rationals in $[0,1]$ is $[0,1]$. $\endgroup$
    – Hanul Jeon
    Oct 13, 2015 at 14:40
  • $\begingroup$ Thanks! Somehow I couldn't believe it would be this simple. I thought I had to somehow use the definition of $\varepsilon$-denseness in order to be rigorous. $\endgroup$
    – user279562
    Oct 13, 2015 at 14:44

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