Let $<$ be the usual ordering on $\mathbb{R}.$ If $\gamma$ is an ordinal, then $f:\gamma \to \mathbb{R}$ is ordering preserving if $\forall \alpha \in\gamma \forall \beta \in \gamma [\alpha \in \beta \implies f(\alpha) < f(\beta)].$
Let $\gamma $ be an ordinal. Prove there is an ordering preserving $f:\gamma \to \mathbb{R}$ iff $\gamma < \omega_1.$
Could anyone advise on this problem?
Given $\exists$ order-preserving $f: \gamma \to \mathbb{R},$ suppose $\gamma \geq \omega_1.$ We know that there is a rational number between any two real numbers, so how I derive a contradiction from here?
Hint: If $f : \omega_1 \to \mathbb{R}$ is order preserving, consider the following family of open intervals: $$\{ (\; f ( \alpha ) , f ( \alpha + 1 )\; ) : \alpha \in \omega_1 \}.$$ (Recall that $\mathbb{R}$ is separable.)
Given $\gamma < \omega_1$, to construct an order-preserving function $f : \gamma \to \mathbb{R}$ it is easier to proceed by induction on $\gamma$, and moreover show that such an $f$ can be taken to have bounded range (i.e., $|f(\alpha)| \leq M$ for all $\alpha \in \gamma$ for some $M \in \mathbb{R}$).