Field homomorphisms of $\mathbb{R}$ How to prove that $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ ?
(We treat it as field homomorphisms. )
I know that $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$ and $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.

Hint: Let $\phi: \mathbb R \to \mathbb R$ is a ring homomorphism. Prove the followings: (1). $\phi(r) = r, \forall r \in \mathbb Q.$ (2). $\phi(x) > 0, \forall x >0.$ (3). $|x-y|<\frac{1}{m} \Rightarrow |\phi(x) - \phi(y)|<\frac{1}{m}.$
