Can there be an ordering preserving (1-1) map $f \colon \langle\aleph_1,\in\rangle \rightarrow \langle\mathbb R, \lt\rangle$?
I remember that if $f$ is order preserving then consider the following family of sets of the form $A = \{(f(\alpha),f(\alpha+1) ) \colon \alpha \in \omega_1\}$ Then $\exists q_{\alpha} \in (f(\alpha),f(\alpha+1))$ where $F \colon A_{\alpha} \rightarrow \mathbb Q$ and we map those intervals by $F$ to $q_{\alpha}$. But then we are mapping the whole of from the uncountable into $\mathbb Q$ which is countable.
Can anyone advise on filling in the gaps here and making this more thorough?
You have the key idea of a proof that there is no order-preserving injection of $\omega_1$ into $\Bbb R$ with their usual orders. Suppose that $f$ was such a map. For $\alpha<\omega_1$ let $A_\alpha=\big(f(\alpha),f(\alpha+1)\big)$; then $\mathscr{A}=\{A_\alpha:\alpha<\omega_1\}$ is an uncountable family of pairwise disjoint, non-empty open intervals in $\Bbb R$. For each $\alpha<\omega_1$ there is a rational number $q_\alpha\in A_\alpha$. The family $\mathscr{A}$ is pairwise disjoint, so the map $\alpha\mapsto q_\alpha$ from $\omega_1$ to $\Bbb Q$ must be injective. This, however, is impossible, since $|\omega_1|=\omega_1>\omega=|\Bbb Q|$. Therefore no such map exists.