If $f:\mathbb R \to \mathbb R$ is an additive function whose graph is $G_{\delta}$ in $\mathbb R^2$ , then the graph is closed in $\mathbb R^2$?

Question

If $f:\mathbb R \to \mathbb R$ is an additive function i.e. $f(x+y)=f(x)+f(y) ,\forall x,y \in \mathbb R $ such that $G(f):\{(x,f(x)) : x\in \mathbb R\}$ is a countable intersection of open sets , then is it true that $G(f)$ is closed in $\mathbb R^2$ with usual Euclidean metric ? (If I can prove this , I can conclude that if the graph of an additive function on the real line is $G_{\delta}$ , then $f$ is linear) . Please help . Thanks in advance

Answer 1

Let $G=G(f)$ and $X=\overline{G}$. Since $\Bbb R\times \Bbb R$ is a topological group, both $G$ and $X$ are topological groups (under addition). Let $G=\bigcap G_n$ be an intersection of a non-increasing sequence $\{G_n\}$ of dense open subsets of $X$. Assume that there exists an element $x\in X\setminus G$. Since sets $G_n\cap (x+G_n)$ are dense open subsets of $X$, by induction we can build a sequence of points $x_n\in X$ such that $\overline{B_{n+1}}\subset G_n\cap (x+G_n)\cap B_n$, where $B_n=\{y\in X:|y-x_n|<r_n\}$, and $r_n<1/n$. Since $X$ is a complete metric space, the sequence $\{x_n\}$ converges to some point $x_\infty\in X$. Then $x_\infty\in\bigcap \overline{B_n}\subset G\cap (x+G),$ a contradiction. Indeed, if $x_\infty\in G$ and $x_\infty=x+z$ where $z\in G$ then $x=x_\infty-z\in G$, but $x\in X\setminus G$.