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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.

Martin Sleziak
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August
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  • I'm not voting to close, but this question is essentially a duplicate of http://math.stackexchange.com/q/75192/ – t.b. Oct 30 '11 at 21:06
  • Ok, closed it , if you want, I did not see that post – August Oct 30 '11 at 21:10
  • 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 :S – August Oct 30 '11 at 21:17
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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. – t.b. Oct 30 '11 at 21:21
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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 proof – August Oct 30 '11 at 21:31
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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) – t.b. Oct 30 '11 at 21:36
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    Jeeze. how can cambridge set something like this for an analysis II question... – Lost1 Jan 08 '14 at 19:12
  • See also: http://math.stackexchange.com/questions/541619/proof-that-a-sequence-of-continuous-functions-f-n-cannot-converge-pointwise – Martin Sleziak Dec 30 '15 at 09:21
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    Does this answer your question? [Pointwise limit of continuous functions is 1) measurable and 2) pointwise discontinuous](https://math.stackexchange.com/questions/75192/pointwise-limit-of-continuous-functions-is-1-measurable-and-2-pointwise-discon) – Another User May 17 '23 at 07:09

2 Answers2

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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.

Community
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ˈjuː.zɚ79365
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Here is a simple solution to this old posting.

Suppose $(f_n:n\in\mathbb{N})\subset \mathcal{C}(\mathbb{R})$ is such that $f_n\xrightarrow{n\rightarrow\infty}\mathbb{1}_{\mathbb{Q}}$ pointwise. Define the sets $$E_n:=\bigcap_{m\geq n}\{x\in\mathbb{R}:|f_m(x)|\leq\frac12\}$$ Each $E_n$, being the intersection of closed sets, is closed.

If $x\in E_n$, then $|f_m(x)|\leq\frac12$ for all $m\geq n$ and so, $\mathbb{1}_{\mathbb{Q}}(x)=\lim_nf_n(x)\leq\frac12<1$, each means that $x$ is irrational. Hence $E_n$ is a nowhere dense closed set.

Conversely, $\mathbb{R}\setminus\mathbb{Q}\subset \bigcup_nE_n$ since for each irrational $x$, $f_n(x)\xrightarrow{n\rightarrow\infty}0$. Hence $$\mathbb{R}\setminus\mathbb{Q}=\bigcup_nE_n$$ and so, $\mathbb{R}\setminus\mathbb{Q}$ is a set of first category. This implies that $\mathbb{R}=\big(\mathbb{R}\setminus\mathbb{Q}\big)\cup\mathbb{Q}$ is of first category. We know from Baire's theorem that $\mathbb{R}$ (with the usual topology) is of second category: contradiction!

Oliver Díaz
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