The set of convergent points of a sequence of continuous functions is Borel 
Suppose that $f_n$ are continuous.  Prove that $E=\{x\in \mathbb{R}:f_n(x)\text{ is convergent}\}$ is Borel.

My first instinct is to take balls around $\lim_{n\to\infty}f_n(x)$, for each $x$ in $E$, and then show that there has to be some ball small enough so that everything inside that ball is a member of $E$. But this would imply not only that $E$ is Borel, but moreover that it is open, so I don't think that can be right.  Can someone show me how to do this?
 A: Noting that convergence in $\mathbb{R}$ is equivalent to the Cauchy condition, your set $E$ can be written thus $$E =\bigcap_{\varepsilon\in\mathbb{Q}_{>0}} \bigcup_{N\in\mathbb{N}}\bigcap_{n,m>N}\{x: |f_n(x)-f_m(x)|<\varepsilon\}$$
using the usual definition of Cauchy sequence but written with set notation (for all $\varepsilon >0$ yada yada...). But this is just countable set operations on open sets (since $f_n$ are all continuous) hence Borel measurable. 
A: Certainly your argument won't work - consider $f_n(x)=nx.$ Now $E=\{0\}$.
This is one question where I really think the logical approach is useful - think about how $E$ is defined, and try to turn that definition into a construction of $E$ via countable unions and intersections.
Specifically, $x\in E$
$\iff$ the $f_n$s converge at $x$
$\iff \lim_{n\rightarrow\infty}f_n(x)$ exists
$\iff \exists r$ such that $\forall \epsilon$, $\exists m$ such that $\forall k>m$, we have: $\vert f_k(x)-r\vert<\epsilon$
(Now to get rid of that "$\exists r$," which quantifies over uncountably many things . . .)
$\iff \forall \delta\in\mathbb{Q}_{<0}\exists q, q'\in\mathbb{Q}$ such that $\vert q-q'\vert<\delta$ and $\exists m$ such that $\forall k>m$, we have: $f_k(x)\in (q, q')$.
(That is, we eventually land in smaller-and-smaller rational intervals.)
Now think of quantification over countable sets as like Boolean operations: "$\forall$" as like countable intersections, and "$\exists$" as like countable unions. For example, the set $\mathbb{Q}=\{x: x\text{ is rational}\}=\{x: \exists a, b(x={a\over b})\}$; the set of $x$ such that $x={a\over b}$ for $a, b$ fixed is closed, and so $\mathbb{Q}$ is a countable union of closed sets, or $F_\sigma$ or $\Sigma^0_2$.

There's a subtlety here - obviously we need to use the fact that the $f_n$s are continuous, somewhere. But where?
