Can there exist two functions $f,g: \mathbb R \to \mathbb R$ so that $f$ is continuous, $g$ is discontinuous, and their graphs $\Gamma\_f, \Gamma\_g \subseteq \mathbb R^2$ are related by an isometry? (I think you can assume the isometry is a rotation.)


The graph of a continuous function must be path connected, so a natural intermediate question is, "can a discontinuous function have path connected graph?"



Can the graph of a continuous function be a rotation of the graph of a discontinuous function?
I think the question has an easy answer, and it's essentially a 1st course in analysis type question that reduces pretty quickly to an intermediate value theorem application. Here's a more general statement. Let $f : \mathbb R \to \mathbb R$ be continuous and $h : \mathbb R^2 \to \mathbb R^2$ a homeomorphism. Then if $h(graph(f))$ is the graph of a function $g : \mathbb R \to \mathbb R$, then $g$ is continuous.


Sketch: consider the function $P : \mathbb R \to \mathbb R$ given by $P(x) = \pi\_1 \circ h(x,f(x))$, where $\pi\_1(x,y)=x$. This is a continuous monotone function by the assumptions, so it is an open map by the intermediate value theorem, so it's inverse exists and is continuous. The claimed function $g$ is then $g(x)=\pi\_2 h(P^{-1}(x), f(P^{-1}(x))$ where $\pi\_2(x,y)=y$.


The question gives me the feeling it's a homework problem, but I've never seen it before. This was solved in meta, with a correction by Anton. I wish I had known about this question yesterday -- I would have put it on my analysis class final exam!


