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Aug 3, 2010 at 16:27 history edited Roland Bacher CC BY-SA 2.5
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Aug 3, 2010 at 15:34 comment added Roland Bacher That is disturbing!
Aug 3, 2010 at 15:09 comment added Malik Younsi I don't see how that would be possible... The paper cited below proves the following result : There is a sequence of continuous functions on $[0,1]$ whose pointwise limit if finite on $S$ and infinite on the complement of $S$ if and only if $S$ is a $F_{\sigma}$. But $\mathbb{R} \setminus \mathbb{Q}$ is not a $F_{\sigma}$, so the sequence of functions you're trying to construct cannot exist...
Aug 3, 2010 at 14:31 comment added Roland Bacher True. One has to choose the sequence $s_n$ suitably in order to avoid this.
Aug 3, 2010 at 14:20 comment added Malik Younsi I don't understand. Isn't it possible that $s_n(d(x,\mathbb Q_n))$ tends to infinity for some irrational $x$?
Aug 3, 2010 at 13:12 history edited Roland Bacher CC BY-SA 2.5
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Aug 3, 2010 at 13:03 history answered Roland Bacher CC BY-SA 2.5