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Let $\{a_i\}_{i=1}^{\infty}$ be an enumeration of the rationals, $\mathbb{Q}$.

Let $I_{ij}$ be the open interval centered at $a_i$ and having length $1/2^{i+j}$, and define $G_j = \cup_{i=1}^\infty \;I_{ij},\;\forall\;j\in\{1,2,3,\dots\}$.

Finally, let $B = \cap_{j=1}^\infty \; G_j$.

Then it follows from a theorem of Borel, and from the construction above, that $B \supseteq \mathbb{Q}$ is a nullset.

What's in $B\,\backslash \mathbb{Q}\,$?

Is possible to exhibit any elements of $B\,\backslash \mathbb{Q}\,$ ?

Is possible to characterize the elements of $B\,\backslash \mathbb{Q}\,$ any further?

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You want to look at this question: math.stackexchange.com/q/61100/462 –  Andres Caicedo May 1 '13 at 0:29
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A quick summary: You want irrational numbers that are extremely well approximable by rational numbers. So write an infinite, non-repeating decimal (so it'll be irrational) containing, every once in a while, a really huge string of consecutive zeros. You get good rational approximations to this irrational number by truncating its decimal expansion at the beginning of one of these huge strings of zeros. –  Andreas Blass May 1 '13 at 0:45
    
@AndreasBlass: "...write an infinite, non-repeating decimal (so it'll be irrational) containing, every once in a while, a really huge string of consecutive zeros." Are you saying that such an irrational belongs to $B\,\backslash \mathbb{Q}\,$? –  kjo May 1 '13 at 12:21
    
@kjo No. Not all irrational numbers have arbitrarily large strings of consecutive zeroes. In fact, irrational numbers don't need to have any $0$s. Your construction seems related to Liouville numbers, which are also, by the way, a decomposition of the line into a nullset and set of first category. –  Andrew Salmon May 6 '13 at 6:06
    
@AndrewSalmon: in that case the guess is incorrect; the set produced by the construction I asked about is neither the Liouville numbers not its complement. –  kjo May 6 '13 at 13:30

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