Let $ f:(X, d) \mapsto (Y,d) $ be an mapping such that $ Graph (f) $ is connected.

Question

Where $ X $ is connected. Does it imply $ f $ to be continuous?

Answer 1

No. In On a connected dense proper subgroup of $\Bbb R^2$ whose complement is connected Ryuji Maehara gives a relatively simple proof of the title result, which is due to F. Burton Jones. The main theorem is that there is an additive function $f:\Bbb R\to\Bbb R$ such that both the graph of $f$ and its complement are dense, connected subsets of $\Bbb R^2$. Such a function cannot be continuous: it’s well-known that if such a function is continuous, it has the form $f(x)=cx$ for some constant $c\in\Bbb R$.