We say that a function $f$ is Baire Class $1$ if there is a sequence of functions $f_i \to f$ pointwise where each $f_i$ is continuous.

The set of discontinuities of a Baire Class $1$ function $f$ must be a first-category (meagre) set of points.

The Dirichlet function $f(x) = \begin{cases} 1 & x \in \mathbb Q \\ 0 & x \notin \mathbb Q \end{cases}$ is not Baire class $1$ since it is discontinuous everywhere.

But can't we write the function as $f_i(x) = \lim_{i\to\infty} \cos(i!\pi x)^{2i}$, each of which is a continuous function that converges pointwise to the Dirichlet function? What is wrong with this?

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    $\begingroup$ Can you really prove that convergence for irrational $x$? $\endgroup$ – Thomas Andrews Dec 26 '12 at 4:43
  • $\begingroup$ it does not converge infact $\endgroup$ – Koushik Dec 26 '12 at 4:58
  • $\begingroup$ @Thomas Andrews I guess not. I don't even think it converges at $e$ since $\cos(i!\pi e) = \cos(n\pi + \sum_{n=i+1}^\infty \frac{1}{n!} \pi )$. $\endgroup$ – A.S Dec 26 '12 at 5:07
  • $\begingroup$ Vertainly, something like $e$, anyway. Not sure if $e$ works, but it is definitely possible to do something like it. $\endgroup$ – Thomas Andrews Dec 26 '12 at 5:27

Let $0\le n_1<n_2<...<n_k<\dots$ and define $$x=2\sum_{j=1}^\infty \frac{1}{n_j!}$$

Basically, we will pick $\{n_i\}$ to grow "fast enough" so that $\limsup_i \left(\cos n_i!\pi x\right)^{2n_i}>0$

Now $n_i!x = 2K + 2\sum_{j>i} \frac{n_i!}{n_j!}$ for some integer $K$. And:$$0\leq \sum_{j>i} \frac{n_i!}{n_j!} < \frac{2}{n_i^{n_j-n_i}}(1+\frac{1}{n_j}+\frac{1}{n_j^2}+..) = \frac{2}{n_i^{n_j-n_i}}\frac{n_j}{n_j-1}$$

So $n_i!x\pi = 2K\pi + z$ where $0\leq z < \frac{2\pi}{n_i^{n_j-n_i}}\frac{n_j}{n_j-1}$.

It's gonna be a little messy, but if you choose a reasonable sequence of $n_i$, then you'll be able to show that $$\limsup_{i\to\infty}\left(\cos n_i!\pi x\right)^{2n_i}>0 $$

In particular then, $\left(\cos j!\pi x\right)^{2j}$ cannot converge to zero.


I think there may be some confusion here. The sequence of functions you've proposed don't actually depend on the index (and, as people have pointed out, the limit doesn't converge everywhere.)

Do you instead mean to write $$f_k(x) = \lim_{i \to \infty} \cos(k!\pi x)^{2i}$$

If this is what you mean, then certainly, $\forall x \in \mathbb{R}$, $$f(x) = \lim_{k \to \infty} f_k(x) = \lim_{k \to \infty} \lim_{i \to \infty} \cos(k!\pi x)^{2i} $$

but the individual $f_k(x)$ are not continuous.

Added to elaborate:

What's continuous here are each $\cos(k! \pi x)$. This makes each $f_k(x)$ above of Baire class 1, but not continuous themselves: each $f_k(x)$ is $0$ on all the irrationals and rationals with sufficiently large denominators, and $1$ on the others (where "sufficiently large denominators" depends on $k$; as $k \to \infty$, we 'sweep out' all the rationals.) So what you have here is a sequence $f_k(x)$ of Baire class 1 functions converging pointwise to the Dirichlet function, where each of those $f_k(x)$ are themselves pointwise limits of a sequence of continuous functions.

  • $\begingroup$ I took his error as being that $f_i$ was defined without the limit, and that he was asserting the limit of these $f_i$ was $f$. $\endgroup$ – Thomas Andrews Dec 26 '12 at 6:32
  • $\begingroup$ Ah, ok, that would make more sense. But then, yes, as you pointed out, it doesn't converge to the Dirichlet function. $\endgroup$ – AndrewG Dec 26 '12 at 6:33

The set of discontinuities of a Baire Class 1 function f must be a first-category (meagre) set of points. Hence Dirichlet is not Baire 1
$f(x)=\lim_{i\to\infty} \lim_{j\to\infty} (\cos(i!πx)^{2j})$. This proves Dirichlet function is Baire 2 function.

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    $\begingroup$ First, this is not a proof that the Dirichlet function is Baire 2, since we haven't precluded the possibility that is Baire Class 1. Also, the question was more about why the limit $\cos(i!\pi x)^{2i}$ does not converge pointwise to the Dirichlet function. $\endgroup$ – A.S Dec 26 '12 at 5:10
  • $\begingroup$ You haven't proven anything yet. You've merely made an assertion. $\endgroup$ – Thomas Andrews Dec 26 '12 at 5:18
  • $\begingroup$ proof is immediate after this $\endgroup$ – Koushik Dec 26 '12 at 5:21
  • $\begingroup$ Maybe you misunderstood the question he's asking. (I also suggest you learn to use LaTeX formulas in your answers, because "lim j->infinity ..." becomes tiresome to read.) $\endgroup$ – Thomas Andrews Dec 26 '12 at 5:22
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    $\begingroup$ He knows from the theorem that it shouldn't be possible, but he thought that he had a counter-example. Using the theorem to disprove the counter-example hardly gives him much assurance, since he doubts the theorem. Instead, you have to find the problem in the counter-example. $\endgroup$ – Thomas Andrews Dec 26 '12 at 5:34

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