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Prove that there does not exist a sequence of continuous functions $ f_n :\left[ {0,1} \right] \to R $ such that converges pointwise, to the function $$f(x)= \begin{cases} 0 & \text{if $x$ is rational},\\\\ 1 & \text{otherwise}. \end{cases}. $$

I have no idea How can I prove this. Prove that there no exist such sequence if the convergence is uniform, it's easy, because the limit would be continuous, but here I don't know How can I do. I suppose that some "nice" properties are "preserved" in the limit, in this kind of convergence, but I don't know any of them.

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    I'm not voting to close, but this question is essentially a duplicate of http://math.stackexchange.com/q/75192/2011-10-30
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    Ok, closed it , if you want, I did not see that post2011-10-30
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    Ok, closed it , if you want, I did not see that post, I think that you are exagerating, but I don´t know what it is measurable functions , and this kind of things :S2011-10-30
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    I said that I **didn't** vote to close and I meant it. Still, the point is that the set of discontinuity of a pointwise limit of continuous functions is a set of [first category](http://en.wikipedia.org/wiki/Meagre_set), while your function is discontinuous everywhere.2011-10-30
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    Wow! That´s nice!!! where can I find a proof of this fact? Or at least the name of the theorem, to find it´s proof2011-10-30
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    It's a theorem of Baire. In his answer in the thread mentioned above LostInMath points to Theorem 1.19 on page 20 of Bruckner, Bruckner & Thomson *[Real analysis](http://books.google.com/books?id=1WY6u0C_jEsC&pg=PA20)* (if you can't access that page, changing google.com to google.cl may help)2011-10-30
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    Jeeze. how can cambridge set something like this for an analysis II question...2014-01-08
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    See also: http://math.stackexchange.com/questions/541619/proof-that-a-sequence-of-continuous-functions-f-n-cannot-converge-pointwise2015-12-30

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