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Thread: show a set has measure 0

  1. #1
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    show a set has measure 0

    Let $\displaystyle \Omega$ denote the set of numbers $\displaystyle x \in [0,1]$ such that there exist infinitely many
    rationals $\displaystyle \frac{p}{q}$, with p and q coprime, with the property
    $\displaystyle |x-\frac{p}{q}|<\frac{1}{q^3}$. Show that $\displaystyle \Omega$ has measure 0.

    I think I should use the Borel Cantelli Lemma for this, however I am struggling to think of a suitable set An so that limsupAn= $\displaystyle \Omega$. Can you help me try to find a suitable set?
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  2. #2
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    Quote Originally Posted by ramdayal9 View Post
    Let $\displaystyle \Omega$ denote the set of numbers $\displaystyle x \in [0,1]$ such that there exist infinitely many
    rationals $\displaystyle \frac{p}{q}$, with p and q coprime, with the property
    $\displaystyle |x-\frac{p}{q}|<\frac{1}{q^3}$. Show that $\displaystyle \Omega$ has measure 0.

    I think I should use the Borel Cantelli Lemma for this, however I am struggling to think of a suitable set An so that limsupAn= $\displaystyle \Omega$. Can you help me try to find a suitable set?
    It seems to work if you let $\displaystyle A_q=\{x\in[0,1] : \left|x-\frac pq\right|<\frac{1}{q^3}$ for some $\displaystyle 0\leq p<q\}$. This excludes the rationals $\displaystyle \frac{p}{q}\notin[0,1]$ but there are only finitely many that fit if $\displaystyle x\in(0,1)$, I think, so this doesn't matter (to be checked). Alternatively, you can allow $\displaystyle 0\leq p<2q$ in the definition to settle this problem more easily (if $\displaystyle |x-p/q|<1/q^3$ then $\displaystyle p/q\leq |x|+1/q^3\leq 1+1=2$)
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