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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.

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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.

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  • $\begingroup$ 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

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