Someone asked me this question. And he said it's an exercise from Rudin's Real and Complex Analysis.

Does there exist a sequence of continuous functions $f_n(x)$, such that $\lim_{n \to \infty} f_n(x)=+\infty$ iff $x \in \mathbb Q$ (or irrationals)?

On the one hand, we know that if the limit of this function exists, then we can't have both as Baire's category theorem applies. But maybe it will happen that the suplim of this sequence is infinity at other points(at which the limit doesn't exists). Because lim equals infinity is the same as inflim equals infinity, so if someone can prove that rationals can't be the (countable)intersections of (countable)unions of G delta set then it's done.

But on the other hand, I suspect that if we let $f_n(x)=\cos(\pi\cdot n!x)^n \cdot n$ is an example.

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  • The set of discontinuities of the pointwise limit of continuous functions must be meagre. Well, $\mathbb R$ is not meagre so I don't think so...but what you have as your pointwise limit isn't exactly a function that's discontinuous everywhere. So I'm not confident enough to offer this as an answer. – A.S Jul 18 '13 at 03:26
  • I'm not if Baire's theorem extends to functions on the extended reals (but if it does, which I would expect, then the answer is no). Perhaps someone else can clarify whether this holds. Related MO post: http://mathoverflow.net/a/32039/20130 – A.S Jul 18 '13 at 03:34
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    I know a sequence of functions that are continuous on the irrationals that gives you what you want. We have $f_n(x) = 0$ for $x$ irrational, and for $x = \frac{p}{q}$ rational with $p, q$ relatively prime integers and $q>0$, then $f_n(x) = \frac{n}{q}$. – Arthur Jul 18 '13 at 03:36
  • @Arthur But this is (if we extend our notion of Baire First Class and Baire Second Class to extended reals) of the second class. The characteristic function for the rationals can also be written as a second class function, but it isn't the pointwise limit of continuous functions. – A.S Jul 18 '13 at 03:40
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    I think the problem here is that we don't ask $f_n(x)$ converge pointwisely, hence we have to ask whether rationals can be intersection of unions of G delta sets. – lee Jul 18 '13 at 13:32
  • @lee: Singletons are $G_\delta$, so the rationals certainly are a countable union of $G_\delta$s. – Nate Eldredge Jul 20 '13 at 18:43
  • $f_n(x) = \cos(\pi\cdot n!x)^n \cdot n$ doesn't work because certain irrational values of $x$ will still diverge to $\infty$. Probably something $x = \sum_n 1/(n!)!$ with extremely good approximations by rationals should work. – Erick Wong Jul 20 '13 at 18:44
  • Just to make sure I'm on the same page, to prove that $f_n(x) = \cos(\pi \cdot n!x)^n \cdot n$ is an example, the only hard step would be to prove that for any irrational $x$, there exists an $\varepsilon > 0$ such that $|\cos(\pi \cdot n!x)| < 1 - \varepsilon$ for infinitely many $n$. This certainly seems true to me. – RghtHndSd Jul 20 '13 at 18:44
  • @AndrewSalmon: Baire's theorem holds in any compact Hausdorff space, which certainly covers the extended reals. The extended reals are actually metrizable, for example using the metric $d(x,y) = |\arctan(x) - \arctan(y)|$. But I don't see how you're going to use Baire's theorem here. – Nate Eldredge Jul 20 '13 at 18:45
  • @rghthndsd It isn't true, though. There are uncountably many $x$ with $\liminf_n \cos(\pi \cdot n!x) = 1$, and they can't all be rational. – Erick Wong Jul 20 '13 at 19:00
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    @Erick: Agreed, and your example $\sum_n 1/(n!)!$ does indeed provide an explicit counter-example. – RghtHndSd Jul 20 '13 at 19:08
  • @Erick Wong , I agree with your example, but I don't know why there are uncountably many x with $liminf_n cos($pi$n!x)=1$, could you explain that to me? – lee Jul 21 '13 at 13:57
  • @lee I had a different argument in mind, but I think any infinite subsequence of $1/(n!)!$ will sum to a similar number. Even if we impose strong regularity conditions on the growth rate of the subsequence, there will be uncountably many choices. – Erick Wong Jul 21 '13 at 16:46
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    A necessary and sufficient condition for a subset $E$ of the reals to be the "diverges to $+\infty$ set" for some sequence of continuous functions is that $E$ is $F_{\sigma \delta}.$ This was proved by Hans Hahn in 1919 and independently by Sierpinski in 1921. See [this math overflow post](http://mathoverflow.net/questions/97654/on-the-set-of-divergence-to-infinity-for-sequences-of-positive-continuous-functi) for a proof, several references, and related issues. [Pings in case others commenting here are interested in this: Andrew Salmon, Erick Wong, rghthndsd, Nate Eldredge, Arthur, Dominik.] – Dave L. Renfro Jul 22 '13 at 21:03
  • @DaveL.Renfro +1 very informative. I should also mention that your thoughtful ping attempts seem to have been ignored by the system, as far as I could tell. – Erick Wong Jul 27 '13 at 16:41
  • Erick Wong: Maybe I need to put "at" in front of each name. I tried that just now and got a message that began "Only one additional @user can be notified ..." – Dave L. Renfro Jul 29 '13 at 20:52

1 Answers1


Since there is a homeomorphism of $\Bbb R $ taking $\Bbb Q $ to the dyadic rationals, that is, rational numbers where the denominator is a power of 2, (see here for an example of such a function), we can restrict out attention to finding a sequence of functions converging to infinity exactly at the dyadics.

Now for $ p, d \in \Bbb R $ we let $ h_{p, d} $ denote a hat function, i. e. a continuous function with $ h_{p, d}(p) =1$ and $ h_{p, d}(x) =0$ for $ x \notin [p-d, p+d]$.

Now define $ f_n = \sum_{k \in \Bbb Z} n h_{k/2^n,2^{-n-2} }$

It is obvious that for dyadic x, $ f_n (x) \to \infty $ as $ n \to \infty $.

If $x$ is not dyadic, there are infinitely many $ n $ with $\{2^n x\} \in [1/4,1/2]$ (here the brackets denote the fractional part of the real number). To see this, consider the binary expansion of $ x $ and note that, since this expansion is not allowed to end in an infinite chain of zeros or in an infinite chain of ones, there are infinitely many integers $ n $ such that the $ n+1$-th digit after the dot is a $0$ and the $ n+2$-th is a $1$.

But with such a choice of $ n $, we get $ f_n (x)=0$ which shows that $ f_n (x )\not \to \infty $ as $ n \to \infty$.

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