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I'm trying to solve this question:

Show that there isn't a sequence of continuous functions $f_n:[0,1]\to \mathbb R$ converges pointwise to the function $f:[0,1]\to \mathbb R$ such that $f(x)=0$ for $x$ rational and $f(x)=1$, for $x$ irrational.

Of course there isn't a sequence $f_n:[0,1]\to \mathbb R$ converges uniformly to the function $f:[0,1]\to \mathbb R$ because $f$ is discontinuous, but when the convergence is pointwise?

I need help here

Thanks in advance

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2 Answers 2

up vote 5 down vote accepted

See my paper, First-class functions, American Mathematical Monthly, Volume 98, Issue 3, March 1991, Pages 237-240 .

See also the variant on this question at MathOverflow where the functions are supposed to go to infinity at the rationals, and then take reciprocals.

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I would like something easier –  user42912 Feb 20 '13 at 6:53
    
Wouldn't we all. Do you have some reason to believe there is something easier? –  Gerry Myerson Feb 20 '13 at 11:50
    
yes, this question is one of the list to prepare to the exam of real analysis in $\mathbb R$, by the way it's the only question I couldn't solve. –  user42912 Feb 20 '13 at 13:08

Suppose that this is the case. Then $$ f^{-1}(1) = \cup_{l=1}^\infty(\cup_{m=1}^\infty(\cap_{n=m}^\infty f_n^{-1}([1-\frac{1}{2^l}, 1+\frac{1}{2^l}]))). $$ Note that this is a countable union of closed sets, since each $f_n$ is continuous (i.e. an $F_\sigma$ set).

On the other hand, we know that $f_n^{-1}(1) = \mathbb{Q}^c\cap[0,1]$, and so cannot be an $F_\sigma$ set by the Baire Category Theorem.

Edit: After thinking about it a little bit more, I realized that easiest thing to say here is that for a sequence of continuous functions, the set of discontinuities will be meagre, which is clearly not the case here. :)

See: http://en.wikipedia.org/wiki/Baire_space#Properties

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I would like to see your solution anyway –  user42912 Feb 20 '13 at 7:03
    
I tried to fix it, hope this works! –  sourmountain Feb 20 '13 at 7:20
    
I don't know what a maigre set is, thank you for your your answer –  user42912 Feb 20 '13 at 8:43
    
I think the union over $l$ in your display should be an intersection, and this breaks your argument. –  Nate Eldredge Jul 20 '13 at 18:50

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