1. ## Real Analysis Help

Let X be a topological space and let ${f_n}$ be a sequence of continuous functions from X to the real numbers. If f(x)= $\lim_{n->\infty} f_n(x)$ for all x in X, show that the set of points where f is continuous is a $G_\delta$ set.

2. ## Hint

Hint: Let $V_n(\epsilon)={x \in X : |f_n(x)-f(x)|<\epsilon}$ and define $O(\epsilon)=\bigcup_{n=1}^\infty Int(V_n(\epsilon))$. Then f is continuous on the $G_\delta$ given by $\bigcap_{k=1}^\infty O(1/k)$

3. Let $U_n = \{x\in X : \exists\text{ op{}en }V_x\ni x \text{ such that }v,w\in V_x\,\Rightarrow\,|f(v)-f(w)|<1/n\}.$

If $x\in U_n$ then $V_x\subseteq U_n$, so $U_n$ is open. Also, f is continuous at x if and only if $x\in\bigcap_nU_n$. Thus the set of points of continuity of f is a $G_\delta$.

What bothers me about that argument is that it seems to apply to an arbitrary function f. (It doesn't assume that f is the pointwise limit of a sequence of continuous functions.) Am I missing something?

4. Originally Posted by Opalg
Let $U_n = \{x\in X : \exists\text{ op{}en }V_x\ni x \text{ such that }v,w\in V_x\,\Rightarrow\,|f(v)-f(w)|<1/n\}.$

If $x\in U_n$ then $V_x\subseteq U_n$, so $U_n$ is open. Also, f is continuous at x if and only if $x\in\bigcap_nU_n$. Thus the set of points of continuity of f is a $G_\delta$.

What bothers me about that argument is that it seems to apply to an arbitrary function f. (It doesn't assume that f is the pointwise limit of a sequence of continuous functions.) Am I missing something?
I think you're right.

But this problem (and the suggested solution) also reminds me of something else, which is somewhat more interesting: under the additional hypothesis that $X$ is a complete metric space (or any Baire space), $f$ is continuous on a dense $G_\delta$ set.

This can be proved by considering $V_n(\varepsilon)=\{x\in X|\forall m\geq n, \forall p\geq n, |f_m(x)-f_p(x)|\leq \varepsilon\}$. This is a closed subset of $X$. Then let $O(\varepsilon)=\bigcup_{n=1}^\infty {\rm Int}(V_n(\varepsilon))$. This is an open subset, and it is dense because of Baire's theorem: if not, then there would be $x\in X$ and a closed ball $B=\overline{B}(x,\delta)$ such that $O(\varepsilon)\cap B=\emptyset$, which means ${\rm Int}(V_n(\varepsilon))\cap B=\emptyset$ for all $n$, hence ${\rm Int}(V_n(\varepsilon)\cap B)=\emptyset$, and Baire's theorem would say ${\rm Int}\left(\bigcup_{n=1}^\infty (V_n(\varepsilon)\cap B)\right)=\emptyset$, in contradiction with $\bigcup_{n=1}^\infty (V_n(\varepsilon)\cap B)=B$ (which holds because the pointwise convergence gives $\bigcup_n V_n(\varepsilon)=X$). Finally, Baire's theorem shows that $O=\bigcap_{k=1}^\infty O(1/k)$ is dense. And one can see that $f$ is continuous on $O$.