Let $\{f_n\}$ be a sequence of continuous functions defined on $\mathbb{R}$. Show that the set of points $x$ at which the sequence $\{f_n(x)\}$ converges to a real number is the intersection of a countable collection of $F_\sigma$ sets.

Continuity of $f_n$ means that for any $x\in\mathbb{R}$ and $\epsilon>0$, there exists $\delta$ such that $|y-x|<\delta$ implies $|f_n(y)-f_n(x)|<\epsilon$.

The sequence $\{f_n(x)\}$ converging to a real number $y$ means that for any $\epsilon>0$, there exists $N$ such that for all $n>N$, $|f_n(x)-y|<\epsilon$.

Intersection of a countable collection of $F_\sigma$ sets... and each $F_\sigma$ set is a countable union of closed set... that seems complicated.


The recipe for these kind of problems is to write down the formula for a point $x$ to be in this set, using countable sets (like $\frac{1}{n}, n \in \mathbb{N}$, instead of arbitrary $\epsilon$ and $\delta$). Also use that $f_n(x)$ converges iff it is a Cauchy sequence in $\mathbb{R}$, as $\mathbb{R}$ is complete.

So $x$ is in this set iff for all $n$ in $\mathbb{N}$ there exists $m$ in $\mathbb{N}$ such that $k,l \ge m$ implies that $|f_k(x) - f_l(x)| \le \frac{1}{n}$.

Now define $A_{k,l,n} = \left\{x: | f_k(x) - f_l(x) | \le \frac{1}{n} \right\}$ which is closed (as the inverse image of the continuous function $|f_k - f_l|$ of the set $[0,\frac{1}{n}]$, eg.).

So the set of convergence points of $(f_n)$ equals $\cap_n \cup_m \cap_{k,l \ge m} A_{k,l,n}$

where the last set is closed, as an intersection of closed sets, and so this set is a countable intersection of a countable union of closed sets.

  • $\begingroup$ Thank you for your very clear explanation. This kind of problems was unfamiliar to me, but I think I'm getting the idea. $\endgroup$ – PJ Miller Jun 29 '13 at 21:13
  • $\begingroup$ I am in the middle of solving this problem. i know that the question was asking long ago. I was wondering what exactly do i need to show in order to solve the problem that is my main question in other words the goal $\endgroup$ – user146269 Oct 1 '15 at 18:32
  • $\begingroup$ @user146269 I have given a full solution, essentially. So what are you missing? $\endgroup$ – Henno Brandsma Oct 1 '15 at 20:03

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.