9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm not voting to close, but this question is essentially a duplicate of math.stackexchange.com/q/75192 $\endgroup$ – t.b. Oct 30 '11 at 21:06
  • 1
    $\begingroup$ 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, while your function is discontinuous everywhere. $\endgroup$ – t.b. Oct 30 '11 at 21:21
  • 1
    $\begingroup$ 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 proof $\endgroup$ – August Oct 30 '11 at 21:31
  • 2
    $\begingroup$ 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 (if you can't access that page, changing google.com to google.cl may help) $\endgroup$ – t.b. Oct 30 '11 at 21:36
  • 1
    $\begingroup$ Jeeze. how can cambridge set something like this for an analysis II question... $\endgroup$ – Lost1 Jan 8 '14 at 19:12
7
$\begingroup$

The reason (given in comments) that $f$ is not a pointwise limit of continuous functions is that $f$ is discontinuous everywhere, while pointwise limits of continuous functions have a comeager set of points of continuity. The latter fact is proved here, additional details are given here, and a textbook reference is: Theorem 1.19 on page 20 of Real analysis by Bruckner, Bruckner & Thomson.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.