Looking at $(\mathbb R,<)$ where $<$ denotes the usual order I understand that $\mathbb R-\{0\}$ will inherit an order (as every subset of $\mathbb R$).
I could not find essential differences so expect this to have the same ordertype as $(\mathbb R,<)$.
- Question1: Am I correct in this?
- Question2: If so, then is there a more or less obvious way to construct an order-preserving bijection?
Thank you in advance.
HINT: $\langle\Bbb R,<\rangle$ is a complete linear order; $\langle\Bbb R\setminus\{0\},<\rangle$ is not, since $(\leftarrow,0)$ is bounded above but has no least upper bound. Use this to show that there can be no such bijection.