Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is motivated by this one on MathOverflow, which contains an example of a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that $$f_n(x) \rightarrow \infty$$ if and only if $x \in \mathbb{Q}$.

My question is the following :

For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ : $$S((f_n)):=\{ x \in \mathbb{R} : f_n(x) \rightarrow \infty \}.$$

Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?

Note that $$S((f_n)) = \bigcap_{N} \bigcup_{k} \bigcap_{n \geq k} \{x: f_n(x)>N\},$$ so a necessary condition is that $S$ must be a countable intersection of countable unions of $G_{\delta}'s$...

share|cite|improve this question
A small improvement: since $S((f_n))=\bigcap_N \bigcup_k \bigcap_{n\ge k}\{x\colon f_n(x)\ge N\}$, the set $S$ is a countable intersection of countable unions of closed sets, so a countable intersection of $F_\sigma$s. Also known as $F_{\sigma\delta}$, I believe. This might also be sufficient... – user31373 May 17 '12 at 16:09
That's right, thank you for the comment! – Kalim May 17 '12 at 17:01
If $g_n:\mathbb R\to[0,1]$ for each $n$, and $f_n= \frac{1}{1-\frac{n}{n+1}g_n}$, then $(f_n(x))\to\infty$ if and only if $(g_n(x))\to 1$. My thought was that it might be easier to work with functions bounded by $1$ and the set where they converge to $1$. – Jonas Meyer May 18 '12 at 5:10
@Jonas Meyer : Good remark, thank you for the comment. – Kalim May 19 '12 at 1:50
Just to note that this question has now been cross-posted to MathOverflow:… – Willie Wong May 22 '12 at 16:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.